Download - Teradata imp

undefined Introduction to Teradata SQL (Module 1)

Objectives After completing this module, you should be able to:

Describe the structure of a Relational Database Management System (RDBMS). Explain the role of Structured Query Language (SQL) in accessing a Relational Database. List the three categories of SQL statements and describe their function.

What is an RDBMS?

Data is organized into tables in a relational database management system (RDBMS). Rows in the table represent instances of an entity, in this case an employee or a department. Columns represent the data fields which comprise the rows. Relations between tables occur when a column in one table also appears as a column in another.

Here is an example of two related tables:

http://adtrack.ministerial5.com/clicknew/?a=637394

Can you answer the following questions using the tables above?

Question 1: What is the department name for employee 1004?

Question 2: Who is the manager of employee 1004?

What is SQL?

Structured Query Language (SQL) is the industry standard language for communicating with Relational Database Management Systems. SQL is used to look up data in relational tables, create tables, insert data into tables, grant permissions and many other things.

Structured Query Language is used to define the answer set that is returned from the Teradata Database.

SQL is a non-procedural language, meaning it contains no procedural-type statements such as those listed here:

GO TO PERFORM DO LOOP

OPEN FILE CLOSE FILE END OF FILE

SQL Commands

SQL statements commonly are divided into three categories:

1. Data Definition Language (DDL) - Used to define and create database objects such as tables, views, macros, databases, and users.

2. Data Manipulation Language (DML) - Used to work with the data, including such tasks as inserting data rows into a table, updating an existing row, or performing queries on the data. The focal point of this course will be on SQL statements in this category.

3. Data Control Language (DCL) - Used for administrative tasks such as granting and revoking privileges to database objects or controlling ownership of those objects. DCL statements will not be covered in detail in this course. For complete Data Control Language coverage, please see the NCR Customer Education "Teradata Database Administration" course.

Data Definition Language (DDL) Examples

SQL statement FunctionCREATE Define a table, view, macro, index, trigger or stored procedure. DROP Remove a table, view, macro, index, trigger or stored procedure. ALTER Change table structure or protection definition.

Data Manipulation Language (DML)

SQL statement Function

SELECT Select data from one or more tables.

INSERT Place a new row into a table.

UPDATE Change data values in one or more existing rows.

DELETE Remove one or more rows from a table. Data Control Language (DCL)

SQL statement Function

GRANT Give user privileges.

REVOKE Remove user privileges.

GIVE Transfer database ownership.

Relational Concepts

The figure below illustrates six rows of a much larger table.

Some interesting things to note about this table:

The intersection of a row and a column is a data value. Data values come from a particular domain. Domains represent the pool of legal values for a column. For example, the data values in the EMPLOYEE NUMBER and MANAGER EMPLOYEE NUMBER come from the same domain because both columns come from the pool of employee numbers. The domain of employee numbers might be defined as the pool of integers greater than zero.

The EMPLOYEE NUMBER column is marked PK, which indicates that this column holds the Primary Key. The next 3 columns are marked FK, which stands for Foreign Key.

The purpose of the Primary Key is to uniquely identify each record. No two values in a primary key column can be identical.

A Foreign Key represents the Primary Key of another table. Relationships between tables are formed through the use of Foreign Keys.

Teradata SQL (Module 2)

Objectives

After completing this module, you should be able to:

Describe the uses of the Teradata SQL SELECT statement. Retrieve data from a relational table using the SELECT statement. Use the SQL ORDER BY and DISTINCT options. Set a default database using the DATABASE command. Write SQL statements using recommended coding conventions. Name database objects according to Teradata SQL rules.

SELECT

Structured Query Language (SQL) consists of three types of statements, previously defined as:

Data Definition Language (DDL)

- Used to create, drop and modify objectsData Manipulation Language(DML)

- Used to add, delete, update and read data rows in a table

Data Control Language(DCL)

- Used to implement security and control on database objects.

Our focus in this course will be mostly on the DML portion of the language. We will first look at the SELECT statement.

The SELECT statement allows you to retrieve data from one or more tables. In its most common form, you specify certain rows to be returned as shown.

SELECT *FROM employeeWHERE department_number = 401;

The asterisk, "*", indicates that we wish to see all of the columns in the table. The FROM clause specifies from which table in our database to retrieve the rows. The WHERE clause acts as a filter which returns only rows that meet the specified condition, in this case, records of employees in department 401.

Note: SQL does not require a trailing semicolon to end a statement but the Basic Teradata Query (BTEQ) utility that we use to enter SQL commands does require it. All examples in this course will include the semicolon.

What if we had not specified a WHERE clause.

SELECT * FROM employee;

This query would return all columns and all rows from the employee table.

Instead of using the asterisk symbol to specify all columns, we could name specific columns

separated by commas:

SELECT employee_number ,hire_date ,last_name ,first_name FROM employee WHERE department_number = 401;

employee_number --------------------

hire_date----------

last_name-----------

first_name-----------

1004 76/10/15 Johnson Darlene 1003 76/07/31 Trader James

1013 77/04/01 Phillips Charles

1010 77/03/01 Rogers Frank

1022 79/03/01 Machado Albert

1001 76/06/18 Hoover William 1002 76/07/31 Brown AlanUnsorted Results

Results come back unsorted unless you specify that you want them sorted in a certain way. How to retrieve ordered results is covered in the next section.

ORDER BY Clause

Use the ORDER BY clause to have your results displayed in a sorted order. Without the ORDER BY clause, resulting output rows are displayed in a random sequence.

SELECT employee_number ,last_name ,first_name ,hire_date FROM employee WHERE department_number = 401 ORDER BY hire_date;

Sort Direction

In the example above, results will be returned in ascending order by hire date. Ascending order is the default sort sequence for an ORDER BY clause. To explicitly specify ascending or descending order, add ASC or DESC, to the end of the ORDER BY clause. The following is an example of a sort using descending sequence.

ORDER BY hire_date DESC;

Naming the Sort Column

You may indicate the sort column by naming it directly (e.g., hire_date) or by specifying its position within the SELECT statement. Since hire_date is the fourth column in the SELECT statement, the following ORDER BY clause is equivalent to saying ORDER BY hire_date.

ORDER BY 4;

Multiple ORDER BY Columns

An ORDER BY clause may specify multiple columns. No single column in an ORDER BY clause should exceed a length of 4096 bytes, otherwise it will be truncated for sorting purposes.

The order in which columns are listed in the ORDER BY clause is significant. The column named first is the major sort column. The second and subsequent are minor sort columns. In the following example, results are sorted by department number in ascending order. Where multiple records share the same department number, those rows are sorted by job_code in ascending order. The following are examples:

SELECT employee_number ,department_number ,job_code FROM employee WHERE department_number < 302 ORDER BY department_number ,job_code;

employee_number--------------------

department_number

----------------------

job_code

----------

801 100 111100

1025 201 211100

1021 201 222101

1019 301 311100

1006 301 312101

1008 301 312102

Note: Each column specified in the ORDER BY clause can have its own sort order, either ascending or descending.

SELECT employee_number ,department_number ,job_code FROM employee WHERE department_number < 302 ORDER BY department_number ASC ,job_code DESC;


department_number

----------------------

job_code

----------

801 100 111100

1021 201 222101

1025 201 211100

1008 301 312102

1006 301 312101

1019 301 311100

DISTINCT

The DISTINCT operator will consolidate duplicate output rows to a single occurrence.

Example Without DISTINCT

SELECT department_number ,job_code FROM employee WHERE department_number = 501;

department_number----------------------

job_code----------

501 512101

501 512101

501 511100

Note: Two people in department 501 have the same job code (512101). If our purpose is simply to find out which job codes exist in department 501, we could use DISTINCT to avoid seeing duplicate rows.

Example With DISTINCT

SELECT DISTINCT department_number ,job_code FROM employee WHERE department_number = 501;


job_code----------

501 511100

501 512101

Note: DISTINCT appears directly after SELECT, and before the first named column. It may appear to apply only to the first column, but in fact, DISTINCT applies to all columns named in the query. Two rows in our result example set both have department_number 501. The combination of department_number and job_code are distinct since the job codes differ.

Naming Database Objects

All Teradata objects, such as tables, must be assigned a name by the user when they are created.

These rules for naming objects are summarized as follows:

Names are composed of:

a-z A-Z 0-9 _ (underscore) $ #

Names are limited to 30 characters.

Names cannot begin with a number.

Teradata names are not case-sensitive.

Examples of valid names:

AccountsAccounts_2005accounts_over_$2000Account#

Equivalence

Accounts and accounts represent the same table. Case is not considered.

Naming Rules

Naming Syntax

Database names and User names must be unique within the Teradata Database. Table, view, macro, trigger, index and stored procedure names must be unique within a

Database or User. Column names must be unique within a Table.

The syntax for fully qualifying a column name is:

Databasename.Tablename.Columnname Example

NAME (unqualified)

EMPLOYEE.NAME (partially qualified)

PAYROLL.EMPLOYEE.NAME (fully qualified)

Note: The amount of qualification necessary is a function of your default database setting.

Coding Conventions

SQL is considered a 'free form' language, that is, it can cover multiple lines of code and there is no restriction on how much 'white space' (i.e., blanks, tabs, carriage returns) may be embedded in the query. Having said that, SQL is syntactically a very precise language.

Misplaced commas, periods and parenthesis will always generate a syntax error.

The following coding conventions are recommended because they make it easier to read, create, and change SQL statements.

Recommended Practice

SELECT last_name ,first_name ,hire_date ,salary_amount FROM employee WHERE department_number = 401 ORDER BY last_name;

Not Recommended Practice

select last_name, first_name, hire_date, salary_amount from employee where department_number = 401 order by last_name;

The first example is easy to read and troubleshoot (if necessary). The second example appears to be a jumble of words. Both, however, are valid SQL statements.

Default Database

Setting the Default Database

As a valid user, you will normally have access rights to your own user database and the objects it contains. You may also have permission to access objects in other databases.

The user name you logon with is usually your default database. (This depends on how you were created as a user.)

For example, if you log on as:

.logon rayc;

password: xyz

then "rayc" is normally your default database.

Note the dot (.) before "logon". Commands that begin with a dot are BTEQ commands, not SQL commands. The BTEQ command processor will read this command; the SQL processor will not see it.

Queries you make that do not specify the database name will be made against your default database.

Changing the Default Database

The DATABASE command is used to change your default database.

For example:

DATABASE payroll; sets your default database to payroll. Subsequent queries (assuming the proper privileges are held) are made against the payroll database.

Lab

For this set of lab questions you will need information from the Database Info document.

These exercises are to be done as a pencil-and-paper workshop. Write SQL statements as you would enter them on-line. Click on the buttons to the left to see the answers.

Question 3:

A. Select all columns for all departments from the department table. B. Using the employee table, generate a report of employee last and first names and

salary for all of manager 1019's employees. Order the report in last name ascending sequence.

C. Modify the previous request to show department number instead of first name. Make it for manager 801's employees instead of manager 1019's.

D. Prepare a report of the department numbers and the manager's employee numbers for everyone in the employee table. Now add the DISTINCT option to the same report.

Note: Save your answers for this Lab, as you actually enter and run these exercises using BTEQ scripts for the Labs in Module 3.

Simple BTEQ (Module 3)

Objectives


Use BTEQ to submit SQL to the Teradata database. Set session parameters to enable

Teradata transaction semantics. The SQL Flagger.

What is BTEQ?

BTEQ is a front end tool for submitting SQL queries. 'BTEQ' stands for Basic Teradata Query program.

BTEQ is client software that resides on your network or channel-attached host. After starting

BTEQ, you will log on to Teradata using a TDP-id (Teradata Director Program id), your user id and password. The TDP-id identifies the instance of Teradata you are going to access.

TDP's come in two varieties - the standard TDP for channel-attached clients, and the Micro-TDP (or MTDP) for network-attached clients.

They are involved in getting your SQL requests routed to a Parsing Engine (PE) which validates your request and passes it to the Access Module Processors (AMPs). AMPs retrieve the answer sets from the disks and send the response set back to the PE which in turn forwards it to the TDP and ultimately back to you at your session.

Where is BTEQ Located?

BTEQ is Teradata client software which is installed on mainframe hosts or network attached clients. It operates under all host systems and local area networks (LANs).

BTEQ Commands

Facts about BTEQ commands:

Must be preceded by a period (.) or terminated by a semi-colon or both. Provide an output listing (an audit trail) of what occurred. Additional commands for report formatting are available. Are not case-sensitive.

Invoking BTEQ varies somewhat depending on the platform. In the UNIX enviroment, simply typing in the command 'bteq' is required.

For purposes of this course, instructions for starting a lab session are contained on each lab page.

Using BTEQ Interactively

Submitting SQL Statements with BTEQ

There are two ways to submit SQL statements to BTEQ in interactive mode:

Type in SQL statements directly to the BTEQ prompt. Open another window with a text editor. Compose SQL statements using the text

editor, then cut and paste the SQL to the BTEQ prompt.

When users log on to BTEQ, they are by default assigned a single session.

Session Parameters

Session parameters include transaction semantics and the SQL Flagger.

Transaction semantics, allows you to set your session in either ANSI or Teradata (BTET)

mode. All features of Teradata SQL will work in either mode, but each mode activates different case-sensitivity and data conversion defaults, and so the same query might return different results in each mode.

For purposes of this course, Teradata-mode sessions will be assumed unless otherwise specified. More on these session parameters is covered in the Teradata SQL Advanced WBT.

Example

.SET SESSION TRANSACTION ANSI; /* Sets ANSI mode*/

.SET SESSION TRANSACTION BTET; /* Sets Teradata mode*/

Note: All text between /* and */ are treated as comments by BTEQ.

You may also activate the ANSI Flagger, which automatically flags non-ANSI compliant syntax with a warning but still returns the expected answer set.

Example

.SET SESSION SQLFLAG ENTRY; /* Causes non-Entry level ANSI syntax to be flagged */

SELECT DATE; /* Causes a warning because keyword DATE is not ANSI standard*/

*** SQL Warning 5821 Built-in values DATE and TIME are not ANSI.*** Query completed. One row found. One column returned.*** Total elapsed time was 1 seconds.

Date--------05/01/12

Example

.SET SESSION SQLFLAG NONE /* Disables ANSI Flagger*/

SELECT DATE; /* DATE keyword is not flagged */

*** Query completed. One row found. One column returned. *** Total elapsed time was 1 second.

Date--------05/01/12

Note: In both cases, the date was returned.

You need to establish your session parameters prior to logging on. Teradata extensions cannot be disabled, but they can be flagged using the ANSI flagger.

The SHOW CONTROL Command

The BTEQ .SHOW CONTROL command displays BTEQ settings. The following output shows the result of this command.

.SHOW CONTROL

Default Maximum Byte Count = 4096 Default Multiple Maximum Byte Count = 2048 Current Response Byte Count = 4096 Maximum number of sessions = 20 Maximum number of the request size = 32000

EXPORT RESET IMPORT FIELD LOGON L7544/tdxxx RUN [SET] ECHOREQ = ON [SET] ERRORLEVEL = ON [SET] FOLDLINE = OFF ALL [SET] FOOTING = NULL [SET] FORMAT = OFF [SET] FORMCHAR = OFF [SET] FULLYEAR = OFF [SET] HEADING = NULL [SET] INDICDATA = OFF [SET] NOTIFY = OFF [SET] NULL = ? [SET] OMIT = OFF ALL [SET] PAGEBREAK = OFF ALL [SET] PAGELENGTH = 55 [SET] QUIET = OFF [SET] RECORDMODE = OFF [SET] REPEATSTOP = OFF [SET] RETCANCEL = OFF [SET] RETLIMIT = No Limit [SET] REPEATSTOP = OFF [SET] RETCANCEL = OFF [SET] RETLIMIT = No Limit [SET] RETRY = ON [SET] RTITLE = NULL [SET] SECURITY = NONE [SET] SEPARATOR = two blanks [SET] SESSION CHARSET = ASCII [SET] SESSION SQLFLAG = NONE [SET] SESSION TRANSACTION = BTET [SET] SESSIONS = 1 [SET] SIDETITLES = OFF for the normal report. [SET] SKIPDOUBLE = OFF ALL [SET] SKIPLINE = OFF ALL [SET] SUPPRESS = OFF ALL [SET] TDP = L7544 [SET] TIMEMSG = DEFAULT [SET] TITLEDASHES = ON for the normal report. [SET] UNDERLINE = OFF ALL

[SET] WIDTH = 75

BTEQ Scripts

A BTEQ script is a file that contains BTEQ commands and SQL statements. A script is built for sequences of commands that are to be executed on more than one occasion, i.e. monthly, weekly, daily.

How to create a BTEQ Script

To create and edit a BTEQ script, use an editor on your client workstation. For example, on a UNIX workstation, use either the vi or text editor.

How to Submit a BTEQ Script

Start BTEQ, then enter the following BTEQ command to submit a BTEQ script:

.run file = <scriptname>

BTEQ Comments -- Teradata Extensions

The Teradata RDBMS supports comments that span multiple lines by using the "/*" and "*/" as delimiters.

Example of a BTEQ comment:

/* You can include comments in a BTEQ script by enclosing the comment text between “/*” and “*/” and the comment may span multiple lines */

Script Example

Let's look at an example of a BTEQ script using ANSI standard comments. The ANSI comment delimiter is the double dash, --.

(This script is named mod2exer.scr )

.SET SESSION TRANSACTION ANSI .LOGON L7544/tdxxx,;

-- Obtain a list of the department numbers and names -- from the department table.

SELECT department_number ,department_name FROM department;

.QUIT

Note: Both Teradata style (/* */) and ANSI style (--) comments may be included in any SQL command script.

To Execute The Script:

.run file=mod2exer.scr

(The following is the output generated by the above command.)

Script started on Sat Feb 15 10:48:19 1997$ bteq

Teradata BTEQ 04.00.01.00 for UNIX5. Enter your logon or BTEQ command:

.run file=mod2exer.scr (this is the only user input; the rest of the commands came from file mod2exer.scr)

.run file=mod2exer.scrTeradata BTEQ 04.00.01.00 for UNIX5. Enter your logon or BTEQ command:

.LOGON L7544/tdxxx,

*** Logon successfully completed. *** Transaction Semantics are ANSI. *** Character Set Name is 'ASCII'. *** Total elapsed time was 1 second.

BTEQ -- Enter your DBC/SQL request or BTEQ command: -- Obtain a list of the department numbers and names -- from the department table.

SELECT department_number ,department_name FROM department;

*** Query completed. 9 rows found. 2 columns returned.*** Total elapsed time was 1 second.

department_number department_name

------------------ -----------------401 customer support

<rest of result rows are not shown in this screen capture>

BTEQ -- Enter your DBC/SQL request or BTEQ command:

.QUIT*** You are now logged off from the DBC.*** Exiting BTEQ...

*** RC (return code) = 0# exit

Using BTEQ in BATCH Mode

If you prefer to write all SQL statements in a single script file, there are two ways you may execute the script:

Start BTEQ in interactive mode. Use the following BTEQ command to execute the script:

.run file = mod2exer.scr

Start BTEQ and redirect standard input to use the script file.

$bteq < mod2exer.scr

(Note: The dollar sign ‘$’ corresponds to the UNIX prompt.)

How to Capture BTEQ Session Output

From a UNIX workstation, use the script <filename> command to capture the output of your session :

$script <filename> Logs all input and output to <filename>.$bteq Starts BTEQ in interactive mode..run file = <scriptname>

Submits pre-written script which contains BTEQ commands and SQL statements.

The contents of <scriptname> and responses from the Teradata database will be output to both the screen and the designated <filename> if one has been designated.

How to Stop Capturing BTEQ Session Output

From a UNIX workstation, issue the following UNIX command to stop the script command:

$exit This closes the logging file but does not terminate the UNIX session.

Identifying Syntax Errors

BTEQ output results will display a "$" to indicate the location where the error was detected in either the script or request. The following is an example of captured output showing a syntax error:

Script started on Thu Oct 24 16:21:21 1996# bteq …

Teradata BTEQ 04.00.00.00 for UNIX5. Enter your logon or BTEQ command:

.logon L7544/tdxxx,

*** Logon successfully completed.*** Transaction Semantics are ANSI.*** Character Set Name is 'ASCII'.*** Total elapsed time was 1 second.

BTEQ -- Enter your DBC/SQL request or BTEQ command:

SELECT

FROM

department_number,DISTINCT department_namedepartment;

,DISTINCT department_name

$ *** Failure 3708 Syntax error, 'DISTINCT' that follows the ',' should be deleted.

Statement# 1, Info =36 *** Total elapsed time was 1 second

.BTEQ -- Enter your DBC/SQL request or BTEQ command:

.quit *** You are now logged off from the DBC. *** Exiting BTEQ... *** RC (return code) = 8 # exit

script done on Thu Oct 24 16:21:39 1996

The dollar sign points to the command DISTINCT. The error text tells us that DISTINCT should be deleted. The DISTINCT command must directly follow the SELECT command in your request.

Lab For this set of lab questions you will need information from the Database Info document.

To start the online labs, click on the Telnet button in the lower left hand screen of the course. Two windows will pop-up: a BTEQ Instruction Screen and your Telnet Window. Sometimes the BTEQ Instructions get hidden behind the Telnet Window. You will need these instructions to log on to Teradata.

Be sure to change your default database to the Customer_Service database in

order to run these labs.

Question 4:

A. See Labs A, B, and C from Module 2 that you did on paper. You can use NotePad or another text editor to create your SQL requests again, and then simply copy and paste them into the Telnet window after you've logged on with BTEQ.

B. Use BTEQ to submit your two reports from Lab D in Module 2. Observe the number of rows returned in your output. Which rows were eliminated due to the DISTINCT option?

Click on the Lab B button to see the answer for this Lab.

C. Set the ANSI Flagger ON and observe the differences it produces.

Do this by:

Logging off of Teradata (.logoff). Setting the flagger on (.SET SESSION SQLFLAG ENTRY).Relogging on to Teradata

Resubmit your query from Lab A.1. What differences do you observe? Why?

(Note: the ANSI Flagger is covered in more detail in SQL Level 3.)

Click on the Lab C button to see the answer for this Lab.

D. 1) Determine the default system session transaction mode. What command did you use?

2) Set the session transaction mode to the opposite setting (e.g., set it to ANSI if the system default is BTET) then verify the new setting.

Click on the Lab D button to see the answer for this Lab.

Lab Answer A.A

DATABASE Customer_Service;SELECT * FROM department ORDER BY 1;

*** Query completed. 9 rows found. 4 columns returned. *** Total elapsed time was 1 second.

department_number department_name budget_amount manager_employee_number ----------------------- --------------------- ------------------ --------------------------------- 100 president 400000.00 801 401 customer support 982300.00 1003 403 education 932000.00 1005 402 software support 308000.00 1011 302 product planning 226000.00 1016

501 marketing sales 308000.00 1017 301 research and development 465600.00 1019 201 technical operations 293800.00 1025 600 None ? 1099

Note: Order is random. And user input is highlighted in red

Lab Answer A.B

SELECT last_name,first_name,salary_amountFROM employeeWHERE manager_employee_number = 1019ORDER BY last_name ;

*** Query completed. 2 rows found. 3 columns returned. last-name KanieskiStein

first-nameCarolJohn

salary-amount 29250.0029450.00

Lab Answer A.C

SELECT last_name ,department_number ,salary_amountFROM employee WHERE manager_employee_number = 801 ORDER BY last_name;

*** Query completed. 8 rows found. 3 columns returned. last_name department_number salary_amount

Daly 402 52500.00

Kubic 301 57700.00

Rogers 302 56500.00

Runyon 501 66000.00

Ryan 403 31200.00

Short 201 34700.00

Trader 401 37850.00

Trainer 100 100000.00

Lab Answer B

SELECT

FROM;

department_number,manager_employee_numberemployee

*** Query completed. 26 rows found. 2 columns returned.

department_numbe

r

manager_employee_numbe

r

301501100401402301301402401403401403403201201501403501401401401403403401501302

101980180180110118011019801100310051003100580180110251017100510171003100310031005100510031017801

SELECT DISTINCT department_number ,manager_employee_number FROM employee ;


department_numbe

r

manager_employee_numbe

r

100 801

201201301301302401401402402403403501501

801102580110198018011003801101180110058011017

Lab Answer C

Note: Actual warning messages generated will vary depending on how you have written your queries. Remember that ANSI does not support any lower case syntax.

.SET SESSION SQLFLAG ENTRY

.LOGON RPC1042/tdxxx;

SELECT * FROM department ;


FROM department $ *** SQL Warning 5836 Token is not an entry level ANSI Identifier or Keyword.

department_number department_name budget_amount manager_employ----------------- ----------------------- ------------- -------------- 100 president 400000.00 801 401 customer support 982300.00 1003 403 education 932000.00 1005 402 software support 308000.00 1011 302 product planning 226000.00 1016

501 marketing sales 308000.00 1017 301 research and development 465600.00 1019 201 technical operations 293800.00 1025 600 None ? 1099

The warning message is generated because the ANSI flagger is enabledand is reporting that ANSI mode does not support lowercase syntax.

Lab Answer D

.SHOW CONTROL . . . [SET] SESSION SQLFLAG = ENTRY [SET] SESSION TRANSACTION = BTET . . . .LOGOFF .SET SESSION TRANSACTION ANSI .SHOW CONTROL . . . [SET] SESSION SQLFLAG = ENTRY [SET] SESSION TRANSACTION = ANSI . . .

HELP Functions Module 4

Objectives

After completing of this module, you should be able to:

Obtain the definition of an existing Database, Table, View, or Macro using the HELP and SHOW statements.

Determine how the Teradata RDBMS will process a SQL request using the EXPLAIN

statement.

Teradata SQL Extensions

Several commands in our software are Teradata-specific. Here is a list of Teradata extensions covered within this course.

ADD_MONTHS BEGIN/ END TRANSACTION

COLLECT/ DROP STATISTICS

COMMENT ON

CONCATENATION EXPLAIN

FALLBACK FORMAT

HELP INDEX

LOCKING MACRO Facility

• CREATE • REPLACE • DROP • EXECUTE

NAMED NULLIFZERO/ZEROIFNULL

SHOW SUBSTR

TITLE TRIM

WITH WITH . . . BYIn this module, we focus on the following functions:

EXPLAIN HELP SHOW

HELP Commands: Database objects

The HELP Command is used to display information about database objects such as (but not limited to):

Databases and Users Tables Views Macros

HELP retrieves information about these objects from the Data Dictionary. Below are the syntactical options for various forms of the HELP command:

HELP Command

HELP DATABASE databasename; HELP USER username;

HELP TABLE tablename;

HELP VIEW viewname;

HELP MACRO macroname;

HELP COLUMN table or viewname.*; (all columns) HELP COLUMN table or viewname.colname . . ., colname; Some of the other HELP commands which will be seen later in this course include:

HELP INDEX tablename;

HELP STATISTICS tablename;

HELP CONSTRAINT constraintname;

HELP JOIN INDEX join_indexname;

HELP TRIGGER triggername;

The HELP DATABASE Command

The HELP DATABASE command in the following example shows all objects in the Customer_Service database. Objects may be recognized by their 'Kind' designation as follows:

Table = T View = V

Macro = M

Trigger = G

Join Index = J

Stored Procedure = P HELP DATABASE Customer_Service;

Results will be displayed as follows:

Table/View/Macro NameKind Comment

contact T ?

customer T ?

department T ?

employee T ?

employee_phone T ?

job T ?

location T ?

location_employee T ?

location_phone T ?All objects in this database are tables. The '?' is how BTEQ displays a null, indicating that no user comments have been entered.

Note: To return HELP information on an object, a user must either own the object or have at least one privilege on it.

The HELP TABLE Command

The HELP TABLE command shows the name, data type, and comment (if applicable), of all columns in the specified table:

HELP TABLE Customer_Service.employee

ColumnName Type Comment

employee_number I System assigned identification

manager_employee_number I ?

department_number I Department employee works in

job_code I Job classification designation

last_name CF Employee surname

first_name CV Employee given name

hire_date DA Date employee was hired

birthdate DA ?

salary_amount D Annual compensation amountData types are as follows:

Type Description Type DescriptionCF CHARACTER FIXED F FLOATCV CHARACTER VARIABLE BF BYTED DECIMAL BV VARBYTEDA DATE AT TIMEI INTEGER TS TIMESTAMPI1 BYTEINT I2 SMALLINT

HELP Commands: Session Characteristics

Use the HELP SESSION; command to see specific information about your SQL session. The following displays the user name with which you logged in log-on date and time, your default database, and other information related to your current session:

User Name DBC

Account Name DBC

Log on Date 05/02/18

Log on Time 11:54:46

Current DataBase DBC

Collation ASCII

Character Set ASCII

Transaction Semantics Teradata

Current DateForm IntegerDate

Session Time Zone 00:00

Default Character Type LATIN

Export Latin 1

Export Unicode 1

Export Unicode Adjust 0

Export KanjiSJIS 1

Export Graphic 0

Default Date Format NoneTo produce the formatting shown above use:

.SET FOLDLINE ON

.SET SIDETITLES ON

The SHOW Command

The SHOW command displays the current Data Definition Language (DDL) of a database object (e.g., Table, View, Macro, Trigger, Join Index or Stored Procedure). The SHOW command is used primarily to see how an object was created.

Sample Show Commands

Command Returns

SHOW TABLE tablename; CREATE TABLE statement

SHOW VIEW viewname; CREATE VIEW statement

SHOW MACRO macroname; CREATE MACRO statement

BTEQ also has a SHOW command (.SHOW CONTROL) which is different from the SQL SHOW command. It provides information on formatting and display settings for the current BTEQ session.

The SHOW TABLE Command

The SHOW TABLE command seen here returns the used to create the employee table. To perform a

privilege on either the table itself or the containing database.

CREATE SET TABLE CUSTOMER_SERVICE.employee ,FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( employee_number INTEGER, manager_employee_number INTEGER, department_number INTEGER, job_code INTEGER, last_name CHAR(20) CHARACTER SET LATIN NOT CASESPECIFIC NOT NULL, first_name VARCHAR(30) CHARACTER SET LATIN NOT CASESPECIFIC NOT NULL, hire_date DATE NOT NULL, birthdate DATE NOT NULL, salary_amount DECIMAL(10,2) NOT NULL) UNIQUE PRIMARY INDEX (employee_number);

Note: If a table is subsequently altered after it has been created, the will always reflect the most current alterations.

The SHOW VIEW Command

The SHOW VIEW command shows you the CREATE VIEW statement used to create a view:

SHOW VIEW dept;

Result

CREATE VIEW dept (dept_num ,dept_name ,budget ,manager)AS SELECTdepartment_number ,department_name ,budget_amount ,manager_employee_number FROM CUSTOMER_SERVICE.department;

The view may be accessed instead of the table:

SELECT * FROM dept;

Resultdept_num

--------- dept_name -----------

budget --------

manager ---------

301 research & development 46560000 1019302 product planning 22600000 1016501 marketing sales 80050000 1017

403 education 93200000 1005402 software support 30800000 1011401 customer support 98230000 1003201 technical operations 29380000 1025

The SHOW Macro Command

The SHOW MACRO command shows you the statement used to create a macro:

SHOW MACRO get_depts;

Result

CREATE MACRO get_depts AS (SELECT department_number ,department_name ,budget_amount ,manager_employee_number FROM department;);

The EXPLAIN Command

The EXPLAIN function looks at a SQL request and responds in English how the optimizer plans to execute it. It does not actually execute the SQL statement however it is a good way to see what database resources will be used in processing your request.

For instance, if you see that your request will force a full-table scan on a very large table or cause a Cartesian Product Join, you may decide to re-write a request so that it executes more efficiently.

EXPLAIN provides a wealth of information, including the following:

1. Which indexes if any will be used in the query. 2. Whether individual steps within the query may execute concurrently (i.e.

parallel steps). 3. An estimate of the number of rows which will be processed. 4. An estimate of the cost of the query (in time increments).

EXPLAIN SELECT * FROM department;

*** Query completed. Ten rows found. One column returned.

Explanation

1. First, we lock a distinct CUSTOMER_SERVICE."pseudo table" for read on a RowHash to prevent global deadlock for CUSTOMER_SERVICE.department.

2. Next, we lock CUSTOMER_SERVICE.department for read. 3. We do an all-AMPs RETRIEVE step from CUSTOMER_SERVICE.department by way

of an all-rows scan with no residual conditions into Spool 1, which is built locally on the AMPs. The size of Spool 1 is estimated with low confidence to be 4 rows. The estimated time for this step is 0.15 seconds.

4. Finally, we send out an END TRANSACTION step to all AMPs involved in processing the request.The contents of Spool 1 are sent back to the user as the result of statement 1. The total estimated time is 0.15 seconds.



A. Use the HELP DATABASE command to find all tables, views, and macros names in the CS_Views database. What kind of objects do you find there? Do a similar HELP command on the Customer_Service database. What kind of objects do you find there?

B. To see the names of the columns in the department table, use the appropriate HELP command. (Since the table is in the Customer_Service database, not your default database, you'll have to qualify the table name with the database name.) This is the command you may wish to use in the future to research data names.

C. In Lab A, you may have noticed that the CS_Views database includes a view called emp. SHOW that view. Notice the list of short names the emp view uses in place of full column names. To save typing, you may use the emp view with the shortened names in place of the employee table in any labs throughout this course. (All lab solutions shown in this book use the employee table in Customer_Service.)

D. Modify Lab A.1 from the previous module to cause the answer set to appear in department name sequence. To find out how Teradata plans to handle this request, submit it with EXPLAIN in front of it.

E. Use the appropriate SHOW command to see the table definition of the employee_phone table in the Customer Service database.

F. Use the appropriate HELP command to see what kinds of indexes exist on the customer table. To get information about your session, use another HELP command. Change the display settings for sidetitles and foldline to better view the session information. Be sure to reset display attributes once you have seen results.

G. Change your current database setting to Customer_Service using the DATABASE command. Try to do a SELECT of all columns and rows in the emp

view. Does it work? If not, how can you make it work?

Lab Answer A

HELP DATABASE cs_views;

*** Help information returned. 15 rows found. Table/View/Macro name Kind Commentagent_sales V ?contact V ?customer V ?daily_sales V ?department V ?emp V ?employee V ?employee_phone V ?jan_sales V ?job V ?location V ?location_employee V ?location_phone V ?repair_time V ?sales_table V ?

HELP DATABASE customer_service;

*** Help information returned. 14 rows found. Table/View/Macro name Kind Commentagent_sales T ?contact T ?customer T ?daily_sales T ?department T ?employee T ?employee_phone T ?jan_sales T ?job T ?location T ?location_employee T ?location_phone T ?repair_time T ?sales_table T ?

Lab Answer B

HELP TABLE customer_service.department;

*** Help information returned. 4 rows found.

Column Name Type Comment ------------------------------ ---- -----------department_number I2 ? department_name CF ? budget_amount D ? manager_employee_number I ?

Lab Answer C

SHOW VIEW cs_views.emp;

*** Text of DDL statement returned.CREATE VIEW emp (emp ,mgr ,dept ,job ,last ,first ,hire ,birth ,sal) AS SELECT employee_number ,manager_employee_number ,department_number ,job_code ,last_name ,first_name ,hire_date ,birthdate ,salary_amount FROM CUSTOMER_SERVICE.employee;

Lab Answer D

EXPLAIN SELECT * FROM customer_service.department ORDER BY department_name ;

*** Help information returned. 11 rows found. Explanation

1. First, we lock a distinct customer_service."pseudo table" for read on a RowHash to prevent global deadlock for customer_service.department.

2. Next, we lock customer_service.department for read.

3. We do an all-AMPs RETRIEVE step from customer_service.department by way of an all-rows scan with no residual conditions into Spool 1, which is built locally on the AMPs. Then we do a SORT to order Spool 1 by the sort key in spool field1. The size of Spool 1 is estimated with low confidence to be 12 rows. The estimated time for this step is 0.15 seconds.

4. Finally, we send out an END TRANSACTION step to all AMPs involved in processing the request. -> The contents of Spool 1 are sent back to the user as the result of statement 1. The total estimated time is 0.15 seconds.

Lab Answer E

SHOW TABLE Customer_Service.employee_phone;

*** Text of DDL statement returned. CREATE TABLE customer_service.employee_phone ,FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( employee_number INTEGER NOT NULL, area_code SMALLINT NOT NULL, phone INTEGER NOT NULL, extension INTEGER, comment_line CHAR(72) CHARACTER SET LATIN NOT CASESPECIFIC) PRIMARY INDEX( employee_number ) ;

Lab Answer F

HELP INDEX Customer_Service.customer;

*** Help information returned. One row.

Unique Primary or Secondary Column Names

Y P customer_number

.SET SIDETITLES ON

.SET FOLDLINE ON

HELP SESSION; User Name tdxxx Account Name $M_P0623_&D Logon Date 00/11/29 Logon Time 11:52:18 Current DataBase tdxxx Collation ASCII Character Set ASCII Transaction Semantics Teradata Current DateForm IntegerDate Session Time Zone 00:00 Default Character Type LATIN Export Latin 1 Export Unicode 1 Export Unicode Adjust 0 Export KanjiSJIS 1 Export Graphic 0 Default Date Format None

.SET SIDETITLES OFF

.SET FOLDLINE OFF

Lab Answer G

DATABASE Customer_Service;

SELECT * FROM emp;

(Fails because the emp view is not in database customer_service. To make it work, fully qualify the view name.)

SELECT * FROM CS_Views.emp;

(You may also change the default database.)

DATABASE CS_Views;

SELECT * FROM emp;

Logical and Conditional Expressions Module 5

Combine the following types of operators into logical expressions: comparison operators [NOT] IN IS [NOT] NULL LIKE

Form expressions involving NULL values. Qualify a range of values. Search for a partial string within a string expression. Combine multiple logical expressions into a conditional expression using

AND and OR.

Logical Operators

Operators are symbols or words that cause an 'operation' to occur on one or more elements called 'operands'.

Logical expressions combine operands and operators to produce a Boolean (true/false) result. Logical expressions can be combined into conditional expressions which are used in the WHERE clause of a SELECT statement.

Types of Operators in Logical Expressions

Operator Syntax Meaning

= equal

<> not equal

> greater than

< less than

>= greater than or equal to

<= less than or equal to

BETWEEN <a> AND inclusive range

[NOT] IN <expression> is in a list or <expression> is not in a list

IS [NOT] NULL <expression> is null or <expression> is not null

[NOT] EXISTS* table contains at least 1 row or table contains no rows

LIKE partial string operatorThe EXISTS operator is covered in the Teradata Advanced SQL course.

BETWEEN -- Numeric Range Testing

To locate rows for which a numeric column is within a range of values, use the BETWEEN <a> AND operator. Specify the upper and lower range of values that qualify the row.

The BETWEEN operator looks for values between the given lower limit <a> and given upper limit as well as any values that equal either <a> or (i.e., BETWEEN is inclusive.)

Example

Select the name and the employee's manager number for all employees whose job codes are in the 430000 range.

SELECT first_name ,last_name ,manager_employee_number FROM employee WHERE job_code BETWEEN 430000 AND 439999;

An alternative syntax is shown below:

SELECT first_name ,last_name ,manager_employee_number FROM employee WHERE job_code >= 430000 AND job_code <= 439999;



manager_employee_number------------------------------

Loretta Ryan 801

Armando Villegas 1005

BETWEEN -- Character Range Testing

Use the BETWEEN <a> AND operator to locate rows for which a character column

is within a range of values. Specify the upper and lower range of values that qualify the row. BETWEEN will select those values which are greater than or equal to <a> and less or equal to . (BETWEEN is inclusive.)

SELECT last_nameFROM employeeWHERE last_name BETWEEN 'r' AND 's';

last_name------------

RyanNote: 'Stein' is not included because 'S_____' sorts lower than 'Stein'. (Teradata is not case-sensitive by default.)

Set Operator IN Use the IN operator as shorthand when multiple values are to be tested. Select the name and department for all employees in either department 401 or 403. This query may also be written using the OR operator which we shall see shortly. SELECT first_name ,last_name ,department_number FROM employee WHERE department_number IN (401, 403);



department_number---------------------

Darlene Johnson 401

Loretta Ryan 403

Armando Villegas 403

James Trader 401

Set Operator NOT IN

Use the NOT IN operator to locate rows for which a column does not match any of a set of values. Specify the set of values which disqualifies the row.

SELECT first_name

,last_name ,department_number FROM employee WHERE department_number NOT IN (401, 403) ;




Carol Kanieski 301

John Stein 301

NOT IN vs. OR

NOT IN provides a shorthand version of a 'negative OR' request. The following is an example using the OR operator for the query example given above.

Select the name and the department for all employees who are NOT members of departments 401 and 403.

SELECT first_name ,last_name ,department_number FROM employee WHERE NOT (department_number=401 OR department_number=403);




Carol Kanieski 301

John Stein 301

NULL

The following are facts about nulls:

NULL is used to represent the absence of a data value. NULL is not the same as having a value of zero or spaces. NULL indicates that a data value is missing or unknown.

NULL in a comparison operation produces an unknown result. When doing an ascending sort on a column with NULL, NULLs sort before

negative values (numeric) and before blank values (character). To prohibit NULLs, a column must be defined as NOT NULL. Null columns may be compressed to occupy zero row space.

Arithmetic and Comparison Operation on NULL

The following logic applies to operations on NULLs:

Col A Operation Col B Results

10 + NULL NULL

10 - NULL NULL

10 * NULL NULL

10 / NULL NULL

10 > NULL UNKNOWN

10 < NULL UNKNOWN

10 > = NULL UNKNOWN

10 < = NULL UNKNOWN

10 = NULL UNKNOWN

10 < > NULL UNKNOWN

NULL > NULL UNKNOWN

NULL < NULL UNKNOWN

NULL > = NULL UNKNOWN

NULL < = NULL UNKNOWN

NULL = NULL UNKNOWN

NULL < > NULL UNKNOWN

Using NULL in a Select

Use NULL in a SELECT statement, to define that a range of values either IS NULL or IS NOT NULL.

The following table lists two employees with unknown telephone extensions:

To list employee numbers in this table with unknown extensions:

SELECT employee_number FROM employee_phone WHERE extension IS NULL;employee_number--------------------

1025

1005To list employee numbers of people with known extensions:

SELECT employee_number FROM employee_phone WHERE extension IS NOT NULL;employee_number--------------------

1008

1013

LIKE Operator

The LIKE operator searches for patterns matching character data strings.

You must provide two parameters for the LIKE operator: a string expression to be searched and a string pattern for which to search.

The string can contain specific characters, as well as the following "wildcards":

% (indicates zero or more character positions)_ (indicates a single character position)

Here are some examples using the LIKE operator:

String pattern Meaning:

example:

LIKE 'JO%' begins with 'JO'

LIKE '%JO%' contains 'JO' anywhere

LIKE '__HN' contains 'HN' in 3rd and 4th position

LIKE '%H_' contains 'H' in next to last position

LIKE Operator -- Case-Blind Comparison

Case-Sensitivity: ANSI vs. BTET session transaction mode

Case-sensitivity of string comparisons is handled differently in the two different session transaction modes (ANSI or BTET). This can lead to different results for a comparison when run from sessions with different transaction modes. To ensure consistent behavior, either case-sensitive or case-blind compares, use the following:

UPPER or LOWER operators for case-blind comparisonsCASESPECIFIC operator for case-sensitive comparisons

Case-Blind Compare

The UPPER function converts a string to uppercase. Use the UPPER string function on both strings being compared to ensure a case-blind comparison regardless of the session transaction mode. The LOWER function may be similarly used.

The NOT CASESPECIFIC data attribute can also be used to force case-blind compares, however it is not an ANSI-compliant operator.

Problem

Display the full name of employees whose last name contains the letter "r" followed by the letter "a".

Teradata Mode Solution

SELECT first_name ,last_name FROM employee WHERE last_name LIKE '%ra%';

first_name last_name------------------------------ --------------------James TraderPeter RabbitI.B. TrainerRobert CraneLarry Ratzlaff

Since we are in Teradata (BTET) mode, the default for comparisons is non-case-

specific. Thus, we could have also used LIKE '%Ra%' or LIKE '%RA%' and produced the same result.

ANSI Mode Solution

SELECT first_name ,last_name

FROM employeeWHERE last_name LIKE '%ra%';

first_name last_name------------------------------ --------------------James TraderI.B. TrainerRobert Crane

Since we are in ANSI mode, the default for comparisons is case-specific. Thus, we will not pick up Ratzlaff or Rabbit, unless we use 'case blind' testing, seen in the next example.


FROM employeeWHERE LOWER(last_name) LIKE LOWER('%ra%'); first_name last_name------------------------------ --------------------James TraderPeter RabbitI.B. TrainerRobert CraneLarry Ratzlaff By applying the LOWER function to each side of the comparison, all testing is done in a 'case blind' mode. UPPER could have also been used. We could also have used LIKE '%ra%' (NOT CASESPECIFIC) to produce this result, however this is not an ANSI standard approach.

LIKE Operator - Case-Sensitive Comparison

Use the CASESPECIFIC data attribute on one or both of the string expressions being compared to ensure a case-sensitive comparison, regardless of the session transaction mode. CASESPECIFIC is a Teradata extension.

ProblemDisplay the full name of employees whose last name contains "Ra". This is a case-sensitive test.

Teradata Mode Solution


FROM employee WHERE last_name (CASESPECIFIC) LIKE '%Ra%';

first_name last_name------------------------------ --------------------Peter Rabbit

Larry Ratzlaff

The default comparison for Teradata mode is not case-specific.

Use of the Teradata extension (CASESPECIFIC) forces a case-specific comparison.

Because we used the case-specific designator, we don't get James Trader in our answer set but only get answers that contain an uppercase "R". The name 'LaRaye' would have also appeared in this answer set, if it existed in the table. Using LIKE 'Ra%' will return only names that begin with "Ra".

ANSI Mode Solution

SELECT first_name ,last_name FROM employee

WHERE last_name LIKE '%Ra%';

first_name last_name------------------------------ --------------------Peter Rabbit

Larry Ratzlaff

The default comparison for ANSI mode is case-specific.

Use of the Teradata extension (CASESPECIFIC) is not required. This could have also been submitted with LIKE 'Ra%' and produced the same result, however it would have missed a name like 'LaRaye'.

LIKE Operator -- Using Quantifiers

To extend the pattern matching functions of the LIKE operator, use quantifiers.

There are three such quantifiers:

ANY — any single condition must be met (OR logic)SOME — same as ANYALL — all conditions must be met (AND logic)

ANY and SOME are synonyms. Using LIKE ANY and LIKE SOME will give the same result.

Problem

Display the full name of all employees with both "E" and "S" in their last name.

Solution

SELECT first_name ,last_name FROM employee WHERE last_name LIKE ALL ('%E%', '%S%');first_name----------

last_name----------

John Stein

Carol Kanieski

Arnando Villegas

Problem

Display the full name of all employees with either an "E" or "S" in their last name.

Solution

SELECT first_name ,last_name FROM employee WHERE last_name LIKE ANY ('%E%', '%S%');first_name-----------


John Stein

Carol Kanieski

Arnando Villegas

Darlene Johnson

James Trader

LIKE with ESCAPE Character

The "_" and "%" symbols are used as 'wildcards' in the string expression of the LIKE construct. However, what if one of the wildcard symbols is in the expression you are

evaluating. For example, to search for the substring "95%", you must define an ESCAPE character to instruct the Parser to treat the '%' as a non-wildcard character.

To Summarize:

The ESCAPE feature of LIKE lets wildcard characters be treated as non-wildcards.

Characters following the escape character are not treated as wildcards.

Problem

List all objects defined in the Teradata Database Data Dictionary whose names contain "_" (underscore) as the second character.

Solution

SELECT tablenameFROM dbc.tablesWHERE tablename LIKE '_Z_%' ESCAPE 'Z';

TableName------------------------------M_7_Bt_wt_erd1

Things To Notice

The defined escape character is the letter 'Z'. The first "_" (underscore) seen represents a wildcard - i.e., any single arbitrary

character. The "Z_" sequence tells the Parser that the second "_" (underscore) is treated as

a character, not as a wildcard.

When you define an ESCAPE character, any use of the ESCAPE character in the string expression must be immediately followed by either the "_", "%", or the ESCAPE character itself.

Assume the escape character is defined to be the letter 'G'.

LIKE 'G_' means the underscore is treated as an underscore, not as a wildcard.LIKE 'G%' means the percent sign is treated as a percent sign, not as a wildcard.LIKE 'GG' means the two consecutive 'G' characters should be treated as a single 'G', not as an escape character.

Example

LIKE '%A%%AAA_' ESCAPE 'A'

Searches for the following pattern:- any number of arbitrary characters, followed by - the "%" single character, followed by- any number of arbitrary characters, followed by- the "A" single character, followed by- the "_" single character, which is the last character in the string.

Logical Operator -- AND

Logical operators AND, OR and NOT allow you to specify complex conditional expressions by combining one or more logical expressions.

Logical Operator AND

The logical operator AND combines two expressions, both of which must be true in a given record for them to be included in the result set.

Boolean Logic Table True AND True = True

False AND True = False

True AND False = False

False AND False = False

True OR True = True

True OR False = True

False OR True = True

False OR False = False

NOT True = False

NOT false = True

Problem

Display the name and employee number of employees in department 403 who earn less than $35,000 per year.

Solution

SELECT first_name ,last_name ,employee_number FROM employee WHERE salary_amount < 35000.00 AND department_number = 403 ;




Loretta Ryan 1005

Logical Operator -- OR

The logical operator OR combines two expressions. At least one must be true in a given record for it to be included in the result set.

Problem

Display the name and the employee number for employees who either earn less than $35,000 annually or work in department 403.

Solution

SELECT first_name ,last_name ,employee_number FROM employee WHERE salary_amount < 35000.00 OR department_number = 403;

Result

first_name----------

last_name----------

employee_number----------------

Arnando Villegas 1007

Carol Kanieski 1008

Loretta Ryan 1005

John Stein 1006

Multiple AND . . . OR

You want to find all employees in either department 401 or department 403 whose salaries are either under $35,000 or over $85,000.

SELECT last_name ,salary_amount ,department_number FROM employee WHERE (salary_amount < 35000 OR salary_amount > 85000) AND (department_number = 401 OR department_number = 403) ;

Logical Operators -- Combinations

Operator Procedures

Parentheses can be used to force an evaluation order. In the presence of parentheses, expressions are evaluated from the inner-most to outer-most set of parenthetical expressions.

In the absence of parentheses, SQL uses the default precedence of the Boolean operators. By default, NOT has higher precedence than AND which has higher precedence than OR. Operators that have the same precedence level are evaluated from left to right (e.g., multiple ORs or multiple ANDs).

If we remove the parentheses from the example above, our conditional expression gets evaluated in a totally different order. If we were to use it in a query, it would yield different results.

NOT has the highest precedence of the three operators.

Logical Operators -- Using Parentheses

Problem

Select the name, department number, and job code for employees in departments 401 or 403 who have job codes 412101 or 432101.

Solution

SELECT last_name ,department_number ,job_code FROM employee WHERE (department_number = 401 OR department_number = 403) AND (job_code = 412101 OR job_code = 432101) ;

Result



job_code----------

Villegas 403 432101

Johnson 401 412101If we accidentally left the parentheses out of our statement, would we get the same results?

SELECT last_name ,department_number ,job_code FROM employee WHERE department_number = 401 OR department_number = 403 AND job_code = 412101 OR job_code = 432101;last_name----------


job_code---------

Villegas 403 432101

Johnson 401 412101

Trader 401 411100QuestionJames Trader does not have either job code. Why was he selected?

AnswerWithout parentheses, the default precedence causes the expression to be evaluated as follows:

SELECT last_name ,department_number ,job_code FROM employee WHERE department_number = 401 OR (department-number = 403 AND job_code = 412101) OR job_code = 432101;

Remember, if any of the expressions that OR combines are true, the result is true. Since James Trader is in department_number 401, that expression evaluates to true. Therefore the row qualifies for output.

Multiple AND

Use multiple ANDs if you need to locate rows which meet all of two or more sets of specific criteria.

For example, what happens if we take the previous example, and replace all the ORs with ANDs?

SELECT last_name ,department_number ,job_code FROM employee WHERE department_number = 401 AND department_number = 403 AND job_code = 412101 AND job_code = 432101;

No rows will be found. Why?

Answer

For a row to be displayed in this answer set, an employee would have to be in department numbers 401 and 403. While we could conceive of an employee who works in two departments, the employee table has only one field for department number. There is no way that any row in this table can contain more than one department number. This query never will return any rows. A more realistic example follows.

Problem

Find all employees in department number 403 with job_code 432101 and a salary greater than $25,000.

Solution

SELECT last_name ,department_number ,job_code ,salary FROM employee WHERE department_number = 403 AND job_code = 432101 AND salary_amount > 25000.00 ;

Result



job_code----------

salary_amount----------------

Lombardo 403 432101 31000.00

Villegas 403 432101 49700.00

Charles 403 432101 39500.00

Hopkins 403 432101 37900.00

Brown 403 432101 43700.00

Logical NOT

Place the NOT operator in front of a conditional expression or in front of a comparison operator if you need to locate rows which do not meet specific criteria.

Problem

Select the name and employee number of employees NOT in department 301.

Solution for NOT Operator

SELECT first_name ,last_name ,employee_number FROM employee WHERE department_number NOT = 301;

Solution for NOT Condition

SELECT first_name ,last_name ,employee_number FROM employee WHERE NOT (department_number = 301);

Result (Note that both approaches return the same result.)

first_name------------

last_name------------


Arnando Villegas 1007

James Trader 1003

Loretta Ryan 1005

Darlene Johnson 1004

SQL Lab Databases Info Document:Your assigned user-id has SELECT access to the ‘Customer_Service’ and the ‘Student’ databases. These databases are used for labs during this SQL course. When selecting from tables, either set your default database to Customer_Service, Student, or your own user-assigned database (depending on the lab requirements) or alternatively, qualify your table names to reflect the appropriate database. The following pages include table definitions for the ‘Customer_Service’ database and the ‘Student’ database.You may want to print this document for reference throughout this course.

‘Customer_Service’ Table Definitions

CREATE SET TABLE agent_sales ,NO FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( agent_id INTEGER, sales_amt INTEGER) UNIQUE PRIMARY INDEX ( agent_id ); --CREATE TABLE clob_files (Id INTEGER NOT NULL,text_file CLOB(10000))UNIQUE PRIMARY INDEX ( Id ); --CREATE TABLE contact, FALLBACK(contact_number INTEGER

,contact_name CHAR(30) NOT NULL,area_code SMALLINT NOT NULL,phone INTEGER NOT NULL,extension INTEGER,last_call_date DATE NOT NULL)UNIQUE PRIMARY INDEX (contact_number);--CREATE TABLE customer, FALLBACK(customer_number INTEGER,customer_name CHAR(30) NOT NULL,parent_customer_number INTEGER,sales_employee_number INTEGER)UNIQUE PRIMARY INDEX (customer_number);--CREATE SET TABLE daily_sales ,NO FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( itemid INTEGER, salesdate DATE FORMAT 'YY/MM/DD', sales DECIMAL(9,2)) PRIMARY INDEX ( itemid );

--CREATE SET TABLE daily_sales_2004 ,NO FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( itemid INTEGER, salesdate DATE FORMAT 'YY/MM/DD', sales DECIMAL(9,2)) PRIMARY INDEX ( itemid );

--CREATE TABLE department, FALLBACK(department_number SMALLINT,department_name CHAR(30) NOT NULL,budget_amount DECIMAL(10,2),manager_employee_number INTEGER)UNIQUE PRIMARY INDEX (department_number),UNIQUE INDEX (department_name);--CREATE TABLE employee, FALLBACK(employee_number INTEGER,manager_employee_number INTEGER,department_number INTEGER,job_code INTEGER,last_name CHAR(20) NOT NULL,first_name VARCHAR(30) NOT NULL,hire_date DATE NOT NULL,birthdate DATE NOT NULL

,salary_amount DECIMAL(10,2) NOT NULL)UNIQUE PRIMARY INDEX (employee_number);--CREATE TABLE employee_phone, FALLBACK(employee_number INTEGER NOT NULL,area_code SMALLINT NOT NULL,phone INTEGER NOT NULL,extension INTEGER,comment_line CHAR(72))PRIMARY INDEX (employee_number);--CREATE SET TABLE Jan_sales ,NO FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( itemid INTEGER, salesdate DATE FORMAT 'YY/MM/DD', sales DECIMAL(9,2)) PRIMARY INDEX ( itemid );

--CREATE TABLE job, FALLBACK(job_code INTEGER,description VARCHAR(40) NOT NULL,hourly_billing_rate DECIMAL(6,2),hourly_cost_rate DECIMAL(6,2))UNIQUE PRIMARY INDEX (job_code),UNIQUE INDEX (description);--CREATE TABLE location, FALLBACK(location_number INTEGER,customer_number INTEGER NOT NULL,first_address_line CHAR(30) NOT NULL,city VARCHAR(30) NOT NULL,state CHAR(15) NOT NULL,zip_code INTEGER NOT NULL,second_address_line CHAR(30),third_address_line CHAR(30))PRIMARY INDEX (customer_number);--CREATE TABLE location_employee, FALLBACK(location_number INTEGER NOT NULL,employee_number INTEGER NOT NULL)PRIMARY INDEX (employee_number);--CREATE TABLE location_phone, FALLBACK(location_number INTEGER,area_code SMALLINT NOT NULL,phone INTEGER NOT NULL

,extension INTEGER,description VARCHAR(40) NOT NULL,comment_line LONG VARCHAR)PRIMARY INDEX (location_number);

--CREATE TABLE phonelist( LastName CHAR(20),FirstName CHAR(20),Number CHAR(12) NOT NULL,Photo BLOB(10000))UNIQUE PRIMARY INDEX ( Number );

--CREATE TABLE repair_time( serial_number INTEGER ,product_desc CHAR(8),start_time TIMESTAMP(0),end_time TIMESTAMP(0))UNIQUE PRIMARY INDEX(serial_number);

--CREATE SET TABLE salestbl,NO FALLBACK , NO BEFORE JOURNAL, NO AFTER JOURNAL ( storeid INTEGER, prodid CHAR(1),sales DECIMAL(9,2)) PRIMARY INDEX ( storeid );

‘Student’ Database Table Definitions

CREATE TABLE city ,NO FALLBACK ,NO BEFORE JOURNAL,NO AFTER JOURNAL(cityname CHAR(15) NOT CASESPECIFIC,citystate CHAR(2) NOT CASESPECIFIC,citypop INTEGER)PRIMARY INDEX ( cityname );

CREATE TABLE customers ,NO FALLBACK ,NO BEFORE JOURNAL,NO AFTER JOURNAL(cust_id integer not null,cust_name char(15),cust_addr char(25) compress)

PRIMARY INDEX ( cust_id);

CREATE TABLE orders ,NO FALLBACK ,NO BEFORE JOURNAL,NO AFTER JOURNAL(order_id INTEGER NOT NULL,order_date DATE FORMAT 'YYYY-MM-DD',cust_id INTEGER,order_status CHAR(1))UNIQUE PRIMARY INDEX ( order_id );

CREATE TABLE state ,NO FALLBACK ,NO BEFORE JOURNAL,NO AFTER JOURNAL(stateid CHAR(2) NOT CASESPECIFIC NOT NULL,statename CHAR(15) NOT CASESPECIFIC,statepop INTEGER NOT NULL,statecapitol CHAR(15) NOT CASESPECIFIC)PRIMARY INDEX ( stateid );

Lab


To start the online labs, click on the Telnet button in the lower left hand screen of the course. Two windows will pop-up: a BTEQ Instruction Screen and your Telnet Window.

Sometimes the BTEQ Instructions get hidden behind the Telnet Window. You will need these instructions to log on to Teradata.

Be sure to change your default database to the Customer_Service database in order to run these labs.

A. Management needs a list with employee number, last name, salary and job code of all employees earning between $40,001 and $50,000. Produce this report using the comparison operators and sort by last name.

Modify the query to use the BETWEEN operator and sort by salary descending.

B. Management requests a report of all employees in departments 501, 301, and 201 who earn more than $30,000. Show employee number, department number, and salary. Sort by department number and salary amount within department.

C. Display location number, area code, phone number, extension, and description from the location_phone table if the description contains the string ‘manager'. Sort by location number.

D. List customers from the customer table identified by:1. An unknown parent customer number; and

2. A known sales employee number.

Display all columns except parent customer number. Order results by customer number.

E. Create a report that lists all location numbers in area code 415 not dedicated to any manager. Include all columns except comment and extension. Sequence by phone number.

F. List all job codes in the 400000 series that are management and all Analyst positions regardless of job code value. List job code and description for the selected rows. Sort by description. Submit again using EXPLAIN to look at how the Teradata RDBMS handled this request.

G. List the location numbers in area codes 415 or 617 which have no phone extensions. Order by area code and phone number. Include all columns of the table except for description and comment line columns. Notice how NULLs appear in your output.

Lab Answer A

DATABASE Customer_Service;SELECT employee_number

,last_name,salary_amount,job_code

FROM employeeWHERE salary_amount >= 40001AND salary_amount <= 50000ORDER BY last_name;


employee_number

last_name

salary_amount

job_code

1024Brown 43700.00 432101

1002Brown 43100.00 4132011010Rogers 46000.00 412101

1007Villegas

49700.00 432101

SELECT employee_number,last_name,salary_amount,job_code

FROM employeeWHERE salary_amount BETWEEN 40001 AND 50000ORDER BY salary_amount DESC;


employee_number

last_name

salary_amount

job_code

1007 Villegas 49700.00 432101

1010 Rogers 46000.00 4121011024 Brown 43700.00 4321011002 Brown 43100.00 413201

Lab Answer B

SELECT employee_number,department_number,salary_amount

FROM employeeWHERE salary_amount > 30000AND department_number IN (501,301,201)ORDER BY department_number

,salary_amount;


employee_number

department_number

salary_amount

1025 201 34700.001021 201 38750.001019 301 57700.001015 501 53625.001018 501 54000.001017 501 66000.00

Lab Answer C

SELECT location_number,area_code,phone,extension,description

FROM location_phoneWHERE description LIKE '%Manager%'ORDER BY location_number;


location_number area_code phone extension description 5000001 415 4491225 ? System Manager14000020 312 5692136 ? System Manager22000013 617 7562918 ? System Manager31000004 609 5591011 224 System Manager33000003 212 7232121 ? System Manager

Lab Answer D

SELECT customer_number,customer_name,sales_employee_number

FROM customerWHERE parent_customer_number IS NULLAND sales_employee_number IS NOT NULLORDER BY 1;

*** Query completed. 14 rows found. 3 columns returned. customer_number customer_name sales_employee_number -------------- ---------------------------- ----------------------- 1 A to Z Communications, Inc. 1015 3 First American Bank 1023 5 Federal Bureau of Rules 1018 6 Liberty Tours 1023 7 Cream of the Crop 1018 8 Colby Co. 1018 9 More Data Enterprise 1023 10 Graduates Job Service 1015 11 Hotel California 1015 12 Cheap Rentals 1018 14 Metro Savings 1018 15 Cates Modeling 1015

17 East Coast Dating Service 1023 18 Wall Street Connection 1023

Lab Answer E

SELECT location_number,area_code,phone,description

FROM location_phoneWHERE area_code = 415AND description NOT LIKE '%Manager%'ORDER BY phone;


location_number area_code phone description --------------- --------- ----------- ------------- 5000010 415 2412021 Alice Hamm 5000011 415 3563560 J.R. Stern 5000001 415 4491221 FEs office 5000001 415 4491244 Secretary 5000019 415 6567000 Receptionist 5000002 415 9237892 Receptionist

Lab Answer F

SELECT job_code,description

FROM jobWHERE (job_code BETWEEN 400000 AND 499999AND description LIKE '%manager%')OR (description LIKE '%analyst%')ORDER BY description;


job_code description ----------- ----------------------------- 411100 Manager - Customer Support

431100 Manager - Education 421100 Manager - Software Support 422101 Software Analyst 222101 System Analyst 412103 System Support Analyst

Note: EXPLAIN text varies depending on optimizer path.

Lab Answer G

SELECT location_number,area_code,phone,extension

FROM location_phoneWHERE extension IS NULLAND area_code IN (415,617)ORDER BY area_code

,phone;


location_number area_code phone extension --------------- --------- ----------- ----------- 5000010 415 2412021 ? 5000011 415 3563560 ? 5000001 415 4491221 ? 5000001 415 4491225 ? 5000001 415 4491244 ? 5000019 415 6567000 ? 5000002 415 9237892 ? 22000013 617 7562918 ? 22000013 617 7567676 ?

Data Types and Conversions Module:- 6

Objectives


Define the data type for a column in a table. Compute values using arithmetic functions and operators. Convert data from one type to another. Manipulate the DATE data type.

Data Types

Every column in a row is associated with a data type that defines the kind of values it accepts. These data types are associated with the fields when the table is created. Data types fall into one of the four categories shown below:

Data Type Holds Character data Character Strings

Byte data Binary Data Strings

Numeric data Numbers

Date/Time data

Dates,Times,Timestamps,Time Intervals(Note: Time, Timestamps and Intervals are covered in the Teradata SQL Advanced course)

Character Data

There are two data types for holding character data:

CHAR - has fixed-length character strings. VARCHAR - has variable-length character strings.

In both cases, the maximum string length is 64,000 characters. String size is specified in parentheses after the keyword CHAR or VARCHAR as in CHAR(20) or VARCHAR(20). There are subtle differences between the two.

In the case of CHAR, if you insert a character string less than the specified size (20 in the above example), the system will append and store trailing blanks as needed to expand the string to the specified length. The size specified on a VARCHAR field represents a maximum length. The system will not pad the contents with trailing blanks. (It will use two extra bytes to store the length internally however.)

When you know that data will always be a specific length (e.g., a Mail Stop that is a four-character length) use CHAR. You save the two bytes that VARCHAR would use to store the length. When data does not have a fixed length (e.g., last names), VARCHAR makes more efficient use of the space, because it does not pad values with trailing blanks.

Character Data Description Example

CHAR (size) Fixed length string Max: 64,000 characters

last_name CHAR(20) Sample contents:'Ryan__________'

VARCHAR (size) CHAR VARYING (size) CHARACTER VARYING (size)

Variable length string Max: 64,000 characters

first_name VARCHAR(30)

Sample contents: 'Loretta'

LONG VARCHAR Equivalent to VARCHAR (64000)

Note: VARCHAR(size), CHAR VARYING(size), and CHARACTER VARYING(size) are all equivalent synonyms.

LONG VARCHAR is equivalent to a maximum VARCHAR.

Character strings may be defined using any one of the following character sets:

Character Set Max Length

Latin 64,000 (English language default)

Kanji1 64,000

KanjiSJIS 64,000

Unicode 32,000 (two bytes per character)

Graphic 32,000 (two bytes per character)

Examples

first_name VARCHAR(30) CHARACTER SET LATIN

image_abc CHAR(32000) CHARACTER SET GRAPHIC

Byte Data

Teradata supports two data types for holding binary data:

BYTE. VARBYTE.

BYTE is for fixed length binary strings. VARBYTE is for variable length binary strings. These data types are used for storing binary objects such as digital images, executable objects, flat files, etc.

Byte Data Description

BYTE (size) Fixed length Binary string

Default: (1) Max: 64,000 bytes

VARBYTE (size) Variable length Binary string Default: (1) Max: 64,000 bytes

Note: BYTE columns are not covertible to other data types.

Numeric Data

The following numeric data types are available:

Data Type Description Examples BYTEINT* Whole number

Range -128 to 127Storage: 1 byte

basket_countBYTEINT +125

SMALLINT Whole number Range: -32,768 to 32,767Storage: 2 bytes

area_code SMALLINT +00213 (up to 5 digits)

INTEGER Whole number Range: -2,147,483,648 to 2,147,483,647Storage: 4 bytes

phone INTEGER +0006495252 (up to 10 digits)

BIGINT Whole numberRange ± 9,233,372,036,854,775,807Storage: 8 bytes

international_phoneBIGINT+0000010807934842145 (up to 19 digits)

DEC(m, n)DECIMAL(m,n)NUMERIC(m,n)(Small Decimal)

'm' is total number of digits. (m <= 18) 'n' is precision places to the right of decimal point.Max Size =18 digitsStorage: 2 to 8 bytes

salary_amount DEC (10,2) +00035000.00

DEC(m, n)DECIMAL(m,n)NUMERIC(m,n)(Large Decimal)

'm' is total number of digits. (18 < m <= 38) 'n' is precision places to the right of decimal point.Max Size = 38 digitsStorage: 16 bytes

sheet_thicknessDEC (32,30) +01.123498782792787582977278497827

FLOAT Floating Point Format (IEEE)2x10-307 to 2x10+308

Storage: 8 bytes

salary_factor FLOAT 4.35400000000000E-001

FLOAT [(precision)]

Internally represented as FLOAT

REAL Internally represented

as FLOAT

DOUBLE PRECISION

Internally represented as FLOAT

* Non-ANSI standard data type

Date Data

DATE represents a calendar date. It simplifies handling and formatting dates common to business applications. DATE is stored as an integer, using the following formula: (year - 1900) * 10000 + (month * 100) + day.

The DATE data type, shown here, is used in several forthcoming examples.

Data Type Description Examples DATE Specially formatted integer:

YYMMDD thru 1999 YYYMMDD from 2000 onward

hire_date DATE +0000990921 +0001000215

In the first example, the date being represented is September 21, 1999.In the second example, the date being represented is February 15, 2000.

Note that the year is represented as an offset from the year 1900 in both cases. Note, this is how the date is stored internally. How you see the date in a report depends on how the date is formatted.

The recommended display format for DATE values is ‘YYYY-MM-DD’. This format will avoid problems that may occur from formats specifying only two digits for the year. This is a significant issue as dates for the year 2000 and beyond are now commonly entered.

Example

2006-03-14 formatted as ‘YY-MM-DD’ displays ad 01-03-14, which is also the representation for 1901-03-14.2006-03-14 formatted as ‘YYYY-MM-DD’ displays as 2006-03-14, which is unambiguous.

Note: Formatting techniques for dates are discussed in the forthcoming module 'Output Attributes'.

Arithmetic Operators

Teradata provides six binary arithmetic operators and two unary arithmetic operators. They are defined in the tables below:

Binary Operators (require two operands)

Operator Definition

** exponentiation

* Multiply

/ Divide

MOD Modulos (remainders)

+ Add

- SubtractUnary Operators (require a single operand)

Operator Definition

+ Positive value

- Negative valueTeradata supports two arithmetic operators which provide additional functionality beyond ANSI SQL:

** (exponentiation) MOD (modulo arithmetic)

** (exponentiation) is used in the form: <n>**<arg>, where <n> is raised to the power of <arg>.

Example

4**3 = 4 * 4 * 4 = 64

MOD is the modulo operator. It calculates the remainder in a division operation.

Example

60 MOD 7 = 4 Sixty divided by 7 equals 8, with a remainder of 4

Arithmetic Functions

Teradata supports several arithmetic functions which provide additional capabilities beyond those available in ANSI SQL. They are described in the table below:

Function Result

ABS (arg) Absolute value

EXP (arg) Raises e to the power of arg (e ** arg)

LOG (arg) Base 10 logarithm

LN (arg) Base e (natural) logarithm

SQRT (arg) Square rootArg is any constant or variable numeric argument.

Examples

SQRT (16) = 4

ABS (a) is equivalent to a positive a, regardless of whether the actual a is positive or negative.

if a = -600 then ABS (a) = 600 if a = +400 then ABS (a) = 400

Computed Values

Use arithmetic expressions in your SELECT to perform calculations on data in the table. Arithmetic expressions will allow you to do two things:

Re-label the column heading for purposes of displaying the output. Perform the calculation on the data values in that column.

Example

Create an alphabetical list with monthly salary of department 401 employees.

SELECT last_name ,first_name ,salary_amount/12 FROM employee WHERE department_number = 401 ORDER BY last_name ;last_name----------

first_name----------

salary_amount/ 12 --------------------

Johnson Darlene 3025.00

Trader James 3154.17Note: salary_amount/12 is considered a 'computed column', because an arithmetic operation is performed on the underlying value. Also, the column title reflects the computation performed.

SELECTing System Variables

A DEFAULT function exists which permits the retrieval of the default value assigned to a particular column. For example, assume we define a table as follows:

CREATE TABLE TableX (Col1 INTEGER, Col2 INTEGER DEFAULT 10, Col3 INTEGER DEFAULT 20, Col4 CHAR(60));

We may retrieve the default values for any column by using the DEFAULT function as follows:

SELECT DEFAULT(Col2), DEFAULT(Col3) FROM TableX;

Default(Col2) Default(Col3)------------- ------------- 10 20 10 20

Default values are returned in place of the normal column values. Note that two rows are returned. This is because there are currently two rows in the table. One row of output is returned for each existing row in the table. You may return a single row by using the following syntax:

SELECT DISTINCT DEFAULT(Col2), DEFAULT(Col3) FROM TableX;

Default(Col2) Default(Col3)------------- ------------- 10 20

If there are no rows in the table, no default values will be returned.

If no default values are assigned to the specified column, then a null is returned.

Literal, Constant, and Calculator Features

Character Literals

You may add character literals to your SELECT statement:

SELECT 'Employee' ,last_name ,first_name FROM employee ;

Employee last_name first_name-------- --------- ----------Employee Stein JohnEmployee Kanieski CarolEmployee Ryan Loretta

Numeric Constants:

You may also add numeric constants to your SELECT statement:

SELECT 12345 ,last_name ,first_name FROM employee ;

12345 last_name first_name ----- --------- ---------- 12345 Stein John 12345 KanieskiCarol

12345 Ryan Loretta

Calculator Mode:

You may use numeric expressions to do calculations:

Calculation Result Comment

SELECT 2*250; 500 Simple multiplication

SELECT 1.01 + 2.2; 3.2 Uses greatest possible precision

SELECT 10/3.000; 3.333 Rounds off

SELECT 10/6.000; 1.667 Rounds up

MOD Operator

The MODULO or remainder operator is a Teradata extension which calculates the remainder in a division operation. In earlier versions of Teradata SQL, MODULO was used significantly for DATE arithmetic. With the addition of the EXTRACT function (seen in a later module), MODULO is less commonly used for date extraction operations (i.e., extracting the month from a date).

Example

7 ÷ 4 = 1 remainder of 3 Therefore 7 MOD 4 = 3

Problem

Identify zip codes in the location table ending in four zeros.

SELECT customer_number ,zip_code FROM location WHERE zip_code MOD 10000 = 0;

customer_number-------------------

zip_code-----------

10 940710000

14 1210000 You could do this search with string comparisons, but it is more efficient to use MOD because no character conversion is required.

Note: Leading zeros are suppressed by default.

DATE Data Type

The Teradata DATE data type is used to store calendar dates representing year, month

and day. It is stored internally as a four-byte integer.

The Teradata database performs basic DATE handling functions including:

Month-to-month conversions. Year-to-year conversions. Leap-year conversions. Dates prior to 1900. Dates after 2000.

The database treats the DATE data type as an INTEGER value but does not permit invalid calendar dates. The system encodes dates using the formula:

((YEAR - 1900) * 10000) + MONTH * 100) + DAY

For March 31, 1997 the calculation is:

YEAR = (1997 - 1900) * 10000 = 970000

MONTH = (3 * 100) = 300

DAY = = 31

DATE = = 970331

For March 31, 2004 the calculation is:

YEAR = (2004 - 1900) * 10000 = 1040000

MONTH = (3 * 100) = 300

DAY = = 31

DATE = = 1040331

For January 1, 1900 the result is 000101.

For dates before Jan 1, 1900, the integer calculation returns a negative number. Remember, this is only how the date is stored, not how it is represented in a query output.

DATE Arithmetic

SQL permits us to perform arithmetic calculations on DATE. When the DATE keyword is SELECTed it is a system variable which represents the current date. (CURRENT_DATE may also be used and is the preferred ANSI standard). As we saw in the previous example, the DATE keyword is always replaced with the current system date.

Find Syntax

The date 30 days from today SELECT DATE + 30;

Since days is the basic unit, we need to express a year in terms of the number of days it contains. DATE + 365 (or in a leap-year, DATE + 366), gives us the date one year from today.

Find Syntax

The date 1 year from now SELECT DATE + 365;

How would you find a person's age (in rounded years)?

SELECT (DATE-birthdate)/365 FROM employee WHERE last_name = 'Stein';

((Date-birthdate)/365) ---------------------- 45

This example takes today's date, (encoded as an integer) and "subtracts" John Stein's birth date, (also encoded as an integer) and divides by the number of days in a year (365).

Suppose today's date is 980826 and John's birth date is 531015. If you did this as a straight integer subtraction, you would find that 980826 - 531015 = 449811. 449811 is not a legal date! Because these are DATE data types, not INTEGERS, the subtraction operator actually returns the number of days between the two operands instead of subtracting the INTEGERs.

(DATE - birth date) thus provides us with the number of days elapsed between a birth date and today. Dividing that amount by the number of days in a year (365) gives us John's age in years. To account for the leap year, we could divide by 365.25 to get a more precise number of years.

Problem

Find employees with more than 10 years of service.

SELECT last_name ,hire_date FROM employee WHERE (DATE-hire_date)/365 > 10;

last_name hire_date-------------------- ---------Ratzlaff 78/07/15Villegas 77/01/02Crane 78/01/15Machado 79/03/01Trader 76/07/31Runyon 78/05/01Rogers 77/03/01Brown 76/07/31Morrissey 78/10/01Stein 76/10/15Charles 78/10/01Rogers 78/03/01Trainer 73/03/01Phillips 77/04/01

Ryan 76/10/15Wilson 78/03/01Lombardo 77/02/01Short 79/05/01Hoover 76/06/18Hopkins 77/03/15Johnson 76/10/15Rabbit 79/03/01Kubic 78/08/01Brown 79/05/01Kanieski 77/02/01Daly 77/03/15

The example shows that operations using DATE may be included in the WHERE clause.

Date Function ADD_MONTHS

The ADD_MONTHS function allows the addition of a specified number of months to an existing date, resulting in a new date.

The format of the function is:

ADD_MONTHS (date, n)

where n is a number or an expression representing the number of months to be added to the date.

The following examples demonstrate its usage.

Query ResultsSELECT DATE; /* March 20, 2001 */ 01/03/20

SELECT ADD_MONTHS (DATE, 2) 2001-05-20

SELECT ADD_MONTHS (DATE, 12*14) 2015-03-20

SELECT ADD_MONTHS (DATE, -3) 2000-12-20

Note: The results of the ADD_MONTH function are always displayed in YYYY-MM-DD format.

ADD_MONTHS may also be applied to a literal date, which must be specified in the YYYY-MM-DD format.

Query Results

SELECT ADD_MONTHS ('2004-07-31', 2) 2004-09-30



ADD_MONTHS accounts for the Gregorian calendar and knows how many days are in each month, including leap years such as 2004.

Comparison of ADD_MONTHS and simple DATE arithmetic

Because of the variable number of days in a month, ADD_MONTH provides a much higher degree of accuracy when projecting dates based on month increments.

Problem

Show the date two month from today (March 20, 2001) using both ADD_MONTHS and simple arithmetic methods.

SELECT DATE ,ADD_MONTHS(DATE, 2) ,DATE + 60;

Date ADD_MONTHS (Date, 2) Date+60) --------- -------------------- ---------- 01/03/20 2001-05-20 01/05/19

Time Related Data

Teradata supports two data types for holding time-related data:

TIME (WITH ZONE) TIMESTAMP (WITH ZONE)

Time is represented as a TIME data type which in reality carries three different fields of information: hours, minutes and seconds. It also has 'clock intelligence' in its implementation, just as DATE has calendar intelligence, thus making possible complex calculations.

TIMESTAMP is a data type which combines both DATE and TIME into a single data type.

Both TIME and TIMESTAMP have a WITH ZONE option, which allows for time-zone dependent storage and processing of time related information.

Time Data Description TIME Single column - representing 3 fields

6 Bytes Length:- Hour - Byteint- Min - Byteint- Secs. - Dec(8,6)

TIME WITH ZONE Single column - representing 5 fields6 Bytes TIME:- Hour - Byteint- Min - Byteint- Secs. - Dec(8,6)

2 Bytes ZONE:- Hour - Byteint- Minute - Byteint

Timestamp Data Description

TIMESTAMP Single column - representing 6 fields 10 Bytes Length:Date = 4 BytesTime = 6 Bytes

TIMESTAMP WITH ZONE

Single column - representing 8 fields12 Bytes Length:- Date = 4 Bytes- Time = 6 Bytes- Zone = 2 Bytes

Note: TIME and TIMESTAMP data types are covered in greater detail the Advanced Teradata SQL Web-based training.

Interval Data Types

Interval data types represent a displacement between two points in time. Intervals are stored internally as one or multiple numeric fields combined into a single data type.

There are two general categories of INTERVAL data types:

1. Year-Month Intervals 2. Day-Time Intervals

Year-Month Intervals

The following Year-Month interval data types are available:

YEAR YEAR TO MONTH MONTH

Day-Time Intervals

The following Day-Time interval data types are also available:

DAY DAY TO HOUR DAY TO MINUTE DAY TO SECOND HOUR HOUR TO MINUTE HOUR TO SECOND

MINUTE MINUTE TO SECOND SECOND

Note: INTERVAL data types are covered in greater detail the Advanced Teradata SQL Web-based training.

Data Conversions Using CAST

The CAST function allows you to convert a value or expression from one data type to another. For example:

Numeric to Numeric Conversion

SELECT CAST (50500.75 AS INTEGER); Result: 50500 (truncated), SELECT CAST (50500.75 AS DEC (6,0)); Result: 50501. (rounded).

When you cast from a decimal to an integer, the system discards everything to the right of the decimal point. In our example, even though the decimal portion was .75, making the number closer to 50501 than to 50500, the result of the cast is still 50500.

Decimal to Decimal Conversions

Casting from a decimal at one precision level to a decimal at a lesser precision level causes the system to round the result to the nearest value displayed at the new precision level.

SELECT CAST(6.74 AS DEC(2,1)); Result: 6.7 (Drops precision)

SELECT CAST(6.75 AS DEC(2,1)); Result: 6.8 (Rounds up to even number)

SELECT CAST(6.85 AS DEC(2,1)); Result: 6.8 (Rounds down to even number)

Character to Character Conversions

SELECT CAST(last-name AS CHAR (5)) FROM employee WHERE department_number = 401;last_name------------

JohnsTradeAs you can see from the example above, CAST can take the CHAR(20) last_name

data field from the employee table and convert it to a CHAR(5) data type. The values displayed as a result of this query are truncated to five characters or padded with trailing blanks up to five characters (if there are less than five characters to start with).

Alternative Syntax

In current and previous versions of Teradata, data type conversions can also be done without the CAST function as seen here:

SELECT last_name (CHAR (5)) FROM employee;

This type of conversion syntax is considered shorthand and is not ANSI standard.

Casting Attributes

In addition to simple CASTing, Teradata allows you to specify data attributes as well as data types in your CAST.

SELECT CAST(last_name AS CHAR(5) UPPERCASE) FROM employee WHERE department_number = 401;last_name------------

JOHNSTRADEThings To Notice:

The preceding example would return an error in an ANSI Mode session. Truncation of character strings is not permitted in an ANSI Mode CAST

expression.

Lab




A. In a single SELECT statement, display the current DATE, TIME and USER.

Display the date 365 days from today.

B. Create a report that gives the literal value "Math Function" as its first column, the numeric value 1 as its second column, and the calculation 2*3+4*5 as its third column. Note the results. Modify this request to include parentheses around the 3+4.

Resubmit the request. Note the difference in the result.

C. Some employees are getting raises. The department 401 manager is interested in what salaries would be if each employee received a 10 percent increase. List the first 10 characters of the last name, current salary, and salary after the raise for the employees in department 401. Rank them in descending order by salary.

Submit again using EXPLAIN to look at how Teradata will process this query.

Note: This lab is used again for exercise D.

D. Results of exercise C left management a bit shaken. They are considering instead a flat $500.00 per employee raise. Modify lab C to include this new estimated salary as well as the 10% increase for comparison. Include an additional column that tells the net difference in dollars between the two options for each employee. Rename this additional column as "Difference". Take the current salary column out of the report.

E. The government has ordered a special employment demographics study. Generate a list of all employees under 50 years old. Include last name (CAST the first ten characters only), department number and computed age in years.

Sort so that the youngest employees are listed first.

Lab Answer A

SELECT DATE, TIME, USER;

*** Query completed. One row found. 3 columns returned.

Date Time User -------- -------- -------------- 00/11/29 12:40:34 tdxxx

SELECT DATE + 365;

*** Query completed. One row found. One column returned.

(Date+365) ---------- 01/11/29

Lab Answer B

SELECT 'Math Function',1,2*3+4*5

;


'Math Function' 1 ((2*3)+(4*5)) --------------- ---- ------------- Math Function 1 26SELECT 'Math Function'

,1,2*(3+4)*5

;


'Math Function' 1 ((2*(3+4))*5) --------------- ---- ------------- Math Function 1 70

Lab Answer C

DATABASE Customer_Service;SELECT CAST (last_name AS CHAR(10))

,salary_amount,salary_amount * 1.1

FROM employeeWHERE department_number = 401ORDER BY salary_amount DESC;


last_name salary_amount (salary_amount*1.1) ---------- ------------- -------------------

Rogers 46000.00 50600.000 Brown 43100.00 47410.000 Trader 37850.00 41635.000 Johnson 36300.00 39930.000 Machado 32300.00 35530.000 Hoover 25525.00 28077.500 Phillips 24500.00 26950.000


Lab Answer D

SELECT CAST (last_name AS CHAR(10)),salary_amount * 1.1,salary_amount + 500,(salary_amount*1.1)-(salary_amount+500) AS

DifferenceFROM employeeWHERE department_number = 401ORDER BY salary_amount DESC;

*** Query completed. 7 rows found. 4 columns returned.last_name (salary_amount*1.1) (salary_amount+500) DifferenceRogers 50600.000 46500.00 4100.000Brown 47410.000 43600.00 3810.000Trader 41635.000 38350.00 3285.000Johnson 39930.000 36800.00 3130.000Machado 35530.000 32800.00 2730.000Hoover 28077.500 26025.00 2052.500Phillips 26950.000 25000.00 1950.000

Lab Answer E

SELECT CAST (last_name AS CHAR(10) ),department_number,(date - birthdate) / 365

FROM employeeWHERE (date - birthdate) / 325.25 < 50ORDER BY birthdate DESC;*** Query completed. 7 rows found. 3 columns returned.

last_name department_number ((Date-birthdate)/365) ---------- ----------------- ---------------------- Phillips 401 41

Rabbit 501 42 Crane 402 44 Rogers 302 45 Kanieski 301 46 Machado 401 47 Wilson 501 47Ryan 403 49

Note: Ages of employees will vary over time.

SubQueries Module:- 7

Lab




Click on the buttons to the left to see the answers for each lab question.

A. Use EXISTS to determine if any job_codes are unassigned to employees. Return ‘TRUE’ if there are unassigned job codes

B. List the last_name, first_name, and job code, for employees with invalid job codes. Use a subquery to establish a referential integrity check.

Note: You may resolve this problem by using quantifiers or by not using quantifiers. See if you can execute it both ways.

C. Write a query to find employees who have invalid Department Numbers. List the employee number, last name, and the invalid department number. Note that department numbers are valid only if they appear in the department table, or they are NULL. This query performs a check for referential integrity violations. If no rows are found, what does that tell you?

D. List the customer number, customer name, and sales representative's number for customers in California. Sort by sales representative's number. Now do an EXPLAIN on the request and note the total estimated time.

Hint: The location table contains both the customer number and the state.

Note: This lab is used again for Lab B in Module 10.

E. Find the names of any department manager whose job code is 211100. List their employee number, last name, department number, and job code. You will need to use a subquery to identify which employees are department managers.

Note: This lab is used again for Lab F below.

F. In Lab E, you found information about department managers whose job code is 211100. In this lab, you need to find information about employees who work for the department managers with job code 211100. List the employee's employee number, last name, department number, and job code. To solve this problem, you'll need to use nested subqueries. Try to do it without any AND's. You may satisfy this request more than one way.

Lab Answer A

DATABASE Customer_Service;SELECT ‘TRUE’WHERE EXISTS

(SELECT job_code FROM jobWHERE job_code NOT IN

(SELECT job_code FROM employee));

‘TRUE’ TRUE

Lab Answer B

(This is the solution NOT using a quantifier)

SELECT last_name,first_name,job_code

FROM employeeWHERE job_code NOT IN

(SELECT job_codeFROM job)

;


last_name first_name job_code -------------------- ------------------------------ ----------- Short Michael 211100 (This is the solution using a quantifier)

SELECT last_name,first_name,job_code

FROM employeeWHERE job_code NOT = ALL

(SELECT job_codeFROM job)

;


last_name first_name job_code -------------------- ------------------------------ ----------- Short Michael 211100

Lab Answer C

SELECT employee_number,last_name

,department_numberFROM employeeWHERE department_number IS NOT NULLAND department_number NOT IN

(SELECT department_numberFROM department)

;

*** Query completed. No rows found.

Lab Answer D

SELECT customer_number,customer_name,sales_employee_number

FROM customerWHERE customer_number IN

(SELECT customer_numberFROM locationWHERE state = 'California')

ORDER BY sales_employee_number;


customer_number customer_name sales_employee_number --------------- ------------------------------ --------------------- 0 Corporate Headquarters ? 2 Simple Instruments Co. 1015 15 Cates Modeling 1015 11 Hotel California 1015 1 A to Z Communications, Inc. 1015 19 More Data Enterprise 1015 10 Graduates Job Service 1015

NOTE: EXPLAIN text will vary depending on optimizer path.

Lab Answer E

SELECT employee_number,last_name,department_number,job_code

FROM employeeWHERE job_code = 211100

AND employee_number IN(SELECT manager_employee_numberFROM department)

;

*** Query completed. One row found. 4 columns returned.employee_number last_name department_number job_code

1025 Short 201 211100

Lab Answer F

Solution NOT Using a Quantifier


FROM employeeWHERE manager_employee_number IN(SELECT manager_employee_numberFROM departmentWHERE manager_employee_number IN

(SELECT employee_numberFROM employeeWHERE job_code = 211100) )

;


employee_number last_name department_number job_code1021 Morrissey 201 222101

Solution Using a Quantifier


FROM employeeWHERE manager_employee_number = ANY

(SELECT manager_employee_numberFROM departmentWHERE manager_employee_number = ANY

(SELECT employee_numberFROM employeeWHERE job_code = 211100) )

;

*** Query completed. One row found. 4 columns returned.employee_number last_name department_number job_code

1021 Morrissey 201 222101

Output Attributes Module:- 8


Use TITLE to add a heading to your output that differs from the column or expression name.

Use AS to specify a name for a column or expression in a SELECT statement.

Use CHARACTERS to determine the number of characters in a string. Use TRIM to Trim blank characters or binary zeroes from data. Use FORMAT to alter the display of a column or expression.

Attributes and Functions

Attributes are characteristics which may be defined for columns, such as titles and formats.

Functions are performed on columns to alter their contents in some way.

Expressions are columns and/or values combined with mathematical operators. (i.e. Col1 + Col2 + 3).

Attributes for columns and expressions include the following:

AS Provides a new name for a column. ANSITITLE Provides a title for a column. Teradata

ExtensionFORMAT Provides formatting for a column. Teradata

Extension

Functions for columns and expressions include the following:

CHARACTERS

Count the number of characters in a column.

Teradata Extension

TRIM Trim the trailing or leading blanks or binary zeroes from a column.

ANSI

AS: Naming A Column or Expression

The AS clause:

Assigns a temporary name to a column or expression. Can be referenced elsewhere in the query. Will default column headings to the newly assigned name.

Use of Keyword AS is optional and may be omitted.

SELECT last_name,first_name,salary_amount / 12 AS monthly_salary

FROM employeeWHERE department_number = 401ORDER BY monthly_salary;



monthly_salary--------------------


Trader James 3154.17

Note: Once you have renamed a column with the AS clause you may re-use that name within your SQL statement. Renaming does not prevent you from using any of the other methods we have discussed in your statement, for example:

order by salary_amount/12 order by salary_amount order by 3 (using the positional reference)

Teradata Extensions

The Teradata RDBMS still supports the following form of specifying a name for a column or expression:

<col or expression> (NAMED <name>)

This extension was introduced in a previous release.

The Teradata Equivalent to the previous example is:

SELECT last_name,first_name,salary_amount / 12 (NAMED monthly_salary)

FROM employee

WHERE department_number = 401ORDER BY monthly_salary

Warning! NAMED is the old way of renaming a column for output display. AS is the new method. Currently, both attributes work. In the future, support for the NAMED attribute will disappear. Since the Teradata extension described above provides equivalent functionality to the ANSI compliant version, all existing Teradata SQL applications that use this form to name a column or expression should be updated to use the ANSI compliant form (AS).

TITLE Attribute

The TITLE attribute allows you to rename columns for output with headings that may contain multiple lines and/or blanks. Column headings may stack up to three levels.

SELECT last_name ,first_name ,salary_amount / 12 (TITLE 'MONTHLY // SALARY') FROM employee WHERE department_number = 401 ORDER BY 3;last_name-----------


MONTHLY SALARY------------------


Trader James 3154.17Column heading justification on titles is determined by the data type of the column. In our example, the words "MONTHLY" and "SALARY" are right-justified because that is the default justification for DECIMAL values contained in that column.

The new title must appear between single quotes ( ' ) in the TITLE phrase. The // is used to represent a line break, causing the stacking. You can do the same thing with AS, which is the ANSI standard. Literal strings must be enclosed with double quotes (") when using AS.

SELECT last_name ,first_name ,salary_amount / 12 AS "MONTHLY // SALARY" FROM employee WHERE department_number = 401

ORDER BY 3;last_name-----------


MONTHLY SALARY------------------


Trader James 3154.17

CHARACTERS Function

The CHARACTERS function is a Teradata-specific function which counts the number of characters in a string. It is particularly useful for working with VARCHAR fields where the size of the string can vary from row to row. In the following example, first_name is a VARCHAR field. In order to determine which employees have more than five characters in their name, we apply the CHARACTERS function to the first_name column.

Problem

To find all employees who have more than five characters in their first name.

Solution

SELECT first_name FROM employee WHERE CHARACTERS (first_name) > 5;first_name------------

Loretta Darlene Arnando

Characters also may be abbreviated to CHAR or CHARS (e.g. "WHERE CHAR (first_name) > 5").

The ANSI-standard version of this function (CHARACTER_LENGTH) may also be used.

TRIM Function

Use the TRIM function to suppress leading and/or trailing blanks in a CHAR column or leading and/or trailing binary zeroes in a BYTE or VARBYTE column. TRIM is most useful

when performing string concatenations.

There are several variations of the TRIM function:

TRIM ([expression]) leading and trailing blanks/binary zeroesTRIM (BOTH FROM [expression]) leading and trailing blanks/binary zeroesTRIM (TRAILING FROM[expression]) trailing blanks/binary zeroes

TRIM (LEADING FROM[expression]) leading blanks/binary zeroes

Problem

List the employees who have exactly four characters in their last name. The data type of last_name is CHAR(20).

Solution 1

SELECT first_name ,last_name (TITLE 'last') FROM employee WHERE CHAR (TRIM (TRAILING FROM last_name)) = 4;

Solution 2

SELECT first_name ,last_name(TITLE 'last') FROM employee WHERE CHAR(TRIM(last_name))=4;

Results of either solution:



Loretta Ryan

James DalyNote: When TRIM is specified with no additional directives, both leading and trailing blanks are trimmed.

Using TRIM with Concatenation Operator

The || (double pipe) symbol is the concatenation operator that creates a new string from the combination of the first string followed by the second.

Example 1:

Concatenating of literals without the TRIM function:

SELECT ' Jones ' || ',' || ' Mary ' AS Name;

Name------------------------

Jones , Mary Example 2:

Concatenating of literals with the TRIM function:

SELECT TRIM(BOTH FROM ' Jones ')|| ',' || TRIM(BOTH FROM ' Mary ')AS Name;

Name------------

Jones,Mary

Using TRIM with Other Characters

Using TRIM with Other Characters

TRIM may also be used to remove characters other than blanks and binary zeroes. Using the TRIM function format in the following examples, any defined character(s) may be trimmed from a character string.

Example 1:

SELECT TRIM(BOTH '?' FROM '??????PAUL??????') AS Trim_String;

Trim_String----------------PAUL

Example 2:

SELECT TRIM(LEADING '?' FROM '??????PAUL??????') AS Trim_String;

Trim_String----------------PAUL??????

Example 3:

SELECT TRIM(TRAILING '?' FROM '??????PAUL??????') AS Trim_String;

Trim_String----------------??????PAUL

FORMAT Phrase

The FORMAT phrase can be used to format column output and override the default

format. For example:

SELECT salary_amount (FORMAT '$$$,$$9.99') FROM employee WHERE employee_number = 1004;

salary_amount---------------$36,300.00

The five dollar signs ($) in front of the FORMAT operand indicate that we want to float the leading dollar sign until we hit the first significant digit. For example, if the amount had been 6300.00, it would have been displayed as $6,300.00.

Additional examples using the format (FORMAT '$$$,$$9.99)

Amount Display Notes

100105.25 ********* Mask cannot be applied - value covers format completely.

6300.00 $6,300.00 Dollar sign appears to left of the most significant digit.

.63 $0.63 Leading zero, because the 9's in the mask force the display of a digit (or a zero, if there is not one).

3,336,300.00 ********** Fails, because this number is too large for the mask.

Things To Notice

Formats must provide for enough characters to contain the numeric expression and also the specified format characters.

Overflow conditions are represented by a string of asterisks. FORMAT is not supported by the ANSI standard.

FORMAT allows you to format column output for a calculation. The format character 'Z' is used to represent leading zero suppression. It works similarly to the '$' in that it floats to the first significant digit.

Problem

Management has decided to give employee 1004 a $1,000 raise and wants to know what percent the increase will be:

Solution

SELECT (1000/salary_amount) * 100 (FORMAT 'ZZ9%') AS "Increase Percentage"FROM employee WHERE employee_number = 1004;

Increase Percentage----------------------

3%

FORMAT Characters The following FORMAT characters are available:

$ Fixed or floating dollar sign. 9 Decimal digit (no zero suppress). Z Zero-suppressed decimal digit. , Comma. Inserted where specified. . Decimal point position. - Dash character. Inserted where specified. / Slash character. Inserted where specified. % Percent character. Inserted where specified. X Character data. Each X represents one character. G Graphic data. Each G represents one logical (double byte) character. B Blank data. Insert a space at this point.

Examples

FORMAT '999999' Data: 08777 Result: 008777 FORMAT 'ZZZZZ9' Data: 08777 Result: 8777 FORMAT '999-9999' Data: 6495252 Result: 649-5252 FORMAT 'X(3)' Data: 'Smith' Result: Smi FORMAT '$$9.99' Data: 85.65 Result: $85.65 FORMAT '999.99' Data: 85.65 Result: 085.65 FORMAT 'X(3)' Data: 85.65 Result: Error

The final example produces an error because numeric data cannot be formatted using a character string format. It must first be converted or 'cast' to a character string before an 'X' type of format may be applied.

The $, 9, Z, X, G, and B symbols can be repeated in a FORMAT phrase either by repeating them (e.g., XXX) or by listing them followed by a repetition number (e.g., (X(3)).

The FORMAT mask can contain specifications for up to 18 digits of data plus other insertion characters. If you needed to have 18 zero-suppressed decimal digits, you could note it as Z(18) rather than ZZZZZZZZZZZZZZZZZZ.

DATE Formats When dates are displayed as character strings, appropriate formats are applied.

The Teradata default format is: YY/MM/DD

The ANSI display format is:

YYYY-MM-DD

The recommended display format eliminates uncertainty about the century. For example, in the case of:

02/04/29

It is unclear whether this format represents

April 29, 1902 or April 29, 2002.

Let's look at some unambiguous DATE formats which show all four digits of the year:

SYNTAX RESULT

FORMAT 'YYYY/MM/DD’ 2005/03/27

FORMAT 'DDbMMMbYYYY' 27 Mar 2005

FORMAT 'mmmBdd,Byyyy' Mar 27, 2005

FORMAT 'DD.MM.YYYY' 27.03.2005DATE format syntax is not case-sensitive.

The following formats are available for date displays:

Y4: 4-Digit year (Same as YYYY)YY: 2-Digit yearM4: Long Month - i.e. fully defined name for month (Same as MMMM)M3: Short Month - i.e., abbreviated name for month (Same as MMM)MM: 2-Digit month DD: 2-Digit day of monthD3: 3-Digit day of year (Julian date) (Same as DDD)

Example

SELECT CURRENT_DATE (FORMAT 'Y4-M3-DD');

Date-------------2005-Mar-17

Note: Date formatting options may also use expanded syntax (YYYY-MMM-DD).

Example

SELECT CURRENT_DATE (FORMAT 'Y4:M4:D3');

Date------------------2005:March:076

|Note: The date is the 76th day of the year 2005.

FORMATing DATE in an SQL SELECT Statement Use the FORMAT phrase within a SELECT statement as follows:

SELECT last_name ,first_name ,hire_date (FORMAT 'mmmBdd,Byyyy') FROM employee ORDER BY last_name;

last_name first_name hire_date-------------------- ------------------------------ ------------Johnson Darlene Oct 15, 1976Kanieski Carol Feb 01, 1977Ryan Loretta Oct 15, 1976Stein John Oct 15, 1976Trader James Jul 31, 1976Villegas Arnando Jan 02, 1977

Sorting By Formatted Columns

Problem

A department manager wants to take employees to lunch on their birthdays. We need to know the month and day of each employee's birthday. The manager would like the list sorted in the order in which the birthdays will occur.

Solution 1 - Using FORMAT

SELECT last_name ,first_name ,birthdate (FORMAT 'mmdd') AS birthday ,birthdate AS whole_date FROM employee WHERE department_number = 401 ORDER BY 3;

last_name first_name birthday whole_date-------------------- ------------------------------ -------- ----------Rogers Frank 0423 35/04/23Brown Alan 0809 44/08/09Johnson Darlene 0423 46/04/23Trader James 0619 47/06/19Hoover William 0114 50/01/14Machado Albert 0714 57/07/14Phillips Charles 0810 63/08/10

This solution does not produce the desired sorted result. Note that Brown's birthday sorts between Rogers and Johnson whereas ideally, it should follow them. The ORDER BY 3 sorts on the full internal representation of the birthdate column (i.e., year, month and day). The FORMAT specification suppresses the year in the viewable output but not in the sort.

Solution 2 - Using MODULO

The MODULO function works similarly to long division. Just as when we divide 7 by 3, we get a quotient of 2 and a remainder of 1. MODULO does exactly the opposite. 7 MOD 3 discards the quotient and keeps only the remainder. Thus, 7 MOD 3 = 1.

SELECT last_name ,first_name ,birthdate MOD 10000 (FORMAT '9999') AS birthday ,birthdate AS whole_date FROM employee WHERE department_number = 401 ORDER BY 3;

last_name first_name birthday whole_date-------------------- ------------------------------ -------- ----------Hoover William 0114 50/01/14Johnson Darlene 0423 46/04/23Rogers Frank 0423 35/04/23Trader James 0619 47/06/19Machado Albert 0714 57/07/14Brown Alan 0809 44/08/09Phillips Charles 0810 63/08/10

Why do we see different results in ordering the results? FORMAT is a mask which does not convert the underlying data. MOD is an arithmetic function which does convert the data before the FORMAT is applied. Here we get the desired sorted output. This is because when we MOD 440809 BY 10000, the result is a new integer: 0809. The sort happens on these four digits only.

Using FORMAT With ODBC-based Applications The FORMAT option, when used, must be applied to data represented as character strings. Thus, when numerics or dates are concerned, the formatting only takes place if the data is first converted to character. Because this happens automatically in BTEQ, it can be surprising when using ODBC-based applications when a FORMAT appears to be not working. The conversion to character data must be explicitly done when using ODBC-based applications like SQL Assistant.

Using FORMAT With SQL Assistant

Things To Notice:

Any specified FORMAT must always be applied to a character string. Thus, a conversion from any numeric to character is required. BTEQ does automatic character conversion prior to applying a format. ODBC does not do it automatically, thus the user must explicitly request the conversion. If the column is not a character data type, the FORMAT will be ignored. FORMAT syntax must precede the specified conversion data type syntax.

Using the EXTRACT Function

Extracting From Current Date

The EXTRACT function allows for easy extraction of year, month and day from any DATE data type. The following examples demonstrate its usage.

Query Result

SELECT DATE; /* March 20,2001 */ 01/03/20 (Default format)

SELECT EXTRACT(YEAR FROM DATE); 2001

SELECT EXTRACT(MONTH FROM DATE); 03

SELECT EXTRACT(DAY FROM DATE); 20Date arithmetic may be applied to the date prior to the extraction. Added values always represent days.

Query Result

SELECT EXTRACT(YEAR FROM DATE + 365); 2002

SELECT EXTRACT(MONTH FROM DATE + 30); 04

SELECT EXTRACT(DAY FROM DATE + 12); 01Note: CURRENT_DATE is the ANSI standard for today's date and may be used in place of DATE, which is Teradata specific.

Extracting From Current Time

The EXTRACT function may also be applied against the current time. It permits extraction of hours, minutes and seconds. The following examples demonstrate its usage.

Query Result

SELECT TIME; /* 2:42 PM */ 14:42:32 (Default format)

SELECT EXTRACT(HOUR FROM TIME); 14

SELECT EXTRACT(MINUTE FROM TIME); 42

SELECT EXTRACT(SECOND FROM TIME); 32Time arithmetic may be applied prior to the extraction. Added values always represent seconds.

Query Result

SELECT EXTRACT(HOUR FROM TIME + 20); 14

SELECT EXTRACT(MINUTE FROM TIME + 20); 42

SELECT EXTRACT(SECOND FROM TIME + 20); 52

SELECT EXTRACT(SECOND FROM TIME + 30); Invalid TimeNote: CURRENT_TIME is the ANSI standard for current time and may be used in place of TIME, which is Teradata specific. While TIME is a Teradata function, it does not have the intelligence of the DATE data type, thus adding 30 seconds to 32 seconds produces an invalid 62 for the seconds portion. For true time and timestamp arithmetic, the TIME and TIMESTAMP data types are needed. These capabilities are covered in the Advanced SQL course.

Truncating CHARACTER Columns The FORMAT phrase can create the appearance of truncating a character column or expression.

The FORMAT phrase controls the display of the result but does not change the underlying data in any way.

Example

SELECT last_name ,first_name

,first_name (FORMAT 'X') FROM employee WHERE last_name = 'Brown' ORDER BY 3; last_name-----------



Brown Alan A

Brown Allen A

Question

Does this method sort on just the first initial?

Answer

No, it sorts on the full first name value.

Problem

Show the last name, first name, and first initial of all employees named Brown. Sequence by the first initial.

SELECT last_name ,first_name ,CAST (first_name AS CHAR(1)) FROM employee WHERE last_name = 'Brown' ORDER BY 3; last_name-----------



Brown Allen A

Brown Alan A

Question

Does this method sort on the first initial only?

Answer

Yes, because we have actually CAST the first_name to a new data type of CHAR(1).

Note: Both of the above examples would return an error if run in an ANSI mode session. This is because ANSI mode does not support character truncation as part of a CAST operation.

Attribute Functions Attribute functions are Teradata extensions which return descriptive information about the operand, which may be either a column reference or general expression.

The functions available for attribute information are the following:

TYPE TITLE FORMAT NAMED CHARACTERS

Here are some examples of how these functions work:

Query Results

SELECT DISTINCT TYPE (job_code) FROM job; INTEGER

SELECT DISTINCT TITLE (job_code)FROM job; job_code

SELECT DISTINCT FORMAT (job_code)FROM job; (10)9

SELECT DISTINCT NAMED (job_code)FROM job; job_code

SELECT DISTINCT CHARACTERS(last_name) FROM employee;

20

Note: DISTINCT is used in these queries to reduce output to a single row of information per query. If DISTINCT is not used, the information is repeated once for each row in the table.




A. The manager of department 401 needs a list of employees for the department that includes last name, hire date and bi-weekly salary (26 pay periods per year) sorted descending by salary. Title the computed salary as "Salary Amount". Format hire date as an English abbreviation, followed by day and a four-digit year. Show salary with floating dollar signs, commas, and two decimal positions. Remember: The salary column in the employee table is the annual salary.

If you get ****** for salary amount, it indicates an insufficient format for the size of the result. Be sure to provide enough places to the left of the decimal point.

B. Select employee number and phone number from the employee phone table. Edit phone number with a hyphen between the third and fourth digits. Sort by employee number. Use column titles "Employee" and "Phone".

C. Print the employee number, names and ages of department 301 employees. Name the computed column "Age" and title the last name and first name columns "Last Name" and "First Name". Sequence ascending by age. Format first name as nine characters.

D. Show the first 10 characters of the last name and show the department number and hire date. Order results by hire date.

Option 1: (Do if the current month is NOT September, November, or December.) Find only employees whose date-of-hire anniversary is this month, regardless of what year they were hired.

Option 2: (Do if the current month IS September, November or December. Find out which employees had an anniversary month four months ago.

E. Modify Lab D to include an additional column on the report containing the three-lettered English abbreviation for the month. Title this column "month". Notice that the month column is right-justified even though it appears alphabetic. With FORMAT, underlying data controls alignment.

F. Select all employees who work in department 501. Show last and first names (both formatted as 8 positions), and three date columns: one for month of hire as a computed value titled "month"; another for English abbreviation of hire month without a title; and a third for year-of-hire as a 4-digit value with a title of "year".

Sort using numeric designators from the SELECT clause. Do it twice. Note sort results. Submit with sort on the extracted month column. Sort again on the full date.

Lab Answer A

Solution NOT Using a Quantifier

DATABASE Customer_Service; SELECT last_name ,hire_date (FORMAT 'mmmbdd,byyyy') ,salary_amount/26 (FORMAT '$$$,$$9.99') (TITLE 'Salary Amount') FROM employee WHERE department_number = 401 ORDER BY salary_amount DESC ;


last_name hire_date Salary Amount---------- -------------- ---------------Rogers Mar 01,1977 $1,769.23Brown Jul 31, 1976 $1,657.69Trader Jul 31, 1976 $1,455.77Johnson Oct 15, 1976 $1,396.15Machado Mar 01, 1979 $1,242.31Hoover Jun 18, 1976 $981.73Phillips Apr 01, 1977 $942.31

Lab Answer B

SELECT employee_number(TITLE 'Employee'),phone (FORMAT '999-9999') (TITLE 'Phone')

FROM employee_phone

ORDER BY employee_number;

*** Query completed. 78 rows found. 2 columns returned.Employee Phone

801 827-87771001 449-12251001 449-1244

. .

. .

. .1003 377-45341003 827-87771004 224-3283

. .

. .

. .1024 272-47431025 827-87771025 296-4652

Lab Answer C

SELECT employee_number,last_name (TITLE 'Last Name'),CAST(first_name AS CHAR(9)) (TITLE 'First

Name') ,(DATE - birthdate)/365 AS Age

FROM employeeWHERE department_number = 301ORDER BY age;


employee_number Last Name First Name Age --------------- --------- ----------- ------ 1008 Kanieski Carol 41 1006 Stein John 46 1019 Kubic Ron 57

Note: The answer set will vary depending on the date of execution.

Lab Answer D

This solution is for Option 1 - assumes current month is July.

SELECT CAST (last_name AS char(10)),department_number,hire_date

FROM employee

WHERE (hire_date MOD 10000)/100 = (DATE MOD 10000)/100

ORDER BY hire_date;


last_name department_number hire_date ---------- ----------------- --------- Trader 401 76/07/31 Brown 401 76/07/31 Ratzlaff 501 78/07/15

Note: The answer set will vary depending on the current month.

This solution is for Option 2 - assumes current month is November. SELECT CAST (last_name AS char(10)) ,department_number ,hire_date FROM employee WHERE (hire_date MOD 10000)/100 =

(DATE MOD 10000)/100 - 4 ORDER BY hire_date ; *** Query completed. 3 rows found. 3 columns returned.

last_name department_number hire_date ---------- ----------------- --------- Trader 401 76/07/31 Brown 401 76/07/31 Ratzlaff 501 78/07/15

Note: Answer set will vary, depending on current month.

Lab Answer E

This solution is for Option 1 - assumes current month is July.

SELECT CAST (last_name AS CHAR(10)) ,department_number ,hire_date ,hire_date (FORMAT 'mmm') (TITLE 'month')FROM employeeWHERE (hire_date MOD 10000)/100 = (DATE MOD 10000)/100ORDER BY hire_date;


last_name department_number hire_date month---------- ----------------- --------- -----Trader 401 76/07/31 JulBrown 401 76/07/31 JulRatzlaff 501 78/07/15 Jul

Note: The answer set will vary depending on the current month.

This solution is for Option 1 - assumes current month is November.

SELECT CAST (last_name AS char(10)) ,department_number ,hire_date ,hire_date (FORMAT 'mmm') (TITLE 'month') FROM employee WHERE (hire_date MOD 10000)/100 = (DATE MOD 10000)/100 - 4 ORDER BY hire_date ;

*** Query completed. 3 rows found. 4 columns returned. last_name department_number hire_date month ---------- ----------------- --------- ----- Trader 401 76/07/31 Jul Brown 401 76/07/31 Jul Ratzlaff 501 78/07/15 Jul

Note: Answer set will vary, depending on current month.

Lab Answer F

SELECT last_name (FORMAT 'x(8)'),first_name (FORMAT 'x(8)'),EXTRACT (MONTH FROM hire_date) (TITLE

'month'),hire_date (FORMAT 'mmm') (TITLE ' '),hire_date (FORMAT 'yyyy')(TITLE 'year')

FROM employeeWHERE department_number = 501ORDER BY 3;


last_name first_name month year--------- ---------- ----------- --- ----Rabbit Peter 3 Mar 1979Wilson Edward 3 Mar 1978Runyon Irene 5 May 1978Ratzlaff Larry 7 Jul 1978SELECT last_name (FORMAT 'x(8) ' )

,first_name (FORMAT 'x(8) ' ), EXTRACT (MONTH FROM hire_date) (TITLE

'month' ),hire_date (FORMAT 'mmm' ) (TITLE ' '),hire_date (FORMAT 'yyyy' ) (TITLE

'year' )FROM employeeWHERE department_number = 501ORDER BY 4;


last_name first_name month year --------- ---------- ----------- --- ---- Wilson Edward 3 Mar 1978 Runyon Irene 5 May 1978 Ratzlaff Larry 7 Jul 1978 Rabbit Peter 3 Mar 1979

Inner Joins Module:- 9


Perform an Inner Join on two or more tables. Perform two special cases of an Inner Join:

o A Cartesian product join. o A Self join.

Use an alias as a temporary name for a Table or View. Determine when to use a subquery versus a Join.

A join is a technique for accessing data from more than one table in an answer set. Tables are joined according to columns they have in common. A join between the employee table and the department table could be done according to department number. An example of an inner join will be seen on the next screen:

Joins may access data from tables, views or a combination of the two.

Type of Joins:

Inner Rows which match based on join criteria.

Outer* Rows which match and those which do not.

Cross Each row of one table matched with each row of another.

Self Rows matching other rows within the same table.* Outer Joins are discussed in a separate section of the SQL WBT.

Our main focus in this module will be Inner Joins.

Inner Join Problem To get a report that includes employee number, last name, and department name, join the employee table and the department table.

Department number is the common column that determines the way data in these two tables will be joined. Most often joined columns are related to each other as primary and foreign keys.

Inner Join Solution employee_number last_name department_name

-------------------- ----------- --------------------1006 Stein research and development1008 Kanieski research and development1005 Ryan education1004 Johnson customer support1007 Villegas education1003 Trader customer support. . .. . .. . .We fully qualified every column referenced in our SELECT statement to include the table that the column is in ( e.g., employee.employer_number). It is only necessary to qualify columns that have identical names in both tables (i.e., department_number).

The ON clause is used to define the join condition used to link the two tables.

Defining and Using Alias Names An alias is a temporary name for a TABLE or VIEW defined in the FROM clause. It can be useful for abbreviating long table names and is required to join a table to itself.

Once the alias name is defined, it must be used throughout the SQL statement.

SELECT e.employee_number ,e.last_name ,d.department_name FROM employee e INNER JOIN department d ON e.department_number = d.department_number;employee_number-------------------

last_name----------

department_name--------------------

1006 Stein research and development

1008 Kanieski research and development

1005 Ryan education

1004 Johnson customer support

1007 Villegas education

1003 Trader customer supportNote: The "e" in the FROM clause is what sets up "e" as the employee table alias. Similarly, the "d" sets up the department table alias. Alias names can be up to 30 characters.

Aliases can be used in the SELECT part of the statement even though SELECT appears before FROM in a statement. Once an alias has been established, it should be used consistently thoughout the query. Using the original table name as a qualifier instead of the alias may produce unexpected results as we shall see in a later example.

Cross Joins A Cross Join is a join that requires no join condition (Cross Join syntax does not allow an ON clause). Each participating row of one table is joined with each participating row of another table. The WHERE clause restricts which rows participate from either table. SELECT e.employee_number

,d.department_number FROM employee e CROSS JOIN department d WHERE e.employee_number = 1008;employee_number--------------------


1008 301

1008 501

1008 402

1008 201

1008 302

1008 600

1008 401

1008 100

1008 403The employee table has 26 rows. The department table has 9 rows. Without the WHERE clause, we would expect that 26 x 9 = 234 rows in our result set. With the constraint that the employee_number must equal 1008 (which only matches one row in the employee table), we now get 1 x 9 = 9 rows in our result set.

Cross Joins by themselves often do not produce meaningful results. This result shows employee 1008 associated with each department. This is not meaningful output.

Cartesian Product A completely unconstrained Cross Join is called a Cartesian product. Each row of one table is joined to each row of another table. A Cartesian product results when a CROSS JOIN is issued without a WHERE clause.

SELECT employee.employee_number ,employee.department_number FROM employee CROSS JOIN department; Each employee row (26) matched with each department row (9) yields 234 rows of output. An 8,000,000 row table and a 50,000 row table would yield a 400,000,000,000 row answer set. The output of a Cartesian product is often not meaningful however they do have useful application as we shall see.

Cartesian products may also result accidentally from an INNER JOIN with improper aliasing or improper join conditions.

Useful Cartesian Products There are some real-world uses for Cartesian product joins. One important use is to benchmark system performance with large data throughputs. Cartesian products make it easy to generate very large answer sets.

Example

Any query requiring all elements of one table to be compared to all elements of another is a candidate for a Cartesian product using the CROSS JOIN syntax.

As demonstrated in the above diagram, a CROSS JOIN can be useful to an airline for comparing all possible combinations of cities in order to evaluate:

flight plan information, passenger rate structures, mileage awards

Inner Joins on Multiple Tables Historically, a join could have up to 16 participating Tables or Views. Since Teradata V2R2.1, the join limit has been increased to 64 participating Tables or Views. The number of participating Tables/Views determines the number of required join conditions. An n-table join requires n-1 join conditions. In this example, three tables are joined, thus requiring two join conditions (two ON clauses). Omitting any join condition will result in a Cartesian product join.

Example SELECT e.last_name ,d.department_name ,j.description FROM employee e INNER JOIN department d ON e.department_number = d.department_number INNER JOIN job j

ON e.job_code = j.job_code;


department_name--------------------

description------------

Daly software support Manager - Software Supp

Runyon marketing sales Manager - Marketing Sales

Trainer president Corporate President

Brown customer support Dispatcher

… … …

Accidental Cartesian Products An accidental Cartesian product can result from an Inner Join with improper aliasing: SELECT employee.employee_number ,d.department_name FROM employee e INNER JOIN department d ON e.department_number = d.department_number;

We forgot to use the "e" alias in the SELECT statement. The parser sees that "e" needs to be joined with "d". That's two tables and one join. So far, so good. Then the parser sees a reference to employee and views it as a third table. We now have three tables and only one join, resulting in an unintended Cartesian product.

A Cartesian product also may result from an Inner Join with an improper join condition:

SELECT e.employee_number ,d.department_name FROM employee e INNER JOIN department d ON 3 = 3;

In the above example, since 3 = 3 is always true, every row meets this condition. Every row in the department table is joined with every row in the employee table. We probably do not want this result.

EXPLAIN all queries before allowing them to run in a production environment! Please see the section on EXPLAIN at the end of this module.

If your EXPLAIN statement estimates a huge number of rows or hours to complete, it could mean that your query will unintentionally produce a Cartesian product.

Self Joins A self join occurs when a table is joined to itself. Which employees share the same surname Brown and to whom do they report?

SELECT emp.first_name (TITLE 'Emp//First Name') ,emp.last_name (TITLE 'Emp//Last Name') ,mgr.first_name (TITLE 'Mgr//First Name') ,mgr.last_name (TITLE 'Mgr//Last Name') FROM employee emp INNER JOIN employee mgr ON emp.manager_employee_number = mgr.employee_number WHERE emp.last_name = 'Brown';

Result

EmpFirst Name

EmpLast Name

MgrFirst Name

MgrLast Name

Allen Brown Loretta Ryan

Alan Brown James TraderWould the results differ if the last condition was:

WHERE employee.last_name = 'Brown' ?

Next time you log on for Lab, try it and see.

Determining Whether to Use Subquery or Join A Subquery qualifies which rows selected in the main query will be in the answer set. Data selected in the subquery will not be included in the answer set.

A Join qualifies which rows from two or more tables will be matched to create rows of an answer set. The answer set can include data from one or more of the joined tables.

List the first name, last name and department number of all employees who work in research departments.

Using a JOIN

SELECT employee.first_name ,employee.last_name ,employee.department_number FROM employee INNER JOIN department ON employee.department_number = department.department_number WHERE department.department_name LIKE '%Research%';

Using a SUBQUERY

SELECT first_name ,last_name ,department_number FROM employee WHERE department_number IN (SELECT department_number FROM department WHERE department_name LIKE '%Research%');

Result

first_name------------



Carol Kanieski 301

John Stein 301

Ron Kubic 301Note that in some cases, either a join or a subquery may be used to produce the same result.

Use EXPLAIN to Understand a Join Use EXPLAIN to see how the database will execute your query. EXPLAIN is especially useful for checking to see that you are not creating an unintended Cartesian product.

EXPLAIN SELECT employee.first_name ,employee.last_name ,employee.department_number FROM employee INNER JOIN department ON employee.department_number = department.department_number WHERE department.department_name LIKE '%Research%';

Explanation

1. First, we lock a distinct CUSTOMER_SERVICE."pseudo table" for read on a RowHash to prevent global deadlock for CUSTOMER_SERVICE.employee.

2. Next, we lock a distinct CUSTOMER_SERVICE."pseudo table" for read on a RowHash to prevent global deadlock for

CUSTOMER_SERVICE.department. 3. We lock CUSTOMER_SERVICE.employee for read, and we lock

CUSTOMER_SERVICE.department for read. 4. We do an all-AMPs RETRIEVE step from CUSTOMER_SERVICE.employee by

way of an all-rows scan with no residual conditions into Spool 2, which is redistributed by hash code to all AMPs. Then we do a SORT to order Spool 2 by row hash. The size of Spool 2 is estimated with low confidence to be 24 rows. The estimated time for this step is 0.05 seconds.

5. We do an all-AMPs JOIN step from Spool 2 (Last Use) by way of an all-rows scan, which is joined to CUSTOMER_SERVICE.department with a condition of (-1 ). Spool 2 and CUSTOMER_SERVICE.department are joined using a merge join, with a join condition of ("Spool_2.department_number=CUSTOMER_SERVICE.department.department_number"). The result goes into Spool 1, which is built locally on the AMPs. The size of Spool 1 is estimated with no confidence to be 24 rows. The estimated time for this step is 0.18 seconds.

6. Finally, we send out an END TRANSACTION step to all AMPs involved in processing the request. -> The contents of Spool 1 are sent back to the user as the result of statement 1. The total estimated time is 0 hours and 0.23 seconds.

EXPLAIN a Subquery You can also use EXPLAIN against your subqueries.

EXPLAIN SELECT first_name ,last_name ,department_number FROM employee WHERE department_number IN (SELECT department_number FROM department WHERE department_name LIKE '%Research%');

Explanation

1. First, we lock a distinct CUSTOMER_SERVICE."pseudo table" for read on a RowHash to prevent global deadlock for CUSTOMER_SERVICE.employee.

2. Next, we lock a distinct CUSTOMER_SERVICE."pseudo table" for read on a RowHash to prevent global deadlock for CUSTOMER_SERVICE.department.

3. We lock CUSTOMER_SERVICE.employee for read and lock CUSTOMER_SERVICE.department for read.

4. We do an all-AMPs RETRIEVE step from CUSTOMER_SERVICE.employee by way of an all-rows scan with no residual conditions into Spool 2, which is redistributed by hash code to all AMPs. Then we do a SORT to order Spool 2 by row hash. The size of Spool 2 is estimated with low confidence. The estimated time for this step is 0.03 seconds.

5. We do an all-AMPs JOIN step from Spool 2 (Last Use) by way of an all-rows scan, which is joined to CUSTOMER_SERVICE.department with a condition of ("CUSTOMER_SERVICE.department.department_name LIKE

'%Research%'"). Spool 2 and CUSTOMER_SERVICE.department are joined using an inclusion merge join, with a join condition of ("Spool_2.department_number=CUSTOMER_SERVICE.department.department_number"). The result goes into Spool 1, which is built locally on the AMPs. The size of Spool 1 is estimated with index join confidence to be 22 to 24 rows. Estimated time for this step is to 0.21 seconds. The contents of Spool 1 are sent back to the user as the result of statement 1. The total estimated time is 0 hours and 0.23 seconds.

Question

The query on this page and the one from the Join on the previous page produce equivalent results. How could you compare their expected performance on the system?

One way is to compare some of the basic characteristics divulged by the EXPLAIN .The Join query took six steps and estimated that it would return 24 rows and take approximately 0.23 seconds to complete.

In comparison, the Subquery took five steps, estimated that it would return 28 rows and take approximately 0.24 seconds to complete.

The two implementations of this query are roughly equivalent from a performance standpoint.

Lab




A. List the employee number, last name, salary, and job description of all employees except the President. (The President is the only employee assigned to Department 100.) Format last name as 9 characters and description as 20 characters. Sort by employee number within job description.

B. Modify Lab D from Module 8 to use a join instead of the subquery to select information about California customers. EXPLAIN the request and note the total estimated time. Compare with the total estimated time from Lab D in Module 9. Which is more efficient: the join or the subquery?

C. Produce a report showing employee number, department number and manager’s last name for all employees hired before Jan 1, 1978. Sort by department number.

D. Do an EXPLAIN statement using CROSS JOIN on the employee and job tables. Select job code, job description and employee last name.

E. Management has a request. They need the last name, department name, job description, and phone number of all employees in the 213 area code with job code 412101. Use table name aliases for your joins. Order by last name.

You might want to use EXPLAIN with this exercise, to see what processing takes place.

Note: Check your quick reference page to be sure you are accessing the correct tables.

Lab Answer A

DATABASE Customer_Service;SELECT e.employee_number

,CAST (e.last_name AS CHAR(9)),e.salary_amount,CAST (j.description AS CHAR(20))

FROM employee e INNER JOINjob j

ON e.job_code = j.job_codeWHERE e.department_number <> 100ORDER BY j.description

,e.employee_number;


employee_number last_name salary_amount description --------------- --------- ------------- -------------------- 1002 Brown 43100.00 Dispatcher 1001 Hoover 25525.00 Field Engineer 1004 Johnson 36300.00 Field Engineer 1010 Rogers 46000.00 Field Engineer 1008 Kanieski 29250.00 Hardware Engineer 1007 Villegas 49700.00 Instructor 1009 Lombardo 31000.00 Instructor 1012 Hopkins 37900.00 Instructor 1020 Charles 39500.00 Instructor 1024 Brown 43700.00 Instructor 1003 Trader 37850.00 Manager - Customer S 1005 Ryan 31200.00 Manager - Education 1017 Runyon 66000.00 Manager - Marketing 1016 Rogers 56500.00 Manager - Product Pl 1019 Kubic 57700.00 Manager - Research a 1011 Daly 52500.00 Manager - Software S 1013 Phillips 24500.00 Product Specialist 1022 Machado 32300.00 Product Specialist 1015 Wilson 53625.00 Sales Rep 1018 Ratzlaff 54000.00 Sales Rep 1023 Rabbit 26500.00 Sales Rep 1014 Crane 24500.00 Software Analyst 1006 Stein 29450.00 Software Engineer 1021 Morrissey 38750.00 System Analyst

Lab Answer B

SELECT l.customer_number,c.customer_name,c.sales_employee_number

FROM customer c INNER JOINlocation l

ON c.customer_number = l.customer_numberWHERE l.state = 'California'ORDER BY 3;


customer_number customer_name sales_employee_number--------------- --------------------------- --------------------- 0 Corporate Headquarters ? 2 Simple Instruments Co. 1015 15 Cates Modeling 1015 11 Hotel California 1015 1 A to Z Communications, Inc. 1015 19 More Data Enterprise 1015 10 Graduates Job Service 1015Note: EXPLAIN text will vary depending on optimizer path.

Lab Answer C

SELECT e.employee_number,e.department_number,m.last_name

FROM employee e INNER JOIN employee m

ON e.manager_employee_number = m.employee_numberWHERE e.hire_date < 780101ORDER BY 2;

*** Query completed. 14 rows found. 3 columns returned. *** Total elapsed time was 1 second.employee_numbe

rdepartment_numbe

r last_name

-------------------------------

- ---------

801 100 Trainer1008 301 Kubic

1006 301 Kubic1010 401 Trader1003 401 Trainer1004 401 Trader1002 401 Trader1001 401 Trader1013 401 Trader1011 402 Trainer1005 403 Trainer1007 403 Ryan1009 403 Ryan1012 403 Ryan

Lab Answer D

EXPLAINSELECT e.job_code

,description,last_name

FROM employee e CROSS JOIN job j;

Results will vary

Lab Answer E

SELECT e.last_name,d.department_name,j.description,p.phone

FROM employee eINNER JOIN

department dON e.department_number = d.department_numberINNER JOIN

job jON e.job_code = j.job_codeINNER JOIN

employee_phone pON e.employee_number = p.employee_numberWHERE p.area_code = 213AND e.job_code = 412101ORDER BY e.last_name;


last_name department_name description --------- ---------------- -------------- Hoover customer support Field Engineer Rogers customer support Field Engineer

Outer Joins Module:- 10

Lab




A. List the last name, job code and job description of all employees. Cast last name as 9 characters, and job description as 20 characters. Sort by last name. Also, include employees who have an invalid or null job_codes. Use outer join syntax to accomplish this.

B. Show the last name, job description and department name of all employees with valid departments and valid job codes. Order by last name. Format last name as 9 characters, job description as 20 characters and the department name as 13 characters.

C. Modify exercise B to include employees with valid departments but without valid job codes.

D. Modify exercise B to include unassigned job codes.

Lab AnswerA

DATABASE Customer_Service;SELECT CAST (last_name AS CHAR(9))

,e.job_code ,CAST(description AS CHAR(20))

FROM employee e LEFT OUTER JOIN job j

ON e.job_code = j.job_codeORDER BY 1;

*** Query completed. 26 rows found. 3 columns retuned.

last_name job_code description ---------- --------- -------------------- Brown 413201 Dispatcher Brown 432101 Instructor Charles 432101 Instructor Crane 422101 Software Analyst Daly 421100 Manager - Software S Hoover 412101 Field Engineer Hopkins 432101 Instructor Johnson 412101 Field Engineer Kanieski 312102 Hardware Engineer Kubic 311100 Manager - Research a Lombardo 432101 Instructor Machado 412102 Product Specialist Morrissey 222101 System Analyst Phillips 412102 Product Specialist Rabbit 512101 Sales Rep Ratzlaff 512101 Sales Rep Rogers 321100 Manager - Product Pl Rogers 412101 Field Engineer Runyon 511100 Manager - Marketing Sales Ryan 431100 Manager - Education Short 211100 ? Stein 312101 Software Engineer Trader 411100 Manager - Customer S Trainer 111100 Corporate President Villegas 432101 Instructor Wilson 512101 Sales Rep

Lab Answer B

SELECT e.last_name (FORMAT 'X(9)'),j.description (FORMAT 'X(26)'),d.department_name (FORMAT 'X(13)')

FROM employee e INNER JOINdepartment d

ON e.department_number = d.department_numberINNER JOIN job j



last_name description_______________ department_nameBrown Dispatcher customer suppBrown Instructor education

Charles Instructor educationCrane Software Analyst software suppDaly Manager - Software Support software suppHoover Field Engineer customer suppHopkins Instructor educationJohnson Field Engineer customer suppKanieski Hardware Engineer research andKubic Manager - Research and Dev research andLombardo Instructor educationMachado Product Specialist customer suppMorrissey System Analyst technical opePhillips Product Specialist customer suppRabbit Sales Rep marketing salRatzlaff Sales Rep marketing salRogers Manager - Product Planning product plannRogers Field Enginee customer suppRunyon Manager - Marketing Sales marketing salRyan Manager - Education educationStein Software Engineer research andTrader Manager - Customer Support customer suppTrainer Corporate President presidentVillegas Instructor educationWilson Sales Rep marketing sal

Lab Answer C



ON e.department_number = d.department_numberLEFT OUTER JOIN job j

ON e.job_code = j.job_codeORDER BY 1; *** Query completed. 26 rows found. 3 columns returned. *** Total elapsed time was 1 second.last_name description________________ department_nameBrown Dispatcher customer suppBrown Instructor educationCharles Instructor educationCrane Software Analyst software suppDaly Manager - Software Support software suppHoover Field Engineer customer suppHopkins Instructor educationJohnson Field Engineer customer suppKanieski Hardware Engineer research andKubic Manager - Research and Dev research andLombardo Instructor educationMachado Product Specialist customer suppMorrissey System Analyst technical opePhillips Product Specialist customer suppRabbit Sales Rep marketing sal

Ratzlaff Sales Rep marketing salRogers Manager - Product Planning product plannRogers Field Enginee customer suppRunyon Manager - Marketing Sales marketing salRyan Manager - Education educationShort ? technical opeStein Software Engineer research andTrader Manager - Customer Support customer suppTrainer Corporate President presidentVillegas Instructor educationWilson Sales Rep marketing sal

Lab Answer D



ON e.department_number = d.department_numberRIGHT OUTER JOIN job j



last_name description_______________ department_name? System Support Analyst ?? Mechanical Assembler ?? Planning Specialist ?? Electronic Assembler ?Brown Instructor, educationBrown Dispatcher customer suppCharles Instructor educationCrane Software Analyst software suppDaly Manager - Software Support software suppHoover Field Engineer customer suppHopkins Instructor educationJohnson Field Engineer customer suppKanieski Hardware Engineer research andKubic Manager - Research and Dev research andLombardo Instructor educationMachado Product Specialist customer suppMorrissey System Analyst technical opePhillips Product Specialist customer suppRabbit Sales Rep marketing salRatzlaff Sales Rep marketing salRogers Manager - Product Planning product plannRogers Field Engineer customer suppRunyon Manager - Marketing Sales marketing salRyan Manager - Education educationStein Software Engineer research and

Trader Manager - Customer Support customer suppTrainer Corporate President presidentVillegas Instructor educationWilson Sales Rep marketing sal

Set Operations Module:- 11


Use the UNION operator. Use the INTERSECT operator. Use the EXCEPT operator.

Set Operators

The following are graphic representations of the three set operators, INTERSECT, UNION and EXCEPT. All are ANSI-compliant operators.

INTERSECT

The INTERSECT operator returns rows from multiple sets which share some criteria in common.

UNION

The UNION operator returns all rows from multiple sets, displaying duplicate rows only once.

EXCEPT

The EXCEPT operator subtracts the contents of one set from the contents of another.

Note: Using the Teradata keyword ALL in conjuction with the UNION operator allows

duplicate rows to remain in the result set.

UNION Operator

Before the introduction of the OUTER JOIN feature, The UNION operator provides functionality similar to the OUTER JOIN feature. UNION is still a viable feature, however. Here are some rules for its usage:

All SELECT clauses:

Must have the same number of expressions. Corresponding expressions must have compatible data types.

The first SELECT statement:

Determines output FORMAT. Determines output TITLE.

The last SELECT statement:

Contains the ORDER BY clause for the entire result when applicable. Uses numeric designators for the ORDER BY columns.

Problem

Show manager 1019 identifying him as the manager, show his employees and identify each of them as an employee.

Solution

SELECT first_name ,last_name ,'employee' (TITLE 'employee//type') FROM employee WHERE manager_employee_number = 1019 UNION SELECT first_name ,last_name ,' manager ' FROM employee WHERE employee_number = 1019 ORDER BY 2

;

Result employee first_name last_name type Carol Kanieski employeeRon Kubic manager

John Stein employee

INTERSECT Operator

Problem

Create a list of all department managers who are assigned subordinate employees. Note that not all department managers have subordinates and not all managers with subordinates are department managers.

Solution

SELECT manager_employee_number FROM employee INTERSECT SELECT manager_employee_number FROM department ORDER BY 1 ;

Result

manager_employee_number801100310051011101710191025

EXCEPT Operator

The EXCEPT operator is ANSI intermediate compliant. It will be flagged as non-entry level by the SQL flagger:

*** SQL warning 5811 MINUS of queries is not in entry level ANSI.

This flagged message refers to "MINUS" because the Teradata-compatible synonym for EXCEPT is MINUS. Use the EXCEPT operator rather than MINUS because although both provide the same functionality, EXCEPT is ANSI compliant.

Problem

To list all department managers who do not have subordinates.

Solution

SELECT manager_employee_number FROM department EXCEPT SELECT manager_employee_number FROM employee ORDER BY 1 ; Result

manager_employee_number

10161099

SET Operators - Additional Rules

Set operators may be used in most SQL constructs but cannot contain a WITH or WITH...BY clause. Set operators are evaluated in order of precedence, as follows:

INTERSECT UNION EXCEPT from left to right

Evaluation order may be manipulated with parentheses.

Each SELECT statement must have a FROM table_name. The GROUP BY clause does not apply to the result set as a whole.

Duplicates are eliminated unless the ALL option is used.

Lab





A. Use the UNION operator to generate a complete list of all phone numbers belonging to a location or to an employee in the 415 area code. Designate a category to which each row belongs. Entitle this literal column "owner". Sort the report by phone number. Hyphenate all phone numbers.

B. Find the valid employee numbers for department level managers. You will be doing a referential integrity check between department and employee tables. Accomplish this using a subquery, then again using the INTERSECT operator. Use EXPLAIN to compare efficiency and optimization choices. Which is better?

C. Produce a list of all job codes which are currently assigned to employees but which are invalid. Title the column 'INVALID JOB CODES' and use the EXCEPT operator to solve the problem.

Lab Answer A

DATABASE Customer_Service;SELECT phone (FORMAT '999-9999')

,'Employee' (TITLE 'Owner')FROM employee_phoneWHERE area_code = 415UNIONSELECT phone

,'Location'FROM location_phoneWHERE area_code = 415ORDER BY 1;


phone Owner -------- --------241-2021 Employee241-2021 Location356-3560 Employee356-3560 Location449-1221 Employee449-1221 Location449-1225 Employee449-1225 Location449-1244 Employee449-1244 Location656-7000 Employee656-7000 Location923-4864 Employee923-7892 Employee923-7892 Location

Lab B

SELECT employee_numberFROM employeeWHERE employee_number IN(SELECT manager_employee_numberFROM department);

*** Query completed. 8 rows found. One column returned. employee_number

employee_number --------------- 1025 1011 1016 1003 801 1019 1017 1005

Note: EXPLAIN text will vary depending on optimizer path.

SELECT employee_number FROM employee INTERSECT SELECT manager_employee_number FROM department ;

*** Query completed. 8 rows found. One column returned. employee_number

employee_number --------------- 1011 801 1016 1019 1017 1005 1025 1003

Note: EXPLAIN text will vary depending on optimizer path.

Lab Answer C

SELECT job_code(TITLE 'INVALID JOB CODES')FROM employeeEXCEPTSELECT job_codeFROM job;


INVALID JOB CODES ----------------- 211100

Creating Tables and Indexes Module:-12

Objectives


CREATE, DROP, or ALTER a table. CREATE, DROP, or ALTER an index. CREATE, DROP, or ALTER table attributes.

Data Definition Language

Data Definition Language (DDL) is used by SQL to create, modify, and remove object definitions:

The Teradata Database implements separate DDL statements for different objects.

Object: DDL Statements:

Databases CREATE DATABASEMODIFY DATABASEDROP DATABASE

Users CREATE USERMODIFY USERDROP USER

Tables CREATE TABLEALTER TABLEDROP TABLE

Views CREATE VIEWREPLACE VIEWDROP VIEW

Macros CREATE MACROREPLACE MACRODROP MACRO

Indexes CREATE INDEXDROP INDEX

Join Indexes CREATE JOIN INDEXDROP JOIN INDEX

Triggers CREATE TRIGGERALTER TRIGGERDROP TRIGGER

Stored Procedures CREATE PROCEDUREREPLACE PROCEDUREDROP PROCEDURE

In this module we will look at creating tables and indexes only.

CREATE TABLE Elements

When executed, the CREATE TABLE statement creates and stores the table structure definition in the Data Dictionary (DD).

The CREATE TABLE statement allows you to create:

TABLE options.

COLUMN definitions. TABLE level constraints. INDEX definitions.

Note: We cover Table level constraints in the Advanced SQL course. Our discussion here will focus on table options, column definitions, and index definitions.

CREATE <SET/MULTISET> TABLE employee <Create Table Options> <Column Definitions> <Index Definitions>;

Element

Definition

Create Table Options

Specify physical attributes of table to include:

Fallback.

Journaling.

Column Definitions

Define each column in the table.

Ind Specify

ex Definitions

indexes for physical access to data.

CREATE TABLE Options

Teradata Extensions

Teradata DDL allows you to specify physical implementation options such as fallback and permanent journal protection, datablocksize, and cylinder freespace factor. You may define a primary index (different than the primary key) as well as secondary indexes to improve physical access to data in the table.

Duplicate row options - SET no duplicate rows allowed (Default) - MULTISET duplicate rows allowed

Table protection options (Underlines indicate defaults) - FALLBACK or NO FALLBACK PROTECTION - BEFORE JOURNAL (NO, SINGLE or DUAL) - AFTER JOURNAL (NO, SINGLE (LOCAL or NOT LOCAL) or DUAL) - WITH JOURNAL TABLE (tablename)

FALLBACK specifies that the system builds a duplicate copy of each row of the table and stores it on a different (FALLBACK) AMP within the cluster.

JOURNAL specifies that the system stores BEFORE and /or AFTER images for each changed row of the table in the specified permanent journal, providing disaster recovery capability.

Space Management Options (Defaults are provided if not specified)

FREESPACE - % cylinder freespace to maintain during loading operation DATABLOCKSIZE - maximum block size for multi-row data blocks

Example

CREATE TABLE table_1 ,FALLBACK, NO JOURNAL (field1 INTEGER , field2 CHAR(10) , field3 DEC(8,2)) ;

COLUMN Definitions

Up to 2,048 columns may be defined in a single CREATE TABLE statement.

You may define the following aspects of each column:

Aspect Meaning

Column Name Name of the column

Data TypeColumn data type (INTEGER, CHAR, etc.)

Data Type Attributes Column attributes such as: DEFAULT, FORMAT, TITLE, NULL, CASESPECIFIC, UPPERCASE

Column Storage Attributes Column compression for nulls or for specific values

Column-Level Constraint Attributes

Column range of possible values, uniqueness, use as a primary or foreign key

We will discuss each in detail in the following sections.

COLUMN Name & Data Type Choices

Column Data Types

Character CHAR, VARCHAR,CHAR VARYING,LONG VARCHAR

Numeric INTEGER, SMALLINT, BIGINT, REAL, FLOAT, DOUBLEPRECISION NUMERIC,DECIMAL, BYTEINT,DATE

Byte BYTE, VARBYTE

Example

CREATE TABLE emp_data (employee_number INTEGER ,department_number SMALLINT ,job_code INTEGER ,last_name CHAR (20) ,first_name VARCHAR (20) . . . ,birthdate DATE ,salary_amount DECIMAL (10,2)

. . );

COLUMN Data Type Attributes

DEFAULT Specify default value in place of null

WITH DEFAULT Specify use of system default value in place of null

FORMAT Default display format

TITLE Default column title

NOT NULL Disallow nulls, value required

CASESPECIFIC Treat as case-specific for comparisons

UPPERCASE Shift to uppercase for storage

Example

CREATE TABLE emp_data (employee_number INTEGER NOT NULL ,last_name CHAR(20) NOT NULL WITH DEFAULT ,street_address VARCHAR(30) TITLE'Address' ,city CHAR(15) DEFAULT'Boise' ,state CHAR(2) WITH DEFAULT ,birthdate DATE FORMAT 'mm/dd/yyyy' ,salary_amount DEC(10,2) ,sex CHAR(1) UPPERCASE );

The WITH DEFAULT option is a special case of the DEFAULT option. The WITH DEFAULT phrase is converted by the Teradata database to a DEFAULT option in which the default system value for that data type becomes the value to use in place of null. For example, character strings will be defaulted to spaces and numeric columns will default to zeroes.

Retrieving Default Values

A DEFAULT function exists which permits the retrieval of the default value assigned to a particular column. For example, assume we define a table as follows:

CREATE TABLE TableX (Col1 INTEGER, Col2 INTEGER DEFAULT 10, Col3 INTEGER DEFAULT 20, Col4 CHAR(60));

We may retrieve the default values for any column by using the DEFAULT function as follows:

SELECT DEFAULT(Col2), DEFAULT(Col3) FROM TableX; Default(Col2) Default(Col3)------------- ------------- 10 20 10 20

Default values are returned in place of the normal column values. Note that two rows are returned. This is because there are currently two rows in the table. One row of output is

returned for each existing row in the table. You may return a single row by using the following syntax:

SELECT DISTINCT DEFAULT(Col2), DEFAULT(Col3) FROM TableX;Default(Col2) Default(Col3)------------- ------------- 10 20

If there are no rows in the table, no default values will be returned.

If no default values are assigned to the specified column, then a null is returned.

COLUMN Storage Attributes

Teradata Extensions

The COMPRESSphrase allows values in one or more columns of a permanent table to be compressed to zero space, thus reducing the physical storage space required for a table.

The COMPRESS phrase has three variations:

Variation: What happens:

COMPRESS Nulls are compressed.

COMPRESS NULL Nulls are compressed.

COMPRESS <constant> Nulls and the specified <constant> value are compressed.

Note: COMPRESS & COMPRESS NULL mean the same thing.

Examples

CREATE TABLE emp_data (employee_number INTEGER ,department_number INTEGER COMPRESS . . . CREATE TABLE bank_account_data (customer_id INTEGER ,account_type CHAR(10) COMPRESS 'SAVINGS' . . .

In the first example, null department numbers will be compressed.

In the second example, both null and 'Savings' account types will be suppressed. The value of 'Savings' is written in the table header for bank_account_data on each AMP in the system.

Primary Key vs Primary Index Primary Key (PK) - is defined as one or more columns used to uniquely identify each row in a table. PKs are used in conjunction with foreign keys to define the important column relationships in a database. PKs are always unique and cannot be null. PKs are not known to the Teradata RDBMS as such. The Teradata RDBMS implements a primary key as a unique index.

Primary Index - is defined as one or more columns used to distribute and locate rows in a table. Choice of primary index will affect distribution, access and performance. Oftentimes, but not always, the Primary Index and Primary Key are the same. Indexes (primary or secondary) may be used to enforce uniqueness (as in a PK) or to improve access. They may be unique or non-unique. Uniqueness and primary key constraints will be discussed in more detail in Level 3.

Indexes may be any combination as seen in the following table:

Primary Secondary

UNIQUE UPI USI

NON-UNIQUE NUPI NUSI

Every table must have exactly one primary index.

The following table illustrates performance characteristics for SQL operations that use indices.

Entries are listed in order from most desirable to least desirable in terms of system resource use. They show the number of AMPs involved in each operation and number of rows that could be returned.

#AMPS #ROWS

UPI 1 0 or 1

NUPI 1 Between 0-N where N = any number > 0

USI 2 Between 0-1

NUSI All Between 0-N

Full table scan All Between 0-N

CREATE TABLE Example

The following example shows everything we have learned up to this point about the CREATE TABLE statement.

CREATE TABLE MJ1.emp_data,FALLBACK, NO BEFORE JOURNAL,

NO AFTER JOURNAL ( employee_number INTEGER NOT NULL, department_number SMALLINT, job_code INTEGER COMPRESS , last_name CHAR(20) NOT CASESPECIFIC NOT NULL, first_name VARCHAR(20) NOT CASESPECIFIC, street_address VARCHAR(30) NOT CASESPECIFIC TITLE 'Address', city CHAR(15) NOT CASESPECIFIC DEFAULT 'Boise' COMPRESS 'Boise', state CHAR(2) NOT CASESPECIFIC DEFAULT ' ', birthdate DATE FORMAT 'mm/dd/yyyy', salary_amount DECIMAL(10,2), sex CHAR(1) UPPERCASE NOT CASESPECIFIC)

UNIQUE PRIMARY INDEX ( employee_number )

INDEX ( department_number );

Note: Employee number is a Unique Primary Index (UPI). The department number is a Non-Unique Secondary Index (NUSI).

CREATE TABLE Rules

Any user may create a table in a given database, provided that they have the 'CREATE TABLE' privilege for that database.

For INDEX definition, if UNIQUE is not explicit, then the default is a non-unique index. Similarly, if PRIMARY is not explicit, then the default is a secondary index.

Examples:

UNIQUE PRIMARY INDEX - UPI createdPRIMARY INDEX - NUPI createdUNIQUE INDEX - USI createdINDEX - NUSI created

DROP TABLE

To remove all data associated with a table, as well as the table structure definition from the Data Dictionary, use the DROP TABLE statement.

Example

Drop the employee data table created in the previous example.

DROP TABLE emp_data;

Deletes all data in emp_data. Removes the emp_data definition from the Data Dictionary. You must recreate the

table if you wish to use it again.

Removes all explicit access rights on the table.

To remove all data associated with a table, without dropping the table definition from the Data Dictionary, use the DELETE statement.

Example

Delete the employee data from the emp_date table in the previous example without dropping the table.

(All three examples are valid syntax which perform the operation.) DELETE FROM emp_data ALL;

DELETE FROM emp_data; DELETE emp_data;

Deletes all data in emp_data. Table definition remains in the Data Dictionary, so you can repopulate it. Access rights to the table remain unchanged.

ALTER TABLE Once a table has been created, certain characteristics are not alterable, such as the Primary Index choice. To change them you must CREATE a new table which includes the new characteristics, then populate that table.

Other characteristics are alterable. You can use the ALTER TABLE statement to modify these characteristics.

ALTER TABLE

1. ADDs and/or DROPs columns from an empty or populated table: ALTER TABLE emp_data ADD educ_level CHAR(1), ADD insure_type SMALLINT ;

ALTER TABLE emp_data DROP educ_level, DROP insure_type

2. Changes the attribute options on existing columns:

ALTER TABLE emp_data ADD birthdate FORMAT 'mmmBdd,Byyy";

Note: In this example, the birthday column already exists. We are adding FORMAT to

the birthday column.

3. Makes a NO FALLBACK table into a FALLBACK table, or vice versa:

ALTER TABLE emp_data ,FALLBACK;

4. Combines various changes to the table structure and protection: ALTER TABLE emp_data ,NO FALLBACK DROP insure_type ,ADD educ_level CHAR(1);

Note: Switching a table from FALLBACK to NO FALLBACK can be used to reclaim database space when needed.

CREATE a Secondary Index Indexes may be defined when creating a table with the CREATE TABLE statement, or alternatively, they may be created after the fact using the CREATE INDEX statement.

Primary indexes are always created at table creation time. Secondary indexes may be created at table creation or after the fact.

The CREATE INDEX statement is used to create secondary indexes on an existing table. Primary indexes must be created when you create the table and cannot be changed without dropping and re-creating the table (and specifying a new Primary Index).

Secondary indexes may be created with or without an identifying name.

Example

Create two secondary indexes on the employee table:

A unique secondary index (USI) on employee name (named) . A non-unique secondary index (NUSI) on job code (unnamed).

USI (named)

CREATE UNIQUE INDEX fullname (last_name, first_name)ON emp_data;

Note: We know this is a USI because primary indexes can't be created (except when the table is created). This index is named 'fullname'.

NUSI (unnamed)

CREATE INDEX (job_code)ON emp_data;

Note: Assigning names to indexes is optional. It has some advantages in terms of simplifying the syntax used to identify the index in subsequent SQL commands.

HELP INDEX

HELP INDEX <tablename> displays index definitions for the specified table. If the index is unnamed, the "Index Name" will display as NULL. Naming indexes is optional, however named indexes provide more ease of use in SQL commands we shall see later.

Note: HELP INDEX may sometimes produce a very wide report. In this case, the best way to view the output is to use the .SET FOLDLINE and .SET SIDETITLES features of BTEQ.

What are the indexes on the emp_data table?

.SET FOLDLINE ON;

.SET SIDETITLES ON;

HELP INDEX emp_data;

Unique?: Y UPI on employee_number Primary//or//Secondary?: P

Column Names:employee_numberIndex Id: 1

Approximate Count: 0Index Name: ?

Unique?: N NUSI on department_number

Primary//or//Secondary?: SColumn Name:s department_number

Index Id: 4Approximate Count: 0

Index Name: ?

Unique?: Y USI on last_name,first_name combined

Primary//or//Secondary?: S

Column Names: last_name,first_name

Index Id: 8

Approximate Count: 0

Index Name: fullname The name we gave to the index

Unique?: N NUSI on job_code Primary//or//Secondary?: S

Column Names: job_codeIndex Id: 12

Approximate Count: 0Index Name: ?

.SET FOLDLINE OFF;

.SET SIDETITLES OFF;

The values in the Index Id column correlate to the index numbers referenced in EXPLAIN text. The Primary index always has an Index ID of 1; Secondary indices have Index IDs in multiples of 4.

DROP INDEX

If you are dropping a named index, you can specify the index by either the index name or its column definition.

If you are dropping an unnamed index, you must specify the index by naming the columns which are associated with it.

Example

Delete both secondary indexes from the employee table.

DROP INDEX FullName ON emp_data; DROP INDEX (job_code) ON emp_data; Note: Only one secondary index may be dropped per DROP statement.

Lab



Hint: Prior to doing these labs, it will be helpful to reset your default database to your own user id (i.e. DATABASE tdxxx;).

A. CREATE an employee table in your own database identical to the employee table in the Customer_Service database. Use the following command to assist you in creating the new table in your own database.

SHOW TABLE CUSTOMER_SERVICE.EMPLOYEE;

Note: This table is used in many subsequent labs.

B. Modify the emp_new table created in Lab A to:

Remove fallback protection.Add a numeric column called 'salary_level' as a BYTEINT.Add a unique secondary index on the salary_level column.

C. Do a HELP INDEX to display the indexes for your table.D. If you have completed Labs B and C, ALTER the table to reverse Lab B. (i.e., DROP

the index; DROP the column; Return table to FALLBACK)

Lab Answer A

DATABASE tdxxx;CREATE TABLE emp_new ,FALLBACK

,NO BEFORE JOURNAL,NO AFTER JOURNAL

(employee_number INTEGER,manager_employee_number INTEGER,department_number INTEGER,job_code INTEGER,last_name CHAR(20) NOT NULL,first_name VARCHAR(30) NOT NULL,hire_date DATE NOT NULL,birthdate DATE NOT NULL,salary_amount DECIMAL(10,2) NOT NULL)

UNIQUE PRIMARY INDEX (employee_number);

*** Table has been created.

Lab Answer B

ALTER TABLE emp_new, NO FALLBACKADD salary_level BYTEINT;

*** Table has been modified.CREATE UNIQUE INDEX (salary_level) ON emp_new;

*** Index has been created.

Lab Answer C

HELP INDEX emp_new;

*** Help information returned. 2 rows.

Primary or Unique? Secondary? Column Names Y P employee_number Y S salary_level

Lab Answer D

DROP INDEX (salary_level) ON emp_new;

*** Index has been dropped.ALTER TABLE emp_new ,FALLBACK DROP salary_level ;

*** Table has been modified.

Data Manipulation Module:- 13


Insert data rows into a table. Update existing rows in a table. Delete rows from a table.

Data Manipulation

Data Manipulation consists of four commands:

INSERT -Add a row to a table.

INSERT SELECT -Add rows to a table from another table.

UPDATE -Change column values in existing rows of a table.

DELETE -Remove rows from a table.

INSERT

INSERT allows you to add a new row to a table.

Example

There are two types of insert:

1) Insert a new employee into the employee table:

INSERT INTO employeeVALUES (1210, NULL, 401, 412101, 'Smith', 'James', 890303, 460421, 41000);

EMPLOYEE (Before Inserts)

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT

PK FK FK FK 1006 1019 301 312101 Stein John 761015 531015 2945000

1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000

1007 1005 403 432101 Villegas Arnando 770102 370131 4970000

1003 0801 401 411100 Trader James 760731 470619 3785000

2) Insert a new employee with only partial data:

INSERT INTO employee (last_name, first_name, hire_date, birthday, salary_amount, employee_number)VALUES ('Garcia', 'Maria', 861027, 541110, 76500.00, 1291);

EMPLOYEE (After Inserts)

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT


1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

1210 401 412101 Smith James 890303 460421 4100000

1291 Garcia Maria 861027 541110 7650000

Note: Columns for which no values are inserted, will receive either their assigned default value (as provided by the DEFAULT or WITH DEFAULT options in the CREATE TABLE statement), or a NULL value if no default is specified but nulls are allowed.

INSERT SELECT

Use INSERT SELECT to copy rows from one table to another.

The syntax is as follows:

INSERT INTO target_table SELECT * FROM source_table;

The SELECT portion of the statement may be used to define a subset of rows and/or a subset of columns to be inserted to the target table.

Problem

To create a duplicate of the 'emp' table called 'emp_copy'.

Solution

First create the table 'emp_copy' with the same table definition as 'emp', then do a simple INSERT SELECT.

INSERT INTO emp_copySELECT * FROM emp;

Assumes:

emp and emp_copy have same definition. a complete replica of emp is desired.

Problem

Create a table for tracking birthdays and populate it with data from the employee table.

Solution

First, create a table which will contain the required rows and columns.

CREATE TABLE birthdays (empno INTEGER NOT NULL ,lname CHAR(20) NOT NULL ,fname VARCHAR(30) ,birth DATE) UNIQUE PRIMARY INDEX (empno);

Then, use INSERT SELECT to populate the new table. Only the required columns are selected for insertion. All rows are included.

INSERT INTO birthdays

SELECT employee_number

,last_name

,first_name

,birthdate

FROM employee;

BIRTHDAYS

EMPNO LNAME FNAMEBIRTHDATE

PK

1006 Stein John 531015

1008 Kanieski Carol 580517

1005 Ryan Loretta 550910

1004 Johnson Darlene 460423

1007 Villegas Arnando 370131

1003 Trader James 470619

EMPLOYEE

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT

PK FK FK FK

1006 1019 30131210

1Stein John 761015 531015 2945000

1008 1019 30131210

2Kanieski Carol 770201 580517 2925000

1005 0801 40343110

0Ryan Loretta 761015 550910 3120000

1004 1003 40141210

1Johnson Darlene 761015 460423 3630000

1007 1005 40343210

1Villegas Arnando 770102 370131 4970000

1003 0801 40141110

0Trader James 760731 470619 3785000

Note: .

In the Advanced Teradata SQL course we will see a feature of SQL which permits the CREATE TABLE statement and the INSERT SELECT statement to be combined into a single statement called CREATE TABLE WITH DATA.

UPDATE

UPDATE allows you to modify one or many columns of one or many rows in a single table.

The WHERE condition can include:

Columns from the table being updated. Joins with columns from other tables. Subqueries.

EMPLOYEE (Before modification)

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT


1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

1010 1003 401 412101 Rogers Frank 770301 350423 4600000

Problem

Change employee 1010's department to 403, job code to 432101, and manager to 1005 in the employee table:

UPDATE employeeSET department_number = ,job_code = ,manager_employee_number =WHERE employee_number =;

EMPLOYEE (After)

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT


1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

1010 1005 403 432101 Rogers Frank 770301 350423 4600000

UPDATE Using Subqueries or Joins

Problem

Update the employee table to give everyone in all support departments a 10% raise. Department numbers for all of the support departments are not known.

DEPARTMENT

DEPARTMENTNUMBER

DEPARTMENTNAME

BUDGETAMOUNT

MANAGEREMPLOYEE

NUMBER

PK FK

501 marketing sales 8005000 1017

301 research and development 46560000 1019

302 product planning 22600000 1016

403 education 93200000 1005

402 software support 30800000 1011

401 customer support 98230000 1003

201 technical operations 29380000 1025

EMPLOYEE

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT


1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3993000


1003 0801 401 411100 Trader James 760731 470619 4163500

Solution 1

Using Subquery:

UPDATE employeeSET salary_amount=salary_amount * 1.10WHERE department_number IN (SELECT department_number FROM department WHERE department_name LIKE '%Support%');

Solution 2

Using Join:

UPDATE employeeSET salary_amount=salary_amount * 1.10WHERE employee.department_number = department.department_numberAND department_name LIKE '%Support%';

Note: In an update, you can't use the ON clause, so the join condition is specified in the WHERE clause.

DELETE DELETE allows you to delete rows from a single table. If no WHERE clause is specified, then all rows are deleted.

The WHERE condition can reference:

Column values in the target table. Column values based on a subquery against another table. Column values based on a join with another table.

Problem

Remove all employees in department 301 from the employee table.

EMPLOYEE

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT

PK FK FK FK

1006 1019 301 312101 Stein John 761015 531015 2945000

1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

1010 1005 403 432101 Rogers Frank 770301 350423 4600000

Solution

DELETE FROM employeeWHERE department_number = 301;

Problem

Delete ALL data in the employee table.

EMPLOYEE

EMPLOYEENUMBER

MANAGEREMPLOYEE

NUMBER

DEPARTMENTNUMBER

JOBCODE

LASTNAME

FIRSTNAME

HIREDATE

BIRTHDATE

SALARYAMOUNT

PK FK FK FK 1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

1010 1003 403 432101 Rogers Frank 770301 350423 4600000

Solution

DELETE FROM employee ALL; /* Teradata syntax */

DELETE FROM employee; /* ANSI standard syntax */

Both produce the same result.

DELETE Using Subqueries or Joins

Problem

Remove all employees assigned to a temporary department for which the department name is "Temp".

EMPLOYEE

EMPLOYEE NUMBER

MANAGER EMPLOYEE

NUMBER

DEPARTMENT NUMBER

JOB CODE

LAST NAME

FIRST NAME

HIRE DATE

BIRTH DATE

SALARY AMOUNT


1008 1019 301 312102 Kanieski Carol 770201 580517 2925000

1005 0801 403 431100 Ryan Loretta 761015 550910 3120000

1004 1003 401 412101 Johnson Darlene 761015 460423 3630000


1003 0801 401 411100 Trader James 760731 470619 3785000

DEPARTMENT

DEPARTMENT NUMBER

DEPARTMENT NAME

BUDGET AMOUNT

MANAGER EMPLOYEE

NUMBER

PK FK

501 marketing sales 8005000 1017

301 research and development 46560000 1019

302 product planning 22600000 1016

403 education 93200000 1005

402 software support 30800000 1011

401 customer support 29230000 1003

600 Temp 1099

Solution 1

Using Subquery:

DELETE FROM employee WHERE department_number IN (SELECT department_number FROM department WHERE department_name ='Temp');

Solution 2

Using Join:

DELETE FROM employee WHERE employee.department_number=department.department_number AND department.department_name='Temp';

Note: In either case, no rows are removed from the 'employee' table because no current employees are assigned to the 'Temp' department.

Lab




The following labs are interdependent. Be sure to do them in sequence.

A. Use an INSERT SELECT statement to copy all rows from the Customer_Service employee table into the emp_new table created in the previous module.

B. Write SQL code to accomplish the following:

Select all department 301 employees to include employee number, last name, job code, and salary from the table created in Lab A in the preceding Data Definition module.

1. Note salary amounts. 2. Update your emp_new table to give a 15% raise to all employees in

department 301.

3. Verify results.C. Reverse the effects of the update in exercise B. Verify the reversal.

Lab Answer A

User input is highlighted in red.

DATABASE tdxxx;INSERT INTO emp_newSELECT *FROM Customer_Service.employee;** Insert completed. 26 rows added.

Lab Answer B


SELECT employee_number,last_name,job_code,salary_amount

FROM emp_newWHERE department_number = 301;


employee_number last_name job_code salary_amount --------------- ---------- -------- ------------- 1006 Stein 312101 29450.00 1008 Kanieski 312102 29250.00 1019 Kubic 311100 57700.00UPDATE emp_newSET salary_amount = salary_amount * 1.15WHERE department_number = 301;

*** Update completed. 3 rows changed.SELECT employee_number ,last_name ,job_code ,salary_amount FROM emp_new WHERE department_number = 301 ;*** Query completed. 3 rows found. 4 columns returned.

employee_number last_name job_code salary_amount --------------- --------- -------- ------------- 1006 Stein 312101 33867.00

1008 Kanieski 312102 33637.50 1019 Kubic 311100 66355.00

Lab Answer C

UPDATE emp_newSET salary_amount = salary_amount / 1.15WHERE department_number = 301;*** Update completed. 3 rows changed.SELECT employee_number ,last_name ,job_code ,salary_amount FROM emp_new WHERE department_number = 301 ;


employee_number last_name job_code salary_amount --------------- --------- -------- ------------- 1006 Stein 312101 29450.00 1008 Kanieski 312102 29250.00 1019 Kubic 311100 57700.00

Simple Macros Module:- 14

Objectives


CREATE and EXEC a simple macro. REPLACE a macro. DROP a macro.

Overview of Macros

A macro is a Teradata extension to ANSI SQL that contains prewritten SQL statements. The actual text of the macro is stored in a global repository called the Data Dictionary (DD). Macros are database objects and thus they belong to a specified user or database. They may contain one or more SQL statements. Macros have special efficiencies in the Teradata

environment and are highly desirable for building reusable queries.

A macro allows you to name a set of one or more statements. When you need to execute those statements, simply execute the named macro. Macros provide a convenient shortcut for executing groups of frequently-run SQL statements.

The following table is a fairly complete list of commands you may use to manipulate macros.

CREATE MACRO macroname AS ( . . . );

Define a macro and store it in the DD.

EXEC macroname; Execute statements within a macro.

SHOW MACRO macroname; Display a macro.

REPLACE MACRO macroname AS (. . . );

Apply changes to a macro or create a new one.

DROP MACRO macroname; Remove a macro definition from the DD. EXPLAIN EXEC macroname; Display EXPLAIN text for the macro's

execution.

CREATE MACRO The CREATE MACRO command allows you to define a set of SQL statements and put them into an executable statement.

Example

Create a macro to generate a birthday list for department 201:

CREATE MACRO birthday_list AS (SELECT last_name ,first_name ,birthdate FROM employee WHERE department_number =201 ORDER BY birthdate;);

To execute the birthday list macro:

EXEC birthday_list;last_name-----------


birthdate---------------

Morrissey Jim 43/04/29

Short Michael 47/07/07 Notice that there is a semicolon before the closing parenthesis. This is a required element of macro syntax. The CREATE MACRO statement will fail if there is a syntax error anywhere in the body of the SQL statement. If a macro exists, it is guaranteed to have previously been syntax checked. This does not guarantee that it will work without errors because objects referenced within the macro (e.g., the employee table) may not exist any longer or may have changed.

EXECUTE Macro

To execute a macro, simply precede its name with the EXEC command.

EXEC birthday_list;last_name-----------


birthdate---------------

Morrissey Jim 43/04/29

Short Michael 47/07/07

DROP Macro

Use the DROP MACRO command to delete a macro.

DROP MACRO birthday_list;

This command removes the macro from the containing database and also removes its entry from the Data Dictionary.

REPLACE Macro

You need to resubmit the entire contents of a macro to modify it. It is convenient to display current macro contents so you can cut and paste them into a text editor. You may then modify the macro without retyping the entire command. To display the definition of a macro, use the SHOW command as seen here:

SHOW MACRO birthday_list;Result

CREATE MACRO birthday_list AS SELECT last_name ,first_name ,birthdate FROM employee WHERE department_number = 201 ORDER BY birthdate;) ;

In editing the macro, change the CREATE command to the REPLACE command, and make the appropriate changes, such as adding a minor sort to the ORDER BY clause. Comments may optionally be added.

REPLACE MACRO birthday_list AS /* Macro is being updated for sorting sequence */ (SELECT last_name ,first_name ,birthdate

FROM employee WHERE department_number = 201 ORDER BY birthdate, last_name;);

Submit this as a BTEQ script. The old birthday_list macro will be replaced by the new one.

Lab





Click on the buttons to the left to see the answers.

References to 'tdxxx' are to be substituted with your personal userid.

Name MACROs to match the LAB numbers. For example, give the macro for the first lab below the name Lab 7_A.

A. Write a simple macro to create a customer geographical distribution list from the location table. Include location number, customer number, city and state (limit city and state to 14 characters each). Sort by city within state. CREATE the macro. Execute the macro. (The CREATE and EXEC should be separate scripts.)

Note: CUSTOMER_SERVICE is a Read-Only database. You need to create the macros in your own user space. Tell the system to do this either by:

1. DATABASE tdxxx; CREATE MACRO LAB 7_A AS...

or

2. CREATE MACRO tdxxx.Lab 7_A as...

Note: This lab is used again for Lab B and Lab D.

B. Modify the macro from Lab A to take location number out. Add zip code as the last column on the report. Zip code should be the lowest level of sort. Limit your selection to customers based in the state of Illinois. Save the modified macro under a new name. Submit and execute the new macro.

C. Use the appropriate HELP command to find the names of all macros in your database.

D. Replace the macro you created for Lab A, with these changes: Select only those

customers in the city of Chicago; Include their first address line as the last column of the report; Remove the state from the report; Sort the output by customer number in descending order.

E. Drop your Lab 7_A macro.

Lab Answer A

DATABASE Customer_Service;CREATE MACRO tdxxx.LAB7_A AS

(SELECT location_number,customer_number,CAST (city AS CHAR(14)),CAST (state AS CHAR(14))FROM customer_service.locationORDER BY state,city;);

*** Macro has been created.EXEC tdxxx.LAB7_A;

*** Query completed. 22 rows found. 4 columns returned. location_number

customer_number

city state

---------------

---------------

----------------

-------------

99 9999 Redondo CA5000015 15 Culver City California5000000 0 Los Angeles California5000011 11 Menlo Park California5000001 1 Palo Alto California5000010 10 Palo Alto California

5000002 2 San Francisco

California

5000019 19 San Mateo California9000005 5 Washington DC14000012 12 Chicago Illinois14000020 20 Chicago Illinois

22000013 13 Boston Massachusetts

23000008 8 Detroit MI34000009 9 Wilmington NC31000004 4 Trenton New Jersey33000017 17 Brooklyn New York

33000018 18 New York City

New York

33000016 16 New York New York

City

33000003 3 New York City

New York

33000006 6 Schenectady New York47000007 7 Arlington Virginia47000014 14 Richmond Virginia

Lab Answer B

SHOW MACRO tdxxx.LAB7_A;

*** Text of DDL statement returned.CREATE MACRO tdxxx.LAB7_A AS

(SELECT location_number,customer_number,CAST (city AS CHAR(14)),CAST (state AS CHAR(14))FROM locationORDER BY state,city;);

CREATE MACRO tdxxx.LAB7_B AS(SELECT customer_number,CAST (city AS CHAR(14)),CAST (state AS CHAR(14)),zip_codeFROM locationWHERE state = 'Illinois'ORDER BY state,city, zip_code;);

*** Macro has been created.

EXEC tdxxx.LAB7_B;


customer_number city state zip_code --------------- -------------- -------------- ----------- 20 Chicago Illinois 606316166 12 Chicago Illinois 606483930

Lab Answer C

HELP DATABASE tdxxx;

*** Help information returned. 12 rows.Table/View/Macro name Kind CommentLAB7_A M ?LAB7_B M ?

Lab Answer D

REPLACE MACRO tdxxx.LAB7_A AS(SELECT location_number,customer_number,CAST (city AS CHAR(14)),first_address_lineFROM customer_service.locationWHERE city = 'Chicago'ORDER BY customer_number DESC;);

*** Macro has been replaced.

EXEC tdxxx.LAB7_A;


location_number customer_number city first_address_line --------------- --------------- -------------- ------------------- 14000020 20 Chicago 876 Winston St 14000012 12 Chicago 510 Benton Av

Lab Answer E

DROP MACRO tdxxx.LAB7_A;

** Macro has been dropped.

Parameterized Macros Module:- 15

Objectives


Create and execute a parameterized macro.

Macros Revisited

Macros contain one or more prewritten SQL statements that are stored in the Teradata Data Dictionary. They can be executed from any viable SQL front-end, including:

Teradata SQL Assistant BTEQ Preprocessor CLI LOGON Startup Another macro.

Other key points to note about Macros:

Macros are a Teradata extension to SQL. Macros can only be executed with the EXEC privilege. Macros can provide column level security.

Note: If a user has EXEC privileges he may execute the macro. It doesn't matter whether or not he has privileges for the underlying tables or views that the macro uses.

Simple Parameterized Macros

Parameterized macros allow substitutable variables. Values for these variables are supplied at runtime.

Example

CREATE MACRO dept_list (dept INTEGER)AS( SELECT last_name FROM employee WHERE department_number = :dept;

);

In parentheses following the macro name is the parameter list. It names each parameter followed by its data type. When a parameter is used in the body of a macro, it is always preceded by a colon.

Employee Table:

Employee Number

Manager Employee Number

Department Number

Job Code Last Name First Name

1008 1019 301 312102 Kanieski Carol

1010 1003 401 412101 Rogers Frank

1014 1011 402 422101 Crane Robert

1006 1019 301 312101 Stein John

1017 801 501 511100 Runyon Irene

1003 801 401 411100 Trader James

1019 801 301 311100 Kubic Ron

1011 801 402 421100 Daly James

1013 1003 401 412102 Phillips Charles

1009 1005 403 432101 Lombardo Domingus

1025 801 201 211100 Short Michael

1024 1005 403 432101 Brown Allen

1007 1005 403 432101 Villegas Arnando

1005 801 403 431100 Ryan Loretta

1021 1025 201 222101 Morrissey Jim

1020 1005 403 432101 Charles John

1022 1003 401 412102 Machado Albert

801 801 100 111100 Trainer I.B.

1015 1017 501 512101 Wilson Edward

1002 1003 401 413201 Brown Alan

1004 1003 401 412101 Johnson Darlene

1016 801 302 321100 Rogers Nora

1018 1017 501 512101 Ratzlaff Larry

1001 1003 401 412101 Hoover William

1023 1017 501 512101 Rabbit Peter

1012 1005 403 432101 Hopkins Paulene

To Execute

EXEC dept_list (301);

Result

last_name------------

SteinKanieskiKubic

Macros with Multiple Parameters

Macros may have more than one parameter. Each name and its associated type are separated by a comma from the next name and its associated type. The order is important. The first value in the EXEC of the macro will be associated with the first value in the parameter list. The second value in the EXEC is associated with the second value in the parameter list, and so on.

Example

Create the macro:

CREATE MACRO emp_check (dept INTEGER ,sal_amt DEC(9,2)) AS (SELECT employee_number from employee WHERE department_number = :dept AND salary_amount < :sal_amt;) ;

Execute the macro:

EXEC emp_check (301, 50000);

Results:

employee_number-------------------

1006 1008

In this example, 301 is associated with the department number and 50000 is associated with the salary amount.

Using a Parameterized Macro to Insert Data

CREATE MACRO new_dept ( dept INTEGER , budget DEC(10,2) DEFAULT 0 , name CHAR(30) , mgr INTEGER) AS ( INSERT INTO department ( department_number , department_name

, budget_amount , manager_employee_number)

VALUES ( :dept , :name

, :budget , :mgr ) ; SELECT department_number (TITLE ‘Number’) ,department_name (TITLE ‘Name’) ,budget_amount (TITLE ‘Budget’) ,manager_employee_number (TITLE ‘Manager’) FROM department WHERE department_number = :dept; );

EXECUTE the INSERT MACRO (With Positional Parameter Values):

Input data in the order specified in the Macro parameter list. Provide the exact number of parameters specified in the list. Check data types for compatibility with CREATE MACRO data types. Check that inserted nulls are permitted in the target columns. Check if defaulting is specified in the CREATE MACRO statement. Use the keyword NULL to explicitly pass a null to the macro. Use positional commas to implicitly pass a null, or a specified default value.

EXEC new_dept (505 ,610000.00 , 'Marketing Research', 1007);Number--------

Name---------------

Budget--------

Manager----------

505 Marketing Research 610000.00 1007

EXEC new_dept (102 , , 'Payroll', NULL);

Number--------

Name---------------

Budget--------

Manager----------

102 Payroll .00 ?In the example above, the value after 102 has been omitted, hence the two commas in a row. In such cases the parameter value will be set to the default specified for in the CREATE MACRO statement. In this case it is 0.

EXECUTE the INSERT MACRO

Another option available for passing parameters to a macro is to explicitly name the parameters with an associated value as seen in this example. When this approach is selected, the parameters may be specified in any sequence. Note in the following example, implicitly defaulted parameters (like 'mgr') may simply be omitted without accounting for their position.

Example

EXEC new_dept( name = 'accounting' ,budget = 425000.00

,dept = 106 ) ;Number-------

Name-------

Budget--------

Manager----------

106 accounting 425000.00 ?

Performance Note: Parameterized macros may be submitted multiple times with different parameter values each time. Because the body of the query doesn't change, it permits execution of the same optimized 'steps' which are cached for re-use. This feature of macros makes them highly desirable from a performance standpoint.

Lab





Click on the buttons to the left to see the answers.

To do the following labs, you will first need to create and populate your own copy of the department table. This may be accomplished by executing the following commands:

DATABASE tdxxx;

CREATE TABLE dept_new, FALLBACK (department_number SMALLINT

,department_name CHAR(30) NOT NULL ,budget_amount DECIMAL(10,2) ,manager_employee_number INTEGER ) UNIQUE PRIMARY INDEX (department_number) ,UNIQUE INDEX (department_name);

INSERT INTO dept_new SELECT * FROM Customer_Service.department;

A. Create a parameterized MACRO to insert a new row into your dept_new table. Display the newly inserted row for verification. Insert three rows into the table using the MACRO . Use the values below for the columns shown:

department_number department_name budget amount manager_emp_no

601 shipping 800,000701 credit 1,200,000 1018905 president’s staff 5,000,000 801

B. Note: Remember to override your default database. C. Execute the exercise A macro again. Put EXPLAIN in front of the EXECute to see

what the Teradata RDBMS is doing. D. Write a parameterized MACRO to first select, then update your employee table to

give all employees in a given department a flat salary increase. Actual department number and salary increase will be given at execution time. The select should precede the update and should include employee number, last name, and two salary columns: one titled “Current Salary”, and the other titled “New Salary”. Format the salary columns with floating dollar signs. Limit last name to 10 characters, and sort by last name. Execute this MACRO to give department 401 employees a $1000 raise. Execute it again to reverse the effects of the previous run.

E. Write a parameterized MACRO to create a report showing the last name, department and salary for employees within a salary range that will be supplied at run time. Execute the MACRO to find all employees in the over $50,000 under $100,000 income bracket. Execute it again to find all employees in the under $30,000 income bracket.

Lab Answer A

DATABASE tdxxx;

*** New default database accepted.

CREATE MACRO dept_macro( dept INTEGER, name CHAR(30), budget DEC(10,2), mgr INTEGER)

AS( INSERT INTO dept_new

( department_number, department_name, budget_amount, manager_employee_number)

VALUES ( :dept, :name, :budget, :mgr )

;SELECT department_number (TITLE 'Number')

,department_name (TITLE 'Name'),budget_amount (TITLE 'Budget'),manager_employee_number (TITLE 'Manager')

FROM dept_newWHERE department_number = :dept;

);


EXEC dept_macro (601, 'Shipping', 800000, NULL) ;

*** Insert completed. One row added.

Number Name Budget Manager------ ------------------------------ ------------ ----------- 601 Shipping 800000.00 ?

EXEC dept_macro (701, 'Credit', 1200000, 1018) ;


Number Name Budget Manager ------- ---------- ------------ --------701 Credit 1200000.00 1018

EXEC dept_macro (905, 'President''s Staff', 5000000, 801) ;


Number Name Budget Manager ------- ---------------- ----------- -------- 905 President's Staff 5000000.00 801

Lab Answer B

EXPLAIN EXEC dept_macro(601, 'shipping', 800000, NULL) ;


Lab Answer C

CREATE MACRO emp_macro(dept INTEGER,raise DECIMAL (5))

AS(SELECT employee_number

,last_name (char (10) ),salary_amount (format ' $$, $$$, $$9.99' )(title 'Current//Salary' ),salary_amount + :raise(format '$$, $$$, $$9.99' )(title 'New//Salary' )

FROM emp_newWHERE department_number = :deptORDER BY 2;UPDATE emp_newSET salary_amount = salary_amount + :raiseWHERE department_number = :dept;);


EXEC emp_macro (401,1000) ;


Current Newemployee_number last_name Salary Salary--------------- ---------- ------------- ------------- 1002 Brown $43,100.00 $44,100.00 1001 Hoover $25,525.00 $26,525.00 1004 Johnson $36,300.00 $37,300.00 1022 Machado $32,300.00 $33,300.00 1013 Phillips $24,500.00 $25,500.00 1010 Rogers $46,000.00 $47,000.00 1003 Trader $37,850.00 $38,850.00

EXEC emo_macro (401, -1000) ;


Current Newemployee_number last_name Salary Salary--------------- ---------- ------------- ------------- 1002 Brown $44,100.00 $43,100.00 1001 Hoover $26,525.00 $25,525.00 1004 Johnson $37,300.00 $36,300.00 1022 Machado $33,300.00 $32,300.00 1013 Phillips $25,500.00 $24,500.00 1010 Rogers $47,000.00 $46,000.00 1003 Trader $38,850.00 $37,850.00

Lab Answer D

CREATE MACRO hilo_macro(low DECIMAL (10,2),high DECIMAL (10,2))

AS(SELECT last_name

,department_number,salary_amount

FROM emp_newWHERE salary_amount BETWEEN :low AND :highORDER BY 3;);


EXEC hilo_macro (50001,99999);


last_name department_number salary_amount-------------------- ----------------- -------------Daly 402 52500.00Wilson 501 53625.00Ratzlaff 501 54000.00Rogers 302 56500.00Kubic 301 57700.00Runyon 501 66000.00EXEC hilo_macro (high = 29999, low = 0) ;


last_name department_number salary_amount-------------------- ----------------- -------------Phillips 401 24500.00Crane 402 24500.00Hoover 401 25525.00Rabbit 501 26500.00Kanieski 301 29250.00Stein 301 29450.00

Aggregating Groups Module:- 16

Objectives


Aggregate column data using the aggregate functions SUM, COUNT, AVERAGE, MINIMUM, MAXIMUM.

Use the GROUP BY clause to aggregate over defined groups.

Aggregate Operators

Aggregate operators perform computations on values in a specified group. The five aggregate operators are:

ANSI Standard Teradata Supported

COUNT COUNT

SUM SUM AVG AVERAGE, AVG

MAX MAXIMUM, MAX

MIN MINIMUM, MIN

AGGREGATE operations ignore NULLs and produce ONLY single-line answers.

Example

SELECT COUNT ( salary_amount ) (TITLE 'COUNT') ,SUM ( salary_amount ) (TITLE 'SUM SALARY') ,AVG ( salary_amount ) (TITLE 'AVG SALARY') ,MAX ( salary_amount ) (TITLE 'MAX SALARY') ,MIN ( salary_amount ) (TITLE 'MIN SALARY') FROM employee ;

Result

COUNT SUM SALARY AVG SALARY MAX SALARY MIN SALARY

6 213750.00 35625.00 49700.00 29250.00 Note: If one salary amount value had been NULL, the COUNT would have returned a count of 5. In that case, the average would have reflected an average of only five salaries. To COUNT all table rows use COUNT (*), which will count rows regardless of the presence of NULLs.

Why Aggregation is Useful

To find the total amount of money spent by each department on employee salaries. Without the GROUP BY clause, we could attempt to get an answer by running a separate query against each department.

Solution

SELECT SUM (salary_amount) FROM employee WHERE department_number = 401 ;

Sum (salary_amount) 74150.00

SELECT SUM (salary_amount) FROM employee WHERE department_number = 403 ;

Sum (salary_amount) 80900.00

SELECT SUM (salary_amount) FROM employee WHERE department_number = 301 ; Sum (salary_amount) 58700.00

Question: What if there are 1000 departments?

GROUP BY provides the answer with a single query, regardless of how many departments there are.

SELECT department_number ,SUM (salary_amount) FROM employee GROUP BY department_number ;department_number---------------------

Sum(salary_amount)---------------------

401403301

74150.0080900.0058700.00

Hint: Think of GROUP BY as For Each.

GROUP BY and the WHERE Clause

The WHERE clause eliminates some rows before GROUP BY puts them into desired groupings.

Problem

Compute total salaries by department for departments 401 and 403.

Solution

SELECT department_number ,SUM (salary_amount) FROM employee WHERE department_number IN (401, 403) GROUP BY department_number ;department_number----------------------

Sum (salary_amount)-----------------------

403 401

80900.00 74150.00

GROUP BY and ORDER BY

GROUP BY does not imply any ordering of the output. An ORDER BY clause is needed to control the order of the ouput.

Problem

For departments 301, 401 and 403, find the total number of employees in each department as well as the highest, lowest and average salaries for each department. Order results by department number.

SELECT department_number (TITLE 'DEPT') ,COUNT (*) (TITLE '#_EMPS') ,SUM (salary_amount) (TITLE 'TOTAL') (FORMAT 'zz,zzz,zz9.99') ,MAX (salary_amount) (TITLE 'HIGHEST') (FORMAT 'zz,zzz,zz9.99') ,MIN (salary_amount) (TITLE 'LOWEST') (FORMAT 'zz,zzz,zz9.99') ,AVG (salary_amount) (TITLE 'AVERAGE')

(FORMAT 'zz,zzz,zz9.99') FROM employee WHERE department_number IN (301,401,403) GROUP BY department_number ORDER BY department_number ;

GROUP BY on Multiple Columns

It is possible and often desireable to GROUP BY more than one column. This permits the query to compute aggregates of groups within groups. In the following example, we are looking for salaries by job code (one group) within department (second group).

By job code, what are the total salaries for departments 401 and 403?

SELECT department_number ,job_code ,SUM (salary_amount) FROM employee WHERE department_number IN (401, 403) GROUP BY 1, 2 ORDER BY 1, 2 ;


job_code----------

SUM (salary_amount)-----------------------

401401401401403403

411100412101412102413201431100432101

37850.00107825.00

56800.0043100.0031200.00

201800.002 Rules of Aggregations:

1. Whenever both aggregate and non-aggregate columns are selected, a GROUP BY clause is required.

2. All non-aggregated selected columns must be included in the GROUP BY clause.

Note: Repeating values such as '401' above may be suppressed using the BTEQ command:

.SET SUPPRESS ON 1; /* Suppress repeating values on first column */

This will cause the value to be displayed only the first time.

The resulting report would look as follows:

DEPT #_EMPS TOTAL HIGHEST LOWEST AVERAGE

301401403

376

116,400.00245,575.00233,000.00

57,700.0046,000.0049,700.00

29,250.0024,500.0031,000.00

38,800.0035,082.1438,833.33

job_code----------

SUM (salary_amount)-----------------------

401

403

411100412101412102413201431100432101

37850.00107825.00

56800.0043100.0031200.00

201800.00

GROUP BY and HAVING Condition

HAVING is just like WHERE , except that it applies to groups rather than rows. HAVING qualifies and selects only those groups that satisfy a conditional expression.

Problem

Determine which departments have average salaries of less than $36,000. Generate the report using the HAVING clause:

SELECT department_number AS DEPT COUNT (*) AS #_EMPS ,CAST ( SUM (salary_amount) AS FORMAT 'zz,zzz,zz9.99') AS TOTAL ,CAST ( MAX (salary_amount) AS FORMAT 'zz,zzz,zz9.99') AS HIGHEST ,CAST ( MIN (salary_amount) AS FORMAT 'zz,zzz,zz9.99') AS LOWEST ,CAST ( AVG (salary_amount) AS FORMAT 'zz,zzz,zz9.99') AS MEAN FROM employee GROUP BY department_number HAVING AVG (salary_amount) < 36000 ;Since only one department (401) has an average salary less than $36,000, the result is a single row, as follows:

GROUP BY Summary

Here is the order of evaluation within a SQL statement if all four clauses are present:

WHERE

Eliminates some or all rows immediately based on condition.

DEPT #_EMPS TOTAL HIGHEST LOWEST AVERAGE

401 3 245,575.00 46,000.00 24,500.00 35,082.14

Only rows which satisfy a WHERE condition are eligible for inclusion in groups.

GROUP BY

Puts qualified rows into desired groupings.

HAVING

Eliminates some (or all) of the groupings based on condition.

ORDER BY

Sorts final groups for output.

(ORDER BY is not implied by GROUP BY)

Lab

For this set of lab questions, you will need information from the Database Info document.




A. Management needs to know the impact of salary increases if all employees in every department receive a ten percent salary increase. Create a report showing minimum and maximum salaries for each department. Format the report as shown below. Limit department name to 9 characters. Format dollar columns with zero-suppress and commas. Sort by department.

B.C. Dept Dept Minimum Maximum Current AdjustedD. Nbr Name Salary Salary Salary SalaryE. --------- --------- ---------- ---------- -------- ---------F. Management needs a summary report showing all departments, jobs within

departments, the number of employees doing each job, and the average salary for each job category. Sequence the report by department and job code within department. Let the column titles default.

Lab Answer A


DATABASE Customer_Service;SELECT e.department_number (TITLE 'Dept//Nbr')

,d.department_name (TITLE 'Dept//Name')(FORMAT 'X(9)')

,MIN(e.salary_amount) (TITLE 'Minimum//Salary')(FORMAT 'Z,ZZZ,ZZ9.99')

,MAX(e.salary_amount) (TITLE 'Maximum//Salary')(FORMAT 'Z,ZZZ,ZZ9.99')

,SUM(e.salary_amount) (TITLE 'Current//Salary')(FORMAT 'Z,ZZZ,ZZ9.99')

,SUM(e.salary_amount * 1.1) (TITLE 'Adjusted//Salary')(FORMAT 'Z,ZZZ,ZZ9.99')


ON e.department_number = d.department_numberGROUP BY e.department_number

,d.department_nameORDER BY e.department_number;


Dept Dept Minimum Maximum Current Adjusted Nbr Name Salary Salary Salary Salary ---- --------- ---------- ---------- ---------- ---------- 100 president 100,000.00 100,000.00 100,000.00 110,000.00 201 technical 34,700.00 38,750.00 73,450.00 80,795.00 301 research 29,250.00 57,700.00 116,400.00 128,040.00 302 product p 56,500.00 56,500.00 56,500.00 62,150.00 401 customer 24,500.00 46,000.00 245,575.00 270,132.50 402 software 24,500.00 52,500.00 77,000.00 84,700.00 403 education 31,000.00 49,700.00 233,000.00 256,300.00 501 marketing 26,500.00 66,000.00 200,125.00 220,137.50

Lab Answer B

SELECT department_number,job_code,COUNT(job_code),AVG(salary_amount)

FROM employeeGROUP BY department_number

,job_codeORDER BY department_number

,job_code;


department_number job_code Count(job_code) Average(salary_amount)----------------- ----------- --------------- ---------------------- 100 111100 1 100000.00

201 211100 1 34700.00 201 222101 1 38750.00 301 311100 1 57700.00 301 312101 1 29450.00 301 312102 1 29250.00 302 321100 1 56500.00 401 411100 1 37850.00 401 412101 3 35941.67 401 412102 2 28400.00 401 413201 1 43100.00 402 421100 1 52500.00 402 422101 1 24500.00 403 431100 1 31200.00 403 432101 5 40360.00 501 511100 1 66000.00 501 512101 3 44708.33

Totals and Subtotals Module:-17

Objectives


Create subtotals using WITH....BY. Create final totals using WITH. Eliminate duplicate results using the DISTINCT option.

Using WITH...BY for Subtotals

The WITH...BY clause is a Teradata extension that creates subtotal lines for a detailed list. It differs from GROUP BY in that detail lines are not eliminated.

The WITH...BY clause allows subtotal "breaks" on more than one column and generates an automatic sort on all "BY" columns.

Creating a Report Using WITH..BY

Problem

To display employee name and salary with subtotals by department.

Solution

SELECT last_name AS NAME ,salary_amount AS SALARY ,department_number AS DEPT

FROM employee WITH SUM(salary) BY DEPT ;

Note: The "BY DEPT" phrase causes an automatic sort on the Dept column.

Result

NAME SALARY DEPTStein 29450.00 301Kanieski 29250.00 301

--------Sum(Salary) 58700.00

Johnson 36300.00 401Trader 37850.00 401

--------Sum(Salary) 74150.00

Villegas 49700.00 403Ryan 31200.00 403

--------Sum(Salary) 80900.00

Note: "WITH BY" is a non-ANSI Teradata extension. It is supported by BTEQ, but not supported by ODBC clients.

WITH..BY Multiple Aggregate

Multiple aggregates within the same WITH BY clause are possible if all aggregates are based on the same 'break' column. In the following example, both aggregates are generated by department. Note that while WITH BY produces aggregates similar to the GROUP BY functionality, the difference is that GROUP BY does not show the detail rows upon which the aggregates are based.

Problem

Show salary subtotals and averages by department and also show the individual employee salaries.

Solution

SELECT last_name AS NAME ,salary_amount AS SALARY ,department_number AS DEPT FROM employee WHERE employee_number BETWEEN 1003 AND 1008 WITH SUM(salary)(TITLE 'Dept Total') ,AVG(salary)(TITLE 'Dept Avg ')BY DEPT ;

Result

NAME SALARY DEPT

SteinKaniesk

29450.0029250.00

301 301

------------

Dept Total Dept Avg

58700.00 29350.00

JohnsonTrader

36300.0037850.00

401 401

------------

Dept Total Dept Avg

74150.0037075.00

VillegasRyan

49700.0031200.00

403 403

------------

Dept Total Dept Avg

80900.0040450.00

Note: The BY DEPT clause causes results to be sorted by department.

WITH... BY Multiple Subtotal Levels

You may use several "WITH BY" clauses together in an SQL statement.

The order in which you specify them determines the order of the sorted output.

Problem

Show total salaries by department and by job code within department for departments 401 and 501. Also show the detail salary for each employee.

Solution

SELECT department_number AS Dept ,job_code AS Job ,employee_number AS Empl ,salary_amount AS Sal FROM employee WHERE department_number IN (401, 501) WITH SUM(salary_amount)(TITLE 'JOB TOTAL') BY Job WITH SUM(salary_amount)(TITLE 'DEPT TOTAL') BY Dept

;

Result

Dept Job Empl Sal-------- ------------ -------- --------------

401 411100 1003 37850.00 ------------

JOB TOTAL 37850.00

401 412101 1010 25525.00401 412101 1004 36300.00401 412101 1010 46000.00

------------JOB TOTAL 107825.00

401 412102 1022 32300.00401 412102 1013 24500.00

------------JOB TOTAL 56800.00

401 413201 1002 43100.00 ------------

JOB TOTAL 43100.00

------------DEPT TOTAL 245575.00

501 511100 1017 66000.00 ------------

JOB TOTAL 66000.00

501 512101 1018 54000.00501 512101 1023 26500.00501 512101 1015 53625.00

------------JOB TOTAL 134125.00

------------DEPT TOTAL 200125.00

Note: Because there are two BY clauses in this query, their sequence in the query affects how the final report is sorted. Because 'BY Dept' is the last one specified, it becomes the highest level sort column, thus the report is sorted as job code within department.

Creating Final Totals Using WITH

The WITH clause without 'BY' is used to create a grand totals.

Problem

Display employee numbers and salary amounts for department 301 and a final

total for the department.

Solution

SELECT employee_number ,salary_amount FROM employee WHERE department_number = 301 WITH SUM(salary_amount)(TITLE 'GRAND TOTAL') ORDER BY employee_number ;

Result

employee_number salary_amount------------------------ --------------------

1006 29450.001008 29250.001019 57700.00

------------GRAND TOTAL 116400.00

WITH is a non-ANSI Teradata extension and is not supported by ODBC clients.

DISTINCT Modifier

The DISTINCT modifier is used in conjuction with the COUNT aggregate to prevent the same value from being counted more than once.

Problem

Count the number of managers for employees numbered between 1003 and 1008.

Solution

SELECT employee_number ,department_number ,manager_employee_number (NAMED manager) FROM employee WHERE employee_number BETWEEN 1003 AND 1008 WITH COUNT (manager)(TITLE 'TOTAL MANAGERS') ;

Result

employee_number department_number manager--------------- ----------------- ----------- 1003 401 801 1007 403 1005

1004 401 1003 1008 301 1019 1005 403 801 1006 301 1019 ----------- TOTAL MANAGERS 6

Note that even though the count is represented as six, there are really only four managers.

To solve this problem, use the DISTINCT modifier.

Solution

SELECT employee_number ,department_number ,manager_employee_number AS manager FROM employee WHERE employee_number BETWEEN 1003 AND 1008 WITH COUNT (DISTINCT manager)(TITLE 'TOTAL MANAGERS') ;

Result

employee_number department_number manager--------------- ----------------- ----------- 1003 401 801 1004 401 1003 1007 403 1005 1008 301 1019 1005 403 801 1006 301 1019 ----------- TOTAL MANAGERS 4

This gives you an actual count for the number of managers.

Note: When using DISTINCT with an aggregate function only a single column or expression may be used.

Legal Example:

SELECT COUNT(DISTINCT(job_code)) ,COUNT(DISTINCT(employee_number)).....

Illegal Example:

SELECT COUNT(DISTINCT(job_code, employee_number)).......

Combining WITH and WITH BY

WITH and WITH BY may be combined within the same query.

Problem

Show the salary for each employee with subtotals by department, a final total, and results sorted by employee name.

Solution

SELECT last_name AS NAME ,salary_amount AS SALARY ,department_number AS DEPT FROM employee WHERE employee_number BETWEEN 1003 and 1008 WITH SUM (SALARY) BY DEPT WITH SUM (SALARY) (TITLE 'GRAND TOTAL') ORDER BY NAME ;

Results

NAME SALARY DEPTKanieskiStein

29250.0029450.00

301 301

------------ Sum (Salary) 58700.00 JohnsonTrader

36300.0037850.00

401 401

------------ Sum (Salary) 58700.00 RyanVillegas

31200.0049700.00

403 403

------------ Sum (Salary) 58700.00 ------------ GRAND TOTAL 213750.00 Note that the WITH BY clause controls the high level sort by department. The ORDER BY clause is applied as a minor level sort, producing a sorted list of names within sorted departments.

Final Totals Using WITH and GROUP BY WITH and GROUP BY clauses may both appear in the same query. The GROUP BY clause is needed when both aggregates and non-aggregates are in the SELECTed columns. The WITH clause is needed when a final aggregation of all the groups is desired.

Problem

Show the total and average salaries for each department plus a final total and average for

the company.

Solution

SELECT department_number (TITLE 'dept_no') ,SUM (salary_amount) ,AVG (salary_amount) FROM employee GROUP BY department_number WITH SUM (salary_amount) (TITLE 'GRAND TOTAL') ,AVG (salary_amount) (TITLE ' ') ORDER BY department_number ;

Result

dept_no SUM(salary_amount) AVG(salary_amount)------------ ----------------------------- ----------------------------

301 58700.00 29350.00401 74150.00 37075.00403 80900.00 40450.00

----------------------------- ----------------------------GRAND TOTAL 213750.00 35635.00

Summary of WITH...BY and WITH

The WITH...BY clause specifies summary lines and breaks, or grouping conditions, that determine how results are returned. This clause is typically used when you need to produce subtotals and also wish to see the detail rows.

WITH summarylist BY breaklist ASC/DESC:

Will provide subtotals, subcounts, and subaverages, and also show detail rows. Summarylist can specify more than one column. Breaklist can specify more than one column. Implied ORDER BY on the breaklist columns. WITH...BY determines the major sort key(s). ORDER BY specifies any additional minor sorts. An SQL statement can have several WITH...BY clauses. The highest level of sort is the last specified WITH...BY clause. Is not supported by ANSI standard or ODBC clients.

WITH summarylist:

The WITH clause produces grand total results for the entire answer set. This clause is typically used when you need to produce final totals and also wish to see the detail rows.

Will provide a final or grand total, count or average.

Summarylist may specify more than one column. Only a single WITH is allowed in a query. WITH must follow any WITH...BY syntax. Not supported by ANSI standard or ODBC clients.

Lab





A. Use the location table to find the customers located in California.Matching on customer number, list customer name, number, and sales representativesnumber from the customer table. Sort by sales representatives number.Give a count of customers at the end of the report as shown below.

customer_name customer_number sales_employee_number . . .. . .. . .---------------Total Cust X

B. Do a salary study report for departments 401 and 403, excluding job codes 411100, 413201 and 431100. List employee's last and first name (10 characters each), department, job code and salary. Give totals for each department and job code within department. Give a grand total for salary. Give an average figure as well. Sort by last name within department and job. Format the report as follows:

C. last_namefirst_name Dept Job SalaryD. ------------ ------------ ---- ------ ---------E. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xxF. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xxG. Job Total xxxxx.xxH. --------I. Job Avg xxxxx.xxJ. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xxK. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xx L. Job Total xxxxx.xxM. --------N. Job Avg xxxxx.xxO. Dept Total xxxxx.xxP. --------Q. Dept Avg xxxxx.xx

R. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xxS. xxxxxxxxxx xxxxxxxxxx xxxx xxxxxx xxxxx.xxT. Job Total xxxxx.xxU. --------V. Job Avg xxxxx.xxW. Dept Total xxxxx.xxX. --------Y. Dept Avg xxxxx.xxZ. Grand Total xxxxx.xx

Grand Avg xxxxx.xx

Lab Answer A

User input is highlighted in red.DATABASE Customer_Service;SELECT c.customer_name

,l.customer_number,c.sales_employee_number

FROM customer c INNER JOINlocation l

ON c.customer_number = l.customer_numberWHERE l.state = 'California'ORDER BY sales_employee_numberWITH COUNT(l.customer_number) (TITLE 'Total Cust');


customer_name customer_number sales_employee_number ------------------------------ --------------- --------------------- Corporate Headquarters 0 ? Simple Instruments Co. 2 1015 Cates Modeling 15 1015 Hotel California 11 1015 A to Z Communications, Inc. 1 1015 More Data Enterprise 19 1015 Graduates Job Service 10 1015 --------------- Total Cust 7

Lab Answer B

SELECT last_name (FORMAT 'X(10)'),first_name (FORMAT 'X(10)'),department_number (NAMED Dept),job_code (NAMED Job),salary_amount (NAMED Salary)

FROM employeeWHERE department_number IN (401, 403)AND job_code NOT IN (411100, 413201, 431100)WITH SUM(salary) (TITLE 'Job Total') BY Job

WITH AVG(salary) (TITLE 'Job Avg') BY JobWITH SUM(salary) (TITLE 'Dept Total') BY DeptWITH AVG(salary) (TITLE 'Dept Avg') BY DeptWITH SUM(salary) (TITLE 'Grand Total')

,AVG(salary) (TITLE 'Grand Average')ORDER BY last_name;


last_name first_name Dept Job Salary ---------- ---------- ----------- ----------- ------------ Hoover William 401 412101 25525.00 Johnson Darlene 401 412101 36300.00 Rogers Frank 401 412101 46000.00 ------------ Job Total 107825.00 ------------ Job Avg 35941.67 Machado Albert 401 412102 32300.00 Phillips Charles 401 412102 24500.00 ------------ Job Total 56800.00 ------------ Job Avg 28400.00 ------------ Dept Total 164625.00 ------------ Dept Avg 32925.00 Brown Allen 403 432101 43700.00 Charles John 403 432101 39500.00 Hopkins Paulene 403 432101 37900.00 Lombardo Domingus 403 432101 31000.00 Villegas Arnando 403 432101 49700.00 ------------ Job Total 201800.00 ------------ Job Avg 40360.00 ------------ Dept Total 201800.00 ------------ Dept Avg 40360.00 ------------ ------------

Grand Total 366425.00 Grand Average 36642.50

Note: Report spacing and underscores will look slightly different if you combine both aggregates in one WITH...BY statement.

Rankings, Samplings and Built-in Functions Module:-18


Perform simple rankings on rows in a table. Extract samples of data from a table Employ the Teradata built-in system functions. Retrieve the top n rows of an output result.

Simple Rankings

The RANK function permits a column to be ranked, either based on high or low order, against other rows in the answer set. You may specify the order of ranking by use of the ORDER BY clause within the RANK function. Descending order will provide traditional ranking wherein the largest value is assigned a rank of one. Ascending order will associate the largest value with the highest rank value.

Most of the examples we will see are rankings of sales figures, where the larger the amount, the better (i.e. lower number) the ranking.

The syntax for the RANK function is:

RANK( ) OVER (ORDER BY sales DESC)

where 'sales' ' represents the column to be ranked and the descending sort key of the result.

Problem

Show the ranking of product sales for store 1001.

Solution

SELECT storeid ,prodid ,sales ,RANK( ) OVER (ORDER BY sales DESC) AS "Rank" FROM salestbl

WHERE storeid = 1001;

Result

storeid prodid sales Rank----------- ------ ----------- ----------- 1001 F 150000.00 1 1001 A 100000.00 2 1001 C 60000.00 3 1001 D 35000.00 4

Things To Notice:

WHERE clause qualifies rows to be ranked. When the order of sales is DESC, the highest sales amount is rank #1

Problem

Show the lowest ranking of product sales for store 1001.

Solution

SELECT storeid ,prodid ,sales ,RANK( ) OVER (ORDER BY sales ASC) AS "Rank" FROM salestbl WHERE storeid = 1001;

Result

storeid prodid sales Rank----------- ------ ----------- ----------- 1001 D 35000.00 1 1001 C 60000.00 2 1001 A 100000.00 3 1001 F 150000.00 4

Things To Notice:

When the order of sales is ASC, the lowest sales amount is rank #1 Rank #1 always appears at top of list unless overridden.

Rankings With Qualification

The QUALIFY clause allows restriction of which rankings will be output in the final result. In the second query at hand, the QUALIFY clause restricts the result to the top three sales amounts.

QUALIFY performs like the HAVING clause by requesting a specific range in the output.

Problem

Get top 3 sales - any product in any store:

Solution

SELECT storeid ,prodid ,sales ,RANK( ) OVER (ORDER BY sales DESC) AS "Ranking"FROM salestbl QUALIFY Ranking <= 3;

Result

storeid prodid sales Ranking----------- ------ ----------- ----------- 1001 F 150000.00 1 1001 A 100000.00 2 1003 B 65000.00 3

Things To Notice:

QUALIFY shows the ranking for the top 3 sales amounts only. When the order of sales is DESC, the highest sales amount is rank #1

Note, salestbl data is shown here.

SELECT * FROM salestbl ORDER BY sales DESC;

storeid prodid sales----------- ------ ----------- 1001 F 150000.00 1001 A 100000.00 1003 B 65000.00 1001 C 60000.00 1003 D 50000.00 1002 A 40000.00 1001 D 35000.00 1002 C 35000.00

1003 A 30000.00 1002 D 25000.00 1003 C 20000.00

Variations On a Ranking

The RANK function produces an ordered sequence. ORDER BY can override the normal sequencing.

Problem

Reverse the ranking sequence in the previous example.

Solution

SELECT storeid prodid ,sales ,RANK( ) OVER (ORDER BY sales DESC) AS "Ranking" FROM salestbl QUALIFY Ranking <= 3ORDER BY 4 DESC;

Result

storeid prodid sales Ranking----------- ------ ----------- ----------- 1003 B 65000.00 3 1001 A 100000.00 2 1001 F 150000.00 1

Things To Notice:

The ORDER BY clause in the SELECT statement may always be used to control the final order of the result set.

When the order of sales is DESC, the highest sales amount is always rank #1 After the ranking is applied, the output results are produced based on the ORDER BY

sequence.

The following is an alternate way to produce the same result.

SELECT storeid ,prodid ,sales ,RANK( ) OVER (ORDER BY sales DESC) AS "Ranking" FROM salestbl QUALIFY Ranking <= 3ORDER BY 3 ASC;

Result

storeid prodid sales Ranking----------- ------ ----------- ----------- 1003 B 65000.00 3 1001 A 100000.00 2 1001 F 150000.00 1

Things To Notice:

The ORDER BY clause in the SELECT statement uses the sales column for final ordering.

Ranking With PARTITION BY

The PARTITION BY clause may be used in conjunction with a RANK function to change the scope of the ranking. Without a PARTITION BY clause, the scope defaults to the RANK column. In the example on the preceding pages, the scope was the sales amount and rankings were done strictly based on that amount.

In the example seen next, the PARTITION BY clause adds the store id to the scope, thus rankings will be based upon sales within a store. Each store will have its top three ranking sales output.

Whereas the RANK( ) ORDER BY clause controls the default sort key, the PARTITION BY clause adds another level of sort to the output.

Problem

Get the top three selling products in each store.

Solution

SELECT storeid ,prodid ,sales ,RANK( ) OVER (PARTITION BY storeid

ORDER BY sales DESC) AS "Ranking"FROM salestblQUALIFY Ranking <= 3;

Result

storeid prodid sales Ranking----------- ------ ----------- ----------- 1001 F 150000.00 1 1001 A 100000.00 2 1001 C 60000.00 3 1002 A 40000.00 1

1002 C 35000.00 2 1002 D 25000.00 3 1003 B 65000.00 1 1003 D 50000.00 2 1003 A 30000.00 3

Things To Notice:

PARTITION BY clause controls scope, i.e., rank sales within store. Without PARTITION BY, scope would default to sales only. QUALIFY the ranking for the top 3 sales amounts per store. Sort sequence of output is sales descending. Due to PARTITION BY, sort is by sales (DESC) within store (ASC). Note -- no aggregation is done in this query. Aggregations are not permitted with ordered analytic functions.

Extended Ranking Scope

In the following example, an additional level of scope has been added via the PARTITION BY clause. Our scope is now sales by product within store. Because these three elements make up a ranking group and because the distinct elements store and product only yield a single row, there are no rankings below rank one. This is a case where the scope matches the granularity of the table, thus no reasonable ranking can be produced.

Whereas the RANK( ) ORDER BY clause controls the default sort key, the PARTITION BY clause adds another level of sort to the output.

Problem

Produce a ranking of sales by product within store.

Solution

SELECT storeid ,prodid ,sales ,RANK( ) OVER (PARTITION BY storeid, prodid

ORDER BY sales DESC) AS "Ranking"FROM salestblQUALIFY Ranking <= 3;

Result

storeid prodid sales Ranking----------- ------ ----------- ----------- 1001 A 100000.00 1 1001 C 60000.00 1 1001 D 35000.00 1

1001 F 150000.00 1 1002 A 40000.00 1 1002 C 35000.00 1 1002 D 25000.00 1 1003 A 30000.00 1 1003 B 65000.00 1 1003 C 20000.00 1 1003 D 50000.00 1

Things To Notice:

An additional PARTITION BY column is added. Scope is sales by product within store. Each product exists only once in each store. Thus, each row gets a rank of 1.

Note, the RANK function is covered in more detail in the Advanced SQL course.

The SAMPLE Function

The SAMPLE function is used to generate samples of data from a table or view. This may be done in one of two ways. SAMPLE n - where n is an integer will yield a sample of n rows, or if the number n is greater than the number of rows in the table, the sample will consist of the number of rows in the table. Rows are not reused within the same sample.

SAMPLE n - where n is a decimal value less than 1.00 and greater than .00

The SAMPLE function allows sampling of data based on:

A percentage of a table. An actual number of rows.

Problem

Get a sampling of 10 employee numbers.

SELECT department_number FROM employee SAMPLE 10;

Result

department_number----------------- 501 501 402

402 301 403 100 301 501 401

Things To Notice:

This is just the first 10 arbitrarily selected rows.

When a percentage is specified, a fraction of the rows is returned based on the decimal value. Note that calculations resulting in fractional rows must be greater than .4999 to actually return a row.

In the following example, 25% of the rows of the employee table are to be returned. The employee table has 26 rows.

26 * .25 = 6.50 = 7 rows in the sample

Example 2

Get a sampling of employee numbers using 25% of employees.

Solution

SELECT employee_numberFROM employeeSAMPLE .25ORDER BY 1;

Resultemployee_number--------------- 1005 1006 1011 1014 1018 1022 1024

Things To Notice:

7 rows out of 26 are returned. Fractional results greater than .4999 generate an added row. 25% of 26 = 6.5 which rounds to 7.

Using SAMPLEID

Multiple sample sets may be generated in a single query if desired. To identify the specific set, a tag called the SAMPLEID is made available for association with each set. The SAMPLEID may be selected, used for ordering, or used as a column in a new table.

Problem

Get three samples from the department table, one with 25% of the rows, another with 25% and a third with 50%.

SELECT department_number,sampleid

FROM departmentSAMPLE .25, .25, .50ORDER BY sampleid;

Result

department_number SampleId----------------- ----------- 301 1 403 1 402 2 201 2 100 3 501 3 302 3 401 3 600 3

Things To Notice:

Note that all 9 of the 9 rows of the department table are returned. This is due to the individual calculations.

Consider the following calculations

9 *.25 = 2.25 = 29 *.25 = 2.25 = 29 *.50 = 4.50 = 5 ----- 9

Example 2

Get three samples from the department table, one with 27% of the rows, another with 35% and a third with 2%.

Solution

SELECT department_number

,SAMPLEIDFROM departmentSAMPLE .27, .35, .02ORDER BY SAMPLEID;

Result

department_number SampleId----------------- ----------- 402 1 403 1 100 2 302 2 401 2

Things To Notice:

The first two samples are able to return rows. The last sample is too small to return a row. 25% of 26 = 6.5 which rounds to 7.

Example 3

Get three samples from the department table, one with 3 rows, another with 5 and a third with 8.

Solution

SELECT department_number,sampleid

FROM departmentSAMPLE 3, 5, 8ORDER BY sampleid;

*** Query completed. 9 rows found. 2 columns returned. *** Warning: 7473 Requested sample is larger than table rows.

Result

department_number SampleId----------------- ----------- 501 1 402 1 403 1 100 2 302 2 301 2 401 2 201 2 600 3

Things To Notice:

Because the rows are not repeated to different sample sets, the supply of rows is exhausted before the third set can be completed.

This results in a warning that there were not enough rows to populate all samples as requested.

This warning is seen in the BTEQ environment, but not in ODBC. The third sample gets the only remaining row.

Stratified Sampling

Stratified samplings permit differing sampling criteria to be applied to different sets of rows, all within a single query.

Problem

Get a sample which conforms to the following requirements:

33% of people earning less than $30,000 67% of people in the $30,001 - $40,000 range 40% of people in the $40,001 - $50,000 range

Solution

SELECT employee_number AS Empno ,salary_amount AS SalamtFROM employee SAMPLEWHEN salary_amount < 30000 THEN .33WHEN salary_amount BETWEEN 30001 AND 40000 THEN .67 WHEN salary_amount BETWEEN 40001 AND 50000 THEN .40ENDORDER BY 2;

Result

Empno Salamt----------- ------------ 1014 24500.00 1023 26500.00 1009 31000.00 1005 31200.00 1022 32300.00 1025 34700.00 1004 36300.00 1012 37900.00 1024 43700.00 1010 46000.00

Things To Notice:

If no rows can support any stratified requirement then:

No rows will be returned. No warning will be generated.

System Variables

The following system variables are available for use in creating SQL queries.

SESSION - contains the session-id of the requesting session/user.DATABASE

- contains the current default database of the requesting session/user.

ACCOUNT

- contains the user account info of the requesting session/user.

Each is defined internally as a VARCHAR(30).

Problem

Return the current session-id, default database and account string of the current user.

Solution

SELECT SESSION, DATABASE, ACCOUNT;

Result

Session Database Account----------- ------------------------------ ---------- 30788 PED $MUSER

The default database setting may change during the course of a session. These built-in functions allow you to keep track of your current database.

Problem

Show the current database both before and after it has been changed.

Solutions

SELECT DATABASE; /* Show the current database for this user session*/

Database------------------------------PED

DATABASE cs_views; /* Change the default database for this user session */

SELECT DATABASE; /* Show the new current database for this user session*/Database------------------------------cs_views

TOP N Overview

The TOP N function provides the ability to produce:

a. The top (or bottom) N rows of results based on specified criteria. b. The top (or bottom) N percentage of rows based on specified criteria. c. These results either with or without ties (more than one row has identical criteria

values). d. Sample rows arbitrarily without regard to rankings or order.

This improvement has been achieved by implementing the following new syntax to the SQL lexicon:

TOP {decimal | integer} [PERCENT] [WITH TIES]

The following meaning is associated with these options:

TOP 10 – Return the top ten rows according to criteria TOP 15 PERCENT – Return the top 15% of rows according to criteria TOP 10 WITH TIES – If more that one row has the same criteria value, return all TOP 15 PERCENT WITH TIES – If more than one row has the same criteria value,

return all

Show all department budgets ordered by budget amount descending. To answer this query, the TOP N feature is not needed.

SELECT department_number , budget_amount FROM department ORDER BY 2 DESC; department_number budget_amount ----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 402 308000.00 501 308000.00 201 293800.00 302 226000.00 600 ?

Note: Null sorts following the valid budget amounts.

TOP N With and Without TIES

Show the top five budget amounts.

SELECT TOP 5 department_number , budget_amount FROM department ORDER BY 2 DESC; department_number budget_amount ----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 501 308000.00

Things to notice:

ORDER BY defines the sequencing of the result set. It therefore defines the ranking criteria. To get the TOP highest amounts, you must use ORDER with DESC. TOP N where N is an integer up to 18 digits in length.

SELECT TOP 5 WITH TIESdepartment_number, budget_amountFROM department ORDER BY 2 DESC;department_number budget_amount----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 501 308000.00 402 308000.00

Things to notice:

Even though TOP 5 is specified, six rows are returned. Because there is a tie for the fifth position, both rows are returned. This only occurs when WITH TIES is specified. WITH TIES returns multiple tied rows when there is a tie for the 'last' position. It will return all rows containing the 'tied' value, but it will only count it as one row. Tied rows which are not in the last position, are each counted separately toward the

N total.

The same result could have been retuned by specifying TOP 6 without the WITH TIES option.

SELECT TOP 6department_number

, budget_amountFROM department ORDER BY 2 DESC; department_number budget_amount----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 501 308000.00 402 308000.00

Things to notice:

The last two rows have the same amount. Each is counted as a separate row for the top six. By default, rows with the same amount are counted as separate rows toward the

total. The same result would be achieved if the WITH TIES option had been specified.

Add the WITH TIES option and note any difference in the result set.

SELECT TOP 6 WITH TIESdepartment_number, budget_amountFROM department ORDER BY 2 DESC; department_number budget_amount----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 501 308000.00 402 308000.00

Things to notice:

There is no difference. WITH TIES has no effect because the top six rows are requested. If a seventh row existed with the same amount as the sixth, it would be returned. Note, that the same result is returned if we change to count to TOP 5.

SELECT TOP 5 WITH TIESdepartment_number, budget_amountFROM department ORDER BY 2 DESC; department_number budget_amount----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00

501 308000.00 402 308000.00

Things to notice:

The same result set is returned whether TOP 5 WITH TIES or TOP 6 is specified. In this case, there is a tie for the 5th row, hence both rows are returned. If a seventh row existed with the same amount as the fifth and sixth, it would also

be returned.

Getting Bottom Results

Now, consider getting the rows at the bottom of an ordered list. This is accomplished by using the ORDER BY clause.

Show the bottom three employees by salary.

SELECT TOP 3 employee_number, salary_amountFROM employee ORDER BY salary_amount ASC;employee_number salary_amount--------------- ------------- 1014 24500.00 1013 24500.00 1001 25525.00

Things to notice:

ORDER BY ASC reverses the ranking sequence and shows the bottom rankings. Two rows with the same salary are treated as two rows of output.

What would you expect to happen if you add the WITH TIES option?

SELECT TOP 3 WITH TIESemployee_number, salary_amountFROM employee ORDER BY salary_amount ASC;employee_number salary_amount--------------- ------------- 1014 24500.00 1013 24500.00 1001 25525.00

Things to notice:

The WITH TIES option has no effect. Ties must occur in the last row returned to have any effect.

If another row tied with the last row (25525.00), a fourth row would be returned.

Add another 'tie row'.

INSERT INTO employeeVALUES (1026,801,401,201401,'Barnes','John',941230,800315,25525.00);

Get the bottom three employees by salary.

SELECT TOP 3employee_number, salary_amountFROM employee ORDER BY salary_amount ASC;employee_number salary_amount--------------- ------------- 1013 24500.00 1014 24500.00 1026 25525.00

Now add the WITH TIES option. Do you expect it to make a difference this time?

SELECT TOP 3 WITH TIESemployee_number, salary_amountFROM employee ORDER BY salary_amount ASC;employee_number salary_amount--------------- ------------- 1014 24500.00 1013 24500.00 1026 25525.00 1001 25525.00

Things to notice:

The WITH TIES option causes a fourth row to be returned. Ties must occur in the last row returned to have any effect. A tie with the last row (25525.00) returns a fourth row.

Selecting Random Rows

The TOP N function may also be used to return unordered rows. This is accomplished by using TOP N without including an ORDER BY clause. In the example seen here, two rows are requested. Since an unordered set cannot produce the top n rows, the first n rows returned from the database are selected. A second execution of the same query will produce the same two rows, assuming that both the table and the configuration of the system have not changed. This is because the database will return the rows in a consistent sequence and always take the first (or top) n specified.

If a truly random sample is required, it is better to use the SAMPLE function which has more

randomized routines available.

Select two random rows.

SELECT TOP 2 employee_number, salary_amountFROM employee;

employee_number salary_amount--------------- ------------- 1008 29250.00 1012 37900.00

Things to notice:

Absence of an ORDER BY clause means rows are returned without regard to ranking. This is similar to using the SAMPLE function, The SAMPLE function however, produces a more truly randomized result set.

Adding The WITH TIES Option

Add the WITH TIES option to the preceding query. What effect is seen?

SELECT TOP 2 WITH TIESemployee_number, salary_amountFROM employee;

employee_number salary_amount--------------- ------------- 1008 29250.00 1012 37900.00

Things to notice:

Any two rows may be returned by this query, since no ranking criteria is defined. Without an ORDER BY clause, the WITH TIES clause is ignored. This is similar to using the SAMPLE function, however the SAMPLE function

produces a more truly randomized result set.

Compared To BTEQ RETLIMIT

Previously, to return less than a full answer set, a user could use the .SET RETLIMIT N option in BTEQ to control the number of rows returned to the client side. Since the DBS must still generate all the result rows, the performance of this approach could vary dramatically depending on the number of rows returned by the SQL request. With the TOP N option, only the rows needed to satisfy the query requirement are fetched by the DBS, thus making it a better choice when a number of random rows are to be selected. The SAMPLE function continues to be the preferred approach for truly random samples.

Using the PERCENT Option

The TOP N function can also produce a percentage of rows in addition to an absolute number of rows.

Return employees whose salaries represent the top ten percent.

SELECT TOP 10 PERCENTemployee_number,salary_amountFROM employee ORDER BY salary_amount DESC;employee_number salary_amount--------------- ------------- 801 100000.00 1017 66000.00 1019 57700.00

Things to notice:

10% of 26 rows is 2.6 rows rounded to 3. PERCENT must be a number between 0 and 100. At least one row is always returned (if there is at least one row in the table). A percentage resulting in a fractional number of rows is always rounded up: 10% of 6 rows = .6 rows = 1 row output 20% of 6 rows = 1.2 rows = 2 rows output 30% of 6 rows = 1.8 rows = 2 rows output

SELECT TOP 45 PERCENT WITH TIESdepartment_number, budget_amountFROM department ORDER BY 2 DESC;department_number budget_amount----------------- ------------- 401 982300.00 403 932000.00 301 465600.00 100 400000.00 501 308000.00 402 308000.00

Things to notice:

45% of 9 rows is 4.05 rows rounded to 5. Because there is a tie for the fifth position, six rows are returned.

Notes About Using PERCENT:If PERCENT is specified, the value of N must be a number greater than 0 and less than or equal to 100. Fractional portions are permitted.

If PERCENT is not specified, the value of N may be up to the maximum decimal value (18

digits), however a fractional portion may not be specified.

Lab


To start the online labs, click on the Telnet button in the lower left hand screen of the course.Two windows will pop-up: a BTEQ Instruction Screen and your Telnet Window. Sometimes the BTEQ Instructions get hidden behind the Telnet Window. You will need these instructions to log on to Teradata.


A. Rank the top three departments in the ‘department’ table based on their budget_amount. Show department number, budget amount and rank.

B. Create a report showing the store id, product id, sales amount from the ‘salestbl’ and add a sales ranking by product id.

C. Modify Lab B to show the same report but limit it to each product id’s number one ranking i.e., eliminate all but the number one rankings.

D. Select the following information about your session using built-in functions:User NameSession IdCurrent DatabaseAccount Id

Hint: Use the following settings to produce the output:.SET FOLDLINE ON;.SET SIDETITLES ON;When the report has been created, reset the settings:.SET FOLDLINE OFF;.SET SIDETITLES OFF;

E. Create a table using the following description:

CREATE TABLE sqlxxx.empsamp( empno INTEGER, deptno INTEGER, job INTEGER, sampid BYTEINT)UNIQUE PRIMARY INDEX ( empno );

Populate this table with three samples from the employee table, identifiable by SAMPLEID 1, 2 and 3. Let each sample contain 33% of the rows of the table. Select all columns against the newly populated table and order by results by SAMPLEID.

F. Delete the rows from empsamp and repopulate the table with three samples of 15 rows each. Select the resulting table and order by sampid to see the results of this sampling.

G. Using the TOP N function, get the top three daily sales amounts for the month of

August 2004 from the daily_sales_2004 table. Also retrieve the itemid and salesdate.

H. Perform the same query using the WITH TIES option. Is there a difference in the output?

I. Produce the same report as the previous lab, but add a column showing the ranking of each row. Tied amounts should show the same rank number.

Lab Answer A

SELECT department_number, budget_amount ,RANK( ) OVER (ORDER BY Budget_amount DESC)

FROM departmentQUALIFY RANK(budget_amount) <=3;

*** Help information returned. 3 rows.department_number budget_amount Rank(budget_amount)----------------- ------------- ------------------- 401 982300.00 1 403 932000.00 2 301 465600.00 3

Lab Answer B

SELECT storeid, prodid, sales,RANK() OVER (PARTITION BY prodid ORDER BY sales DESC) AS

Rank_SalesFROM salestbl;

*** Query compeleted. 11 rows found. 4 comlumns returned.

storeid prodid sales Rank_Sales----------- ------ ----------- ----------- 1001 A 100000.00 1 1002 A 40000.00 2 1003 A 30000.00 3 1003 B 65000.00 1 1001 C 60000.00 1 1002 C 35000.00 2 1003 C 20000.00 3 1003 D 50000.00 1 1001 D 35000.00 2 1002 D 25000.00 3 1001 F 150000.00 1

Lab Answer C

SELECT storeid, prodid, sales,RANK() OVER (PARTITION BY prodid ORDER BY sales DESC) AS

Rank_SalesFROM salestblQUALIFY rank_sales < 2;

*** Query compeleted. 5 rows found. 4 comlumns returned.

storeid prodid sales Rank_Sales----------- ------ ----------- ----------- 1001 A 100000.00 1 1003 B 65000.00 1 1001 C 60000.00 1 1003 D 50000.00 1 1001 F 150000.00 1

Lab Answer D

.SET FOLDLINE ON;

.SET SIDETITLES ON;SELECT USER, SESSION, DATABASE, ACCOUNT;

*** Query completed. One row found. 4 columns returned.*** Total elapsed time was 1 second.

User SQL00 Session 105234 Database CS_VIEWS Account $M_P0623.SET FOLDLINE OFF;.SET SIDETITLES OFF;

Lab Answer E

CREATE TABLE sql00.empsamp ( empno INTEGER, deptno INTEGER, job INTEGER, sampid BYTEINT)

UNIQUE PRIMARY INDEX ( empno );

INSERT INTO SQL00.empsamp SELECT employee_number, department_number, job_code, SAMPLEID FROM employee SAMPLE .33, .33, .33;

*** Insert completed. 26 rows added.*** Total elapsed time was 1 second.

SELECT * FROM SQL00.empsamp ORDER BY sampid, empno;


empno deptno job sampid----------- ----------- ----------- ------ 1002 401 413201 1 1011 402 421100 1 1015 501 512101 1 1017 501 511100 1 1018 501 512101 1 1019 301 311100 1 1023 501 512101 1 1024 403 432101 1 1025 201 211100 1 801 100 111100 2 1004 401 412101 2 1006 301 312101 2 1007 403 432101 2 1009 403 432101 2 1012 403 432101 2 1013 401 412102 2 1016 302 321100 2 1022 401 412102 2 1001 401 412101 3 1003 401 411100 3 1005 403 431100 3 1008 301 312102 3 1010 401 412101 3 1014 402 422101 3 1020 403 432101 3 1021 201 222101 3 (Note: Results in individual rows may vary)

Lab Answer F

DELETE FROM SQL00.empsamp;INSERT INTO SQL00.empsamp SELECT employee_number, department_number, job_code, SAMPLEID FROM employee SAMPLE 15, 15, 15;***Insert completed. 26 rows added. ***Warning: 7473 Requested sample is larger than table rows. All rows returnedSELECT * FROM SQL00.empsamp ORDER BY sampid, empno;*** Query completed. 26 rows found. 4 columns returned. *** Total elapsed time was 1 second. empno deptno job sampid----------- ----------- ----------- ------ 801 100 111100 1 1003 401 411100 1 1004 401 412101 1 1005 403 431100 1 1007 403 432101 1 1008 301 312102 1 1010 401 412101 1 1012 403 432101 1 1013 401 412102 1 1018 501 512101 1 1019 301 311100 1 1020 403 432101 1 1022 401 412102 1 1024 403 432101 1 1025 201 211100 1 1001 401 412101 2 1002 401 413201 2 1006 301 312101 2 1009 403 432101 2 1011 402 421100 2 1014 402 422101 2 1015 501 512101 2 1016 302 321100 2 1017 501 511100 2 1021 201 222101 2 1023 501 512101 2 (Note: Results in individual rows may vary)

Lab Answer G

SELECT TOP 3 itemid, salesdate, salesFROM daily_sales_2004WHERE salesdate BETWEEN DATE '2004-08-01' AND DATE '2004-08-31'ORDER BY 3 DESC; itemid salesdate sales----------- ---------- ----------- 10 2004-08-10 550.00

10 2004-08-05 350.00 10 2004-08-03 250.00

Lab Anwer H

SELECT TOP 3 WITH TIES itemid, salesdate, salesFROM daily_sales_2004WHERE salesdate BETWEEN DATE '2004-08-01' AND DATE '2004-08-31'ORDER BY 3 DESC;

itemid salesdate sales----------- ---------- ----------- 10 2004-08-10 550.00 10 2004-08-05 350.00 10 2004-08-03 250.00 10 2004-08-04 250.00

Lab Answer I

SELECT TOP 3 WITH TIES itemid, salesdate, sales,RANK() OVER (ORDER BY sales DESC) AS RankingFROM daily_sales_2004WHERE salesdate BETWEEN DATE '2004-08-01' AND DATE '2004-08-31'ORDER BY 3 DESC;

itemid salesdate sales Ranking----------- ---------- ----------- ----------- 10 2004-08-10 550.00 1 10 2004-08-05 350.00 2 10 2004-08-03 250.00 3 10 2004-08-04 250.00 3

Case Expressions Module:-20

Lab: Case Expression


To start the online labs, click on the Telnet button in the lower left hand screen of the course.Two windows will pop-up: a BTEQ Instruction Screen and your Telnet Window. Sometimes the BTEQ Instructions get hidden behind the Telnet Window. You will need these instructions to log on to Teradata.


A. Calculate the fraction of the total company budget represented by departments 401 and 403. Express the result as a percentage as follows:

B. 401/ 403C. Bgt RatioD. ---------

xx%

E. Calculate the fraction of the total company budget represented by departments 401 and 403 after department 403 has been given a 5% budget increase. Express the result as a decimal ratio as follows:

Dept_401_403_Bgt_Ratio------------------------ .xx

F. Create a budget report from the department table.Show the total, average, minimum and maximum budget amounts.Title the columns "Total", Avg", "Min" and "Max".

Do the query twice:

a.) once treating NULL values as zero, b.) once excluding NULL values in aggregates.

Compare the results.

G. Accounting wants to find out which way of slanting the statistics is most beneficial to the company. Do a departmental salary list of the total salaries for each department, and the average departmental salary three times:

- without any changes (treating NULLs as null and zero values as zero)- treating NULL values as zero- treating zero values as NULL

Also list the count of rows that are going into each averaging function. Sequence the report by department. Limit the department number and count columns to four digits each with zero suppress.

Title your columns as follows:

Dept Tot # Tot Sal Avg Sal ZIN # ZIN Avg NIZ # NIZ Avg

Are there differences in these computations? Why or why not?

Lab Answer A

DATABASE Customer_Service;SELECT CAST(SUM(

CASE department_numberWHEN 401 THEN budget_amountWHEN 403 THEN budget_amountELSE 0END) / SUM(budget_amount) * 100 as numeric(2,0)) (title '401/403//Bgt Ratio',FORMAT '99%')

FROM department;

401/403 Bgt Ratio 49%

Lab Answer B CASE Expressions

SELECT SUM(CASEWHEN department_number = 401

THEN budget_amountWHEN department_number = 403

THEN budget_amount * 1.05ELSE 0END) / SUM(CASEWHEN department_number = 403

THEN budget_amount*1.05ELSE budget_amountEND ) (FORMAT 'Z.99')

AS Dept_401_403_Bgt_Ratio FROM department;


Dept_401_403_Bgt_Ratio .50

Lab Answer C: CASE Expressions

This solution is for Option a) - treating NULL values as zero

SELECT SUM (COALESCE (budget_amount,0))(DEC(9,2)) (TITLE 'Total'),AVG (COALESCE (budget_amount,0))(DEC(9,2)) (TITLE 'Avg'),MIN (COALESCE (budget_amount,0))(DEC(9,2)) (TITLE 'Min'),MAX (COALESCE (budget_amount,0))(DEC(9,2)) (TITLE 'Max')

FROM department;;

*** Query completed. One row found. 4 columns returned. Total Avg Min Max

3915700.00 435077.78 .00 982300.00

This Solution is for Option b) - not including NULL values in aggregates.

SELECT SUM(budget_amount) (TITLE 'Total'),AVG(budget_amount) (TITLE 'Avg'),MIN(budget_amount) (TITLE 'Min'),MAX(budget_amount) (TITLE 'Max')

FROM department;

*** Query completed. One row found. 4 columns returned. Total Avg Min Max

3915700.00 489462.50 226000.00 982300.00

Lab Answer D: CASE Expressions

SELECT department_number (TITLE 'Dept' )(FORMAT 'ZZZ9'),COUNT(salary_amount) (TITLE 'Tot #' )(FORMAT 'ZZZ9' ),SUM(salary_amount) (TITLE 'Tot Sal' ),AVG(salary_amount) (TITLE 'Avg Sal' ),COUNT(COALESCE(salary_amount,0))(TITLE 'ZIN #' )(FORMAT

'ZZZ9' ),AVG(COALESCE(salary_amount,0)) (TITLE 'ZIN Avg' )(FORMAT

'ZZZZZZ.99'),COUNT(NULLIF(salary_amount,0)) (TITLE 'NIZ #')(FORMAT

'ZZZ9' ),AVG(NULLIF(salary_amount,0)) (TITLE 'NIZ Avg' )

FROM employeeGROUP BY 1ORDER BY 1;

*** Query completed. 8 rows found. 8 columns returned.Dept Tot# Tot Sal Avg Sal ZIN# ZIN Avg NIZ# NIZ Avg----- ---- ---------- --------- ---- ---------- ---- ----------100 1 100000.00 100000.00 1 100000.00 1 100000.00201 2 73450.00 36725.00 2 36725.00 2 36725.00301 3 116400.00 38800.00 3 38800.00 3 38800.00302 1 56500.00 56500.00 1 56500.00 1 56500.00401 7 245575.00 35082.14 7 35082.14 7 35082.14402 2 77000.00 38500.00 2 38500.00 2 38500.00403 6 233000.00 38833.33 6 38833.33 6 38833.33501 4 200125.00 50031.25 4 50031.25 4 50031.25

Views Module:-20 Objectives


Create a view. Use a view to provide secure access to data. Drop or modify a view. List several reasons for using views.

What is a View?

A view functions like a 'window' into a table. It allows users to customized their access to base tables by:

Restricting which columns are visible from the base table. Restricting which rows are visible from the base table. Combining columns and rows from several base tables.

Restrict Columns from the Base Table

One can restrict columns from the base table(s) by explicitly listing the desired column names from the base table(s) in the view definition.

Restrict Rows from the Base Table

One can restrict which rows may be accessed by using the WHERE clause in the view definition. This will limit the rows returned from a SELECT statement referencing the view.

Creating and Using Views

Problem

Create a view of the employees in department 403 to use for both read and update. Limit the view to an employee’s number, last name, and salary.

Solution

CREATE VIEW emp_403 AS SELECT employee_number ,last_name ,salary_amount FROM employee WHERE department_number = 403;

To Read From This View

SELECT * FROM emp_403;

Result

employee_number last_name salary_amount--------------- ------------- --------------- 1009 Lombardo 31000.00 1007 Villegas 49700.00 1012 Hopkins 37900.00 1005 Ryan 31200.00 1024 Brown 43700.00 1020 Charles 39500.00

To Update Through This View

UPDATE emp_403 SET salary_amount = 100000 WHERE last_name = 'Villegas' ;

What is a Join View?

A 'join view' consists of columns from more than one table.

Example

Create a join view of the employee and call_employee tables for call dispatch use.

CREATE VIEW employee_call AS SELECT employee.employee_number ,last_name ,first_name ,call_number ,call_status_code ,assigned_date ,assigned_time ,finished_date ,finished_time FROM employee INNER JOIN call_employee ON call_employee.employee_number = employee.employee_number;

Using this view permits SELECT of information from both tables. Note that column names which appear in both tables (i.e., employee_number) must be qualified.

Problem

Who is employee 1002 and what calls has he answered?

Solution

SELECT last_name ,first_name ,call_number FROM employee_call WHERE employee_number = 1002;

Result

last_name first_name call_number

Brown Alan 9

Renaming Columns

You may create a view that renames columns. This is often used to shorten long column names, making queries easier to create.

Example

CREATE VIEW shortcut (emp, dept, last, first, sal) AS SELECT employee_number ,department_number ,last_name ,first_name ,salary_amount FROM employee WHERE department_number = 201;

Now you may SELECT rows using this view. SELECT last ,sal shortcut ORDER BY last DESC;

Result

last sal ------------ ---------- Short 38750.00 Morrissey 34700.00

To UPDATE rows using this View:

UPDATE shortcut SET sal = 100000.00 WHERE emp = 1019;

Replacing a View

You may replace a view in order to include or drop specfic columns or values.

Problem

Change the shortcut view to include only department 301.

Solution

First, display the current view definition:

SHOW VIEW shortcut;

You should see the following view definition:

CREATE VIEW shortcut (emp, dept, last, first, sal) AS

SELECT employee_number ,department_number ,last_name ,first_name ,salary_amountFROM employeeWHERE department_number = 201;

Copy and paste the view definition to a text editor. Change the CREATE keyword to REPLACE and make the desired changes.

--> REPLACE VIEW shortcut (emp, dept, last, first, sal) AS SELECT employee_number ,department_number ,last_name ,first_name ,salary_amount FROM employee--> WHERE department_number = 301;

Using Views For Formatting and Titles

The FORMAT clause may be used to set formats and the AS clause may be used to assign new column names. For example, let's create a view for department 201 with shortened titles and appropriate formatting. Also, show the salary amount as a monthly figure.

CREATE VIEW report AS SELECT employee_number (FORMAT '9999') AS Emp ,department_number (FORMAT '999') AS Dept ,last_name (TITLE 'Name') AS Last ,first_name (TITLE ' ') AS First ,salary_amount / 12 (FORMAT '$$$,$$9.99') AS Monthly_SalaryFROM employeeWHERE department_number = 201;

To SELECT rows using this View:

SELECT * FROM report ORDER BY monthly_salary; Emp Dept Name Monthly_Salary---- ---- ------------------ ------------------ --------------1025 201 Short Michael $2,891.671021 201 Morrissey Jim $3,229.17

Aggregate Views

You may create a view that summarizes information by using aggregates. For instance,

let's create a view that summarizes salary information by department.

CREATE VIEW deptsalsAS SELECT department_number AS department

,SUM (salary_amount)AS salary_total,AVG (salary_amount)AS salary_average,MAX (salary_amount)AS salary_max,MIN (salary_amount)AS salary_min

FROM employeeGROUP BY department_number;

Note: Aggregate and derived columns must be assigned names in the view definition.

Select the average salary for all departments using the view we created:

SELECT department

,salary_averageFROM deptsals; department salary_average100 100000.00201 36725.00403 38833.33... ...401 35082.14

Aggregate View Using HAVING

The HAVING clause may be used within a view definition to restrict which groups will participate in the view.

We may modify the deptsals view to include only departments with average salaries of less than $36,000.

REPLACE VIEW deptsals AS SELECT department_number AS department ,SUM(salary_amount) AS salary_total ,AVG(salary_amount) AS salary_average ,MAX(salary_amount) AS salary_max ,MIN(salary_amount) AS salary_minFROM employeeGROUP BY department_numberHAVING AVG(salary_amount) < 36000;

To SELECT all departments with average salaries less than $36,000, we may use the view deptsals as follows:

SELECT department ,salary_averageFROM deptsals;

department salary_average ------------- ----------------- 401 35082.14

Joins on Views with Aggregates Since views are really virtual tables, they may be joined with other tables and/or views in a query.

Problem

First, create a view showing the the aggregate salaries by department number.

Solution

CREATE VIEW dept_salaries (deptnum, salaries)ASSELECT department_number, SUM(salary_amount)FROM employeeGROUP BY 1;

Use the View to SELECT all rows:

SELECT * FROM dept_salaries;

Results

deptnum salaries----------- ------------ 100 100000.00 201 73450.00 301 116400.00 302 56500.00 401 245575.00 402 77000.00 403 233000.00 501 200125.00

Now, let's look at another example, this time requiring a join.

Problem

Show by department name which departments have aggregate salaries greater than $100,000.

Solution

SELECT deptnum

,department_name ,salaries

FROM dept_salaries ds INNER JOIN department deON ds.deptnum = de.department_numberWHERE salaries > 100000ORDER BY 1;

Result

deptnum department_name salaries----------- ------------------------------ ------------ 301 research and development 116400.00 401 customer support 245575.00 403 education 233000.00

501 marketing sales 200125.00

Nested Aggregates An aggregate result may be obtained from a view which has aggregation defined in its creation. In the following example, an aggregated view is created.

CREATE VIEW dept_sals AS SELECT department_number AS DEPT

,SUM (salary_amount) AS SumSalFROM employeeGROUP BY dept;

SELECT all rows from the view:

SELECT * FROM dept_salsORDER BY 1;

Result

dept SumSal ----- --------- 100 100000.00 201 73450.00 301 116400.00 302 56500.00 401 245575.00 402 77000.00 403 233000.00

501 201125.00

You may apply aggregate operators in your SELECT statement against the aggregate view you just created. This represents aggregate functions being applied to an aggregated view or what

is referred to as ' nested aggregates'.

SELECT COUNT(Sumsal) AS #_Depts,SUM(Sumsal) AS Total_Sals,AVG(Sumsal) AS Avg_Dept_Tot,MIN(Sumsal) AS Min_Dept_Tot,MAX(Sumsal) AS Max_Dept_Tot

FROM dept_sals;

Result

#-Depts Total_Sals Avg_Dept_Tot Min_Dept_Tot Max_Dept_Tot------- ---------- ------------ ------------ ------------ 8 1102050.00 137756.25 56500.00 245575.00

Nested Aggregates in a WHERE Clause Nested aggregates in a WHERE clause may be used to limit response rows.

Repeated for convenience:

CREATE VIEW dept_sals AS SELECT department_number AS DEPT,SUM (salary_amount) AS SumSalFROM employeeGROUP BY dept;

Problem

Show the department with the highest total salary. .

Solution

SELECT dept ,Sumsal as TopsalFROM dept_salsWHERE topsal = (SELECT MAX(Sumsal) FROM dept_sals);

Result

dept Topsal---- --------- 401 245575.00

Things to notice:

The aggregated view is used in both the main query and the subquery The subquery determines the maximum salary from the aggregate view.

The main query uses the maximum salary to capture the department number.

Views WITH CHECK OPTION

In a view, the WITH CHECK OPTION controls the update and insert capability of users with those privileges on the view. A user may not insert or update a row if the resulting row will fall outside of the view. Said differently, an insert or update will be denied if the resulting row will violate the constraints of the WHERE clause in the view definition.

For example, let's recreate a view and add a WHERE clause and a WITH CHECK OPTION:

REPLACE VIEW dept_budget ASSELECT department_number AS Dept ,department_name AS Name ,budget_amount AS Budget ,manager_employee_number AS MgrFROM department WHERE Budget <= 1000000 WITH CHECK OPTION;

With the WITH CHECK OPTION in place in a view, the following inserts and updates are affected:

New departments may not be inserted if the budget amount is above $1,000.000. A department budget may not be updated to be more than $1,000,000.

Changing Rows Via the View Users will not be able to update or insert rows whose values violate the constraints of the WHERE clause in the view definition.

Example 1

INSERT INTO dept_budget VALUES (601, 'Support', 2000000, 1018);

**** Failure 3564 Range constraint:Check error in field dept.budget_amount.

Example 2

UPDATE dept_budget SET budget = 2000000 WHERE dept = 401;

**** Failure 3564 Range constraint:Check error in field dept.budget_amount.

Both operations above were disallowed because the WITH CHECK OPTION prevents these changes.

Restrictions and Advantages With Views

Restrictions When Using Views

An index cannot be created on a view. A view cannot contain an ORDER BY clause. The WHERE clause of a SELECT against a view can reference all aggregated columns

of that view. Derived and aggregated columns must be assigned a name using one of the

following techniques: o NAMED clause. o AS clause (ANSI standard). o Creating a list of assigned column names in the CREATE VIEW definition.

A view cannot be used to UPDATE if it contains: o Data from more than one table (Join View). o The same column twice. o Derived columns. o A DISTINCT clause. o A GROUP BY clause.

Advantages When Using Views

An additional level of security. Help in controlling read and update privileges. Simplify end-user's access to data. Are unaffected if a column is added to a table; and Are unaffected if a column is dropped, unless the dropped column is referenced by

the view.

Suggestions For Using Views

Create at least one view for each base table to restrict user's access to data they don't need to see.

Create query views as you need them for ad hoc situations. Use access locks when creating views to maximize data availability to users.

Lab: VIEW




Hint: Prior to doing these labs, it will be helpful to reset your default database to

your own user id (i.e. DATABASE tdxxx;).

A. List the customer number, customer name, and sales representative's number for customers in California. Sort by sales representative's number.

HINT: The location table contains both the customer number and the state.

Now modify the request to use a join instead of a subquery.

Now create a VIEW called "sales_territory".

Do not create it as part of a macro. Is data in this view updatable?

B. Select all columns from the sales_territory view.C. Create an aggregate view called "dept_sal" which accesses the

customer_service.employee table and consists of two columns: a department number and the sum of the salaries for that department. Select all rows from this view.

D. Using the ‘dept_sal’ view, produce a report showing department name and total departmental salary for each department whose total exceeds $100,000.

Remember: Like tables and macros, views must be created in your own database.

E. Create a full view of any table in the customer_service database. Include formatting and titling of your own choosing. Select all rows and columns from your view to observe how the view overrides default formats and column headings.

Lab Answer A: Views

DATABASE tdxxx;CREATE VIEW sales_territoryAS

SELECT l.customer_number,c.customer_name,c.sales_employee_numberFROM Customer_Service.customer c,Customer_Service.location lWHERE c.customer_number = l.customer_numberAND l.state = 'California'

;

*** View has been created.

QUESTION: Is data in this VIEW updatable? ANSWER: No, it cannot be updatable when there is more than one table accessed.

Lab Answer B: Views

SELECT *FROM sales_territory;


customer_number customer_name sales_employee_number --------------- --------------- --------------------- 15 Cates Modeling 1015 11 Hotel California 1015 19 More Data Enterprise 1015 0 Corporate Headquarters ? 2 Simple Instruments Co. 1015 1 A to Z Communications, Inc. 1015 10 Graduates Job Service 1015

Lab Answer C: Views

CREATE VIEW dept_sal (dep, sumsal)ASSELECT department_number

,SUM(salary_amount)FROM Customer_Service.employeeGROUP BY 1;

SELECT * FROM dept_sal;

***Query completed. 8 rows found. 2 columns returned.

Dep Sumsal ------ ------------ 501 200125.00 301 116400.00 201 73450.00 100 100000.00 402 77000.00 403 233000.00 302 56500.00 401 245575.00

Lab Answer D: Views

SELECT department_name, sumsal

FROM Customer_Service.department dINNER JOIN dept_sal dsON ds.dep = d.department_numberWHERE sumsal > 100000;


department_name sumsal --------------- ------------ marketing sales 200125.00 research and development 116400.00 education 233000.00 customer support 245575.00

Lab Answer E: Views

CREATE VIEW cust ASSELECT customer_number(FORMAT '9(4)' )(TITLE 'Cust#' ),customer_name (TITLE 'Name' ),parent_customer_number (FORMAT '9(4)' )(TITLE 'Parent' ),sales_employee_number (FORMAT '9(4)' )(TITLE 'Rep' )FROM Customer_Service.customer

;

*** View has been created. SELECT * FROM cust ;*** Query completed. 21 rows found. 4 columns returned.

Cust# Name Parent Rep ------ ------------------------ ------- ----- 0015 Cates Modeling ? 1015 0013 First American Bank 0003 1023 0019 More Data Enterprise 0009 1015 0000 Corporate Headquarters ? ? 0018 Wall Street Connection ? 1023 0011 Hotel California ? 1015 0005 Federal Bureau of Rules ? 1018 0017 East Coast Dating Service ? 1023 0016 VIP Investments 0003 1023 0014 Metro Savings ? 1018 0003 First American Bank ? 1023 0009 More Data Enterprise ? 1023 0002 Simple Instruments Co. 0001 1015 0008 Colby Co. ? 1018 0001 A to Z Communications Inc. ? 1015 0007 Cream of the Crop ? 1018 0006 Liberty Tours ? 1023 0020 Metro Savings 0014 1018 0004 Sum Bank 0003 1023 0012 Cheap Rentals ? 1018 0010 Graduates Job Service ? 1015Note: Solutions will vary depending on the table chosen.

String Functions Module:-21

Objectives


Extract strings using SUBSTRING. Combine strings using concatenation. Locate the starting position of substrings.

String Expressions

Several functions are available for working with strings in SQL. Also, the concatenation operator is provided for combining strings. The string functions and the concatenation operator are listed here.

String Operator Description

|| (Two vertical bars) Concatenates (combines) character strings together.

String Functions

SUBSTR Obtains a section of a character string. (Teradata extension)

SUBSTRING Obtains a section of a character string. (ANSI standard)

INDEX Locates a character position in a string. (Teradata extension)

POSITION* Locates a character position in a string. (ANSI standard)

TRIM* Trims blanks from a string.

UPPER* Converts a string to uppercase.

LOWER* Converts a string to lowercase.

All functions listed are part of the ANSI standard except for INDEX which has an ANSI equivalent POSITION function.

*These topics are introduced elsewhere in the SQL course sequence.

POSITION, UPPER and LOWER are covered in more detail in the Teradata SQL For Application Development course.

The TRIM function was covered in a previous module of this course.

SUBSTRING Function

The SUBSTRING function is intermediate-level ANSI compliant. Teradata continues to support the alternate SUBSTR syntax as well.

Examples

SELECT SUBSTRING ('catalog' FROM 5 for 3);

Result

'log'SELECT SUBSTR ('catalog',5,3);

Result

'log'

Let's look at several more examples showing the results of a SELECT using either SUBSTRING or SUBSTR . Equivalent results are returned in all cases.

SUBSTRING Result

SUBSTR Result

SUBSTRING(‘catalog’ FROM 5 FOR 4) ‘log’ ‘log’

SUBSTRING(‘catalog’ FROM 0 FOR 3) ‘ca’ ‘ca’

SUBSTRING(‘catalog’ FROM -1 FOR 3) ‘c’ ‘c’

SUBSTRING(‘catalog’ FROM 8 FOR 3) 0 length string 0 length string

SUBSTRING(‘catalog’ FROM 1 FOR 0) 0 length string 0 length string

SUBSTRING(‘catalog’ FROM 5 FOR -2) error error

(without a FOR clause)

SUBSTRING(‘catalog’ FROM 0) ‘catalog’ ‘catalog’

SUBSTRING(‘catalog’ FROM 10) 0 length string 0 length string

SUBSTRING(‘catalog’ FROM -1) 0 length string 0 length string

SUBSTRING(‘catalog’ FROM 3) ‘talog’ ‘talog’

Since the Teradata extension provides equivalent functionality to the ANSI-compliant feature, we recommend that all existing applications replace any use of the SUBSTR function with the SUBSTRING function.

Using a SUBSTRING in a List

Problem

Display the first initial and last name of all employees in department 403. Sort the result by last name.

Solution

SELECT SUBSTRING (first_name FROM 1 FOR 1) AS FI ,last_name

FROM employeeWHERE department_number = 403ORDER BY last_name;

Result

FI last_name-- --------------------A BrownJ CharlesP HopkinsD LombardoL RyanA Villegas

Using SUBSTRING with INTEGERS

A SUBSTRING function can also be performed on numeric data. Internally, the numeric data must be converted to a character string before the SUBSTRING function can occur. Because SUBSTRING relies on the exact position of each character, care must be taken when performing this function with numerics.

The following gives examples of how integers are converted when a SUBSTRING is performed on them.

These three column definitions are used in the example below:

Column Data Type and Conversion Size

no_of_ dependents

BYTEINT - (3 digits) 4 characters including sign

area_code SMALLINT - (5 digits) 6 characters including sign

phone INTEGER - (10 digits) 11 characters including sign

When an integer field is converted to CHAR, a sign character is added and the value is right-aligned as follows:

+ 1 2 7 (no_of_dependents BYTEINT) + 0 0 2 1 3 (area_code SMALLINT)+ 0 0 0 5 5 5 4 1 3 4 (phone INTEGER)

Problem

Produce a list of all 3-digit area codes from the location_phone table.

Solution

SELECT DISTINCT SUBSTRING (area_code FROM 4 FOR 3) AS AREA_CODE

FROM location_phoneORDER BY 1;

Result

AREA CODE202212213312313415609617718804919

Using SUBSTRING in a WHERE Clause

Problem

Find those customers in the location table whose nine-digit zip code ends with four zeros.

Solution

SELECT customer_number ,zip_code (FORMAT '9(9)')

FROM locationWHERE SUBSTRING (zip_code FROM 8 FOR 4) = '0000';

Result

customer number zip code 13 001210000 10 940710000

String Concatenation

Multiple strings may be combined using the string concatenation operator '||'. Only character and byte data types may be concatenated. (Byte data types must be concatenated with another byte data type.)

Use either solid (or broken) vertical bars || as a concatenation operator.

Example

SELECT 'abc' || 'xyz';

combines the two strings into a single character string.

Result

'abcxyz'

Concatenation and the COALESCE Function

Normally, concatenation of any string with a null produces a null result.

The COALESCE function allows values to be substituted for nulls.

Example

Assume col1 = 'a', col2 = 'b'

SELECT col1 || col2 From tblx;

Result

'ab'

If either column contains a null, the result is null.

Example

Assume col1 = 'a', col2 = null

SELECT col1 || (COALESCE (col2,'x')) FROM tblx;

Result

'ax'

Teradata Alternative

The Teradata database also supports the use of exclamation marks (!!) as the concatenation operator. Existing applications using exclamation marks should be re-written using vertical bars to meet ANSI standards.

Problem

Display the first and last names for employees in department 403.

Solution

SELECT first_name || ' ' || last_name AS EMPLOYEE FROM employee WHERE department_number = 403 ;

Result

EMPLOYEE Arnando Villegas Loretta Ryan

Things to notice:

The concatenation operators are used to place a space between first and last name. Even though two columns are concatenated together, it is treated as a single

column of output. There are no commas in this statement, since there is only one column of output.

String Concatenation with SUBSTRING and TRIM

SUBSTRING and concatenation may be used together in a query.

Problem

Display the first initial and last name of employees in department 403.

Solution

SELECT SUBSTRING (first_name FROM 1 FOR 1 ) || '. ' || last_name AS EMPLOYEE FROM employee WHERE department_number = 403 ;

Result

EMPLOYEE-----------------------D. LombardoA. VillegasP. HopkinsL. RyanA. BrownJ. Charles

Problem

Produce the same list, but concatenate last name followed by first name separated by a comma.

Solution

SELECT last_name || ', ' || first_name AS EMPLOYEE FROM employee WHERE department_number = 403 ;

Result

EMPLOYEE-------------------------------------Lombardo , DomingusVillegas , ArnandoHopkins , PauleneRyan , LorettaBrown , AllenCharles , John

Things to notice:

The column last_name is defined as a fixed length CHAR(20).

Consequently, there will be a number of spaces following each last name. This can be corrected with the use of the TRIM function.

Use the TRIM function to remove the extra spaces in your query result.

Problem

Drop the spaces after the last names prior to concatenating with the first name.

Solution

SELECT TRIM (last_name) || ', ' || first_name AS EMPLOYEE FROM employee WHERE department_number = 403 ;

Result

EMPLOYEE-------------------Lombardo, DomingusVillegas, ArnandoHopkins, PauleneRyan, LorettaBrown, AllenCharles, John

The POSITION Function

The POSITION function finds the position within a string where a specified substring begins. The ANSI standard POSITION function has a Teradata equivalent function called INDEX. The INDEX function will continue to be supported in future releases of the product, however the use of POSITION is encouraged for future applications.

Both the POSITION and INDEX functions return a character position in a string.

Examples of the POSITION function

SELECT POSITION( 'b' IN 'abc'); Result:2 SELECT POSITION( 'ab'IN 'abc'); Result:1 SELECT POSITION( 'd' IN 'abc'); Result:0

Comparable examples of the INDEX function.

SELECT INDEX('abc', 'b'); Result:2 SELECT INDEX('abc', 'ab'); Result:1 SELECT INDEX('abc', 'd'); Result:0

POSITION will be used for subsequent examples.

SUBSTRING and POSITION functions may be used together in a query.

Problem

Display a list of departments where the word "SUPPORT" appears in the department name, and list the starting position of the word.

Solution

SELECT department_name,POSITION('SUPPORT' IN department_name)FROM departmentWHERE POSITION('SUPPORT' IN department_name) > 0ORDER BY department_number;

Result

DeptName PositionNum------------------------------ -----------customer support 10software support 10Note: POSITION is discussed in more detail in the Teradata SQL For Application Development course.

Using SUBSTRING and POSITION Use of SUBSTRING and POSITION may be necessary to 'normalize' a character string column which is really two column values combined into a single column. In this case, the column 'contact_name' contains both first and last name. These functions may be used to separate them and to produce a different report format, in this case, first name followed by last.

The Contact table, seen below, is used with this example.

Problem

Display the first name and last names separated by a single space for each person in the contact table.

Solution

SELECT SUBSTRING (contact_name FROM POSITION (',' IN contact_name) +2) || ' ' || SUBSTRING (contact_name FROM 1 FOR POSITION (',' IN contact_name) -1) AS "Contact Names"FROM contact;

Result

Contact Names--------------------------Jack HughesNancy DibbleJames LeblancGinny SmithAlison TorresConnie Brayman

Things to notice:

The column contact_name is defined as a variable length VARCHAR(30). First and last name are stored in the colunm as seen in the table below. To reposition first and last name, the POSITION function is used to search for a

comma. This type of string manipulation is necessary when data is not properly normalized. If first and last name were separate columns, this manipulation would be

unnecessary.

Lab: String Function




Be sure to change your default database to the Customer_Service database in

order to run these labs.

A. Select last and first name, area code and phone number for employees in the 609 area code. Limit first name to 10 characters. You will need to join the employee and the employee_phone tables to accomplish this. Use table name aliasing for your join. Display your output in the format below:

last_name first_name Phone Number ------------------ ---------- ------------- Johnson Darlene (609)559-1011

Hint: Use substringing and concatenation to solve this lab.

Note: This lab is used again in exercise C.

B. Find all jobs that have the word "Analyst" in their descriptions. List the descriptions and positions where the character strings are found.

C. Modify exercise A to concatenate the phone extension with the rest of the phone number. Use COALESCE to transform any NULLs in the extension column into zero. Give the concatenated phone expression the following title, appropriately lined up over the data:

D.E. Phone Number ExtF. -------------------- G. (xxx) xxx-xxxx xxx H. List all contacts from the contact table, breaking the first name out into its own

column and the last name into its own column. Include phone number. Select only contacts in area code 408. Sort by first name. Custom column titles for the name columns are: First Name and Last Name.

Note: The format of the contact table is shown here for your reference.

CREATE TABLE contact, FALLBACK (contact_number INTEGER ,contact_name CHAR(30) NOT NULL ,area_code SMALLINT NOT NULL ,phone INTEGER NOT NULL ,extension INTEGER ,last_call_date DATE NOT NULL) UNIQUE PRIMARY INDEX (contact_number);

Note: The column 'contact_name' is denormalized, meaning that there are two pieces of information in one column. This situation often makes string manipulation necessary.

Lab Answer A: String Functions

DATABASE Customer_Service;

SELECT e.last_name,e.first_name (FORMAT 'X(10)'),'(' || SUBSTRING(p.area_code FROM 4 FOR 3) || ')' ||SUBSTRING(p.phone FROM 5 FOR 3) || '-' ||SUBSTRING(p.phone FROM 8) (TITLE 'Phone Number')

FROM employee e,employee_phonep

WHERE p.area_code = 609AND e.employee_number = p.employee_number;


last_name first_name Phone Number --------- ----------- ------------- Johnson Darlene (609)559-1011 Johnson Darlene (609)559-1011 Johnson Darlene (609)559-1011 Johnson Darlene (609)578-5781 Rabbit Peter (609)559-1011

Lab Answer B: String Functions

SELECT description,POSITION('Analyst' IN description)

FROM jobWHERE POSITION('Analyst' IN description) > 0;


description Index(description,'Analyst')------------------------- ----------------------------System Analyst 8Software Analyst 10System Support Analyst 16

Lab Answer C: String Functions

SELECT e.last_name,e.first_name (FORMAT 'X(10)'),'(' || SUBSTRING(p.area_code FROM 4 FOR 3) || ')' ||SUBSTRING(p.phone FROM 5 FOR 3) || '-' || SUBSTRING(p.phone FROM 8)|| ' ' || SUBSTRING((COALESCE(p.extension,0)) FROM 1)(TITLE 'Phone Number Ext')

FROM employee e,employee_phonep

WHERE p.area_code = 609AND e.employee_number = p.employee_number

;


last_name first_name Phone Number Ext---------- ----------- ---------------- -------Johnson Darlene (609)559-1011 213Johnson Darlene (609)559-1011 224Johnson Darlene (609)559-1011 225Johnson Darlene (609)578-5781 0Rabbit Peter (609)559-1011 213

Lab Answer D: String Functions

SELECT SUBSTRING(contact_name FROM POSITION (', ' IN contact_name)+2)AS "First Name"

,SUBSTRING(contact_name FROM 1 FOR POSITION (',' IN contact_name)-1)

AS "Last Name",phone

FROM contactWHERE area_code = 408ORDER BY 1;

*** Query completed. 2 rows found. 3 columns returned. First Name Last Name phone -------------- ---------- --------Connie Brayman 112345 Ginny Smith 3792152

Test Question:1. Macros are an ANSI standard feature.A. TrueB. False Ö

2. A table is made up of:A. Primary and Foreign KeysB. Rows and columns Ö C. Records and fieldsD. Columns and fieldsE. Rows and records

3. The DISTINCT option eliminates duplicate:A. rows Ö

B. columnsC. values

4. To get the data definition of the employee table, which statement would you use?A. SHOW TABLE employee; ÖB. SHOW employee;C. HELP employee;

5. A NULL representsA. a field with a zero in itB. a data typeC. the absence of a data value Ö

6. The command SELECT TIME gives you the current system time.A. True ÖB. False

7. To execute a macro the privilege(s) required are:A. The CREATE macro privilegeB. The privileges required by the body of the macroC. The EXEC macro privilege Ö

8. A Cartesian Product is a constrained cross join.A. TrueB. False Ö

9.The WHERE clause is used to:A. Eliminate data types from a queryB. Eliminate rows from a query Ö C. Eliminate columns from a query

10. To search for a partial value in a character string use this feature:A. LIKE ÖB. FORMATC. TRIM

11. The HELP command can be used with all database objects.A. True ÖB. False

12. To see how the database would process a request to retrieve all columns from the department table, which SQL statement would you use?A. HELP * department;B. SHOW department table;C. EXPLAIN SELECT * from department; Ö

13. What is the recommended DATE format for Year 2000 and beyond?A. 'YYYY-MM-DD' ÖB. 'MM-DD-YYYY'C. 'YY-MM-DD'

14. Subqueries always produce a DISTINCT list of values.A. True ÖB. False

15. An Inner Join performs which of the following:A. combines all rows from two or more tables.B. combines information from two tables according to a join condition. ÖC. matches all rows of one table with all rows of another table.

16. To run a macro, you use which command?A. EXEC Macro Ö B. SHOW MacroC. CREATE Macro

17. What language do we use to access and maintain data in a Teradata relational database?A. C++B. SQL Ö C. BTEQ

18. The asterisk "*" is used to select all the columns in the table.A. True Ö B. False

19. An alias is required to join a table to itself and is:A. a temporary name for a TABLE or VIEW Ö B. used to ALTER the name of a TABLEC. the DBA's login id

20. The HELP Database command gives you:A. The CREATE statement used to create the databaseB. The objects contained in that database ÖC. The owner of the database

21. Choose the command that would retrieve the name and department number for all employees in either department 401 or 403.A. SELECT last_name, department_number from employee WHERE department_number IN (401, 403);B.SELECT last_name, department_number from employee WHERE department_number IN 401 or 403;C. SELECT last_name, department_number from employee WHERE DEPARTMENT_NUMBER = 401 OR 403;

22. The default order for the output rows of a query is:A. Random ÖB. Ascending on first column specifiedC. Descending on first column specified

23. When using set operators such as UNION, you may not use aA. ORDER BY clauseB. WITH BY clause ÖC. SELECT ALL

24. Which of the following is a parameterized macro?A. CREATE MACRO dept_list (dept INTEGER) AS (SELECT last_name FROM customer_servicemployee WHERE department_number = :dept;); ÖB. CREATE MACRO birthday_list AS (SELECT first_name, birthdate FROM customer_servicemployee WHERE department_number = 201 ORDER BY birthdate;);

25. The SUBSTR syntax is a(n):A. Teradata extension ÖB. ANSI standard feature

26. The ability to explicitly define primary and secondary indexes is:A. an SQL standardB. an ANSI compliant featureC. a Teradata extension Ö

27. The INDEX function does the following:A. obtains a section of a character stringB. trims blanks from a stringC. locates a character position in a string

28. WITH...BY allows you to do which of the following?A. Create subtotal lines for a detail list; eliminate duplicate values; generate automatic sort on all BY columns.B. Generate automatic sort on all BY columns; create a grand total, create subtotal lines for a detail list.C. Create subtotal lines for a detail list; generate an automatic sort on all BY columns; allow subtotal "breaks" on more than one column. Ö

29.Which set operator gives you an answer set containing rows that are common to all the sets generated by each SELECT operation?A. INTERSECT ÖB. EXCEPTC. UNION

30. How many columns can be defined in a single CREATE TABLE statement? A. 256B. 512 C. 1024D. 2048 Ö

31.What are the four keywords used for Data Manipulation?A. INSERT, MOVE, UPDATE, CREATEB. DISCARD, INSERT, CREATE, UPDATEC. INSERT, SELECT, UPDATE, DELETE Ö

32. A view is:A. stored in the Data DictionaryB. a virtual tableC. all of the above Ö

33.Which set operator is similar in function to the NOT IN operator?A. INTERSECTB. UNIONC. EXCEPT Ö

34. What purpose do WHERE and GROUP BY have in computing aggregates?A. Eliminate some rows and put them into desired groupings. ÖB. Group rows according to their location.C. Produce one level of subtotal.

35. Which of the following is NOT true regarding the DROP INDEX command?

A. Indexes may be dropped by column name(s)B. Indexes may be dropped by nameC. Indexes may be dropped either primary or secondary Ö

36. Which condition selects only the groups from a GROUP BY clause that satisfy a conditional expression?A. COUNTB. HAVING ÖC. ORDER BY

37. Which function performs a similar function to an INNER JOIN?A. UNIONB. EXCEPTC. INTERSECT Ö

38.To prevent insertion of rows through a view that would not be seen by the view you would use which option?A. HAVINGB. WITH CHECK OPTION Ö B. FORMAT

39.Of the following aggregate operators, which are ANSI standard?A. MINIMUM, MAXIMUM, SUM, COUNTB. MINIMUM, MAXIMUM, AVERAGEC. MIN, SUM, COUNT, MAX, AVG Ö

40. A join view is made up of columns from several tables.A. TrueB. False Ö

41. Which of the following is legal in a view definition?A. INDEXB. NAMED Ö C. ORDER BY

42. You can control the scope of a RANK function by use of which of the following clauses.A. ORDER BY B. PARTITION BY Ö C. QUALIFYD. WHERE

43. You can restrict which rows participate in a ranking by use of the which of the following clauses.A. OVERB. PARTITION BYC. QUALIFY ÖD. ORDER BY

44. A sampling operation where differing sampling criteria are applied to different sets of rows is called a _________ sampling.A. Stratified Ö B. ComplexC. OverlappingD. Random

45. Which three are possible using the TOP N feature:A. Produce the top 5 sales amounts including duplicates. ÖB. Produce a 10 percent sample of rows ÖC. Produce the bottom 10 rows according to price. ÖD. Produce the sum of the top ten percentile of salaries.

46. When using the TOP N function, unless there is an ORDER BY clause, the WITH TIES option is ignored.A. True ÖB. False

Question/Answer: 1:Customer Support 2:James Trader 3: A: SELECT * FROM department ; B: SELECT last_name ,first_name ,salary_amount FROM employee WHERE manager_employee_number = 1019 ORDER BY last_name ; C: SELECT last_name ,department_number ,salary_amount FROM employee WHERE manager_employee_number = 801 ORDER BY last_name ; D: SELECT department_number ,manager_employee_number FROM employee ;

SELECT DISTINCT department_number ,manager_employee_number FROM employee ;

undefined