PrinciplesofDatabaseSystems...

Principles of Database Systems CSE 544p

Lecture #1 September 28, 2011

1 Dan Suciu -- p544 Fall 2011

Staff

•  Instructor: Dan Suciu – CSE 662, [email protected] – Office hours: Wednesdays, 5:30-‐6:20

•  TAs: – Sandra Fan, [email protected]

Dan Suciu -- p544 Fall 2011 2

CommunicaRons

•  Web page: hSp://www.cs.washington.edu/p544 –  Lectures will be available here –  Homework will be posted here –  Announcements may be posted here

•  Mailing list: – Announcements, group discussions –  If you registered, you are automaRcally subscribed


Textbook(s)

Main textbook: •  Database Management Systems, Ramakrishnan and Gehrke

Second textbook: •  Database Systems: The Complete Book, Garcia-‐Molina, Ullman, Widom


Course Format

•  Lectures Wednesdays, 6:30-‐9:20

•  7 Homework Assignments

•  Take-‐home Final


Grading

•  Homework: 70 %

•  Take-‐home Final: 30%


Homework Assignments

1.  SQL 2.  Conceptual design 3.  JAVA/SQL 4.  TransacRons 5.  Database tuning 6.  XML/XPath/XQuery 7.  Pig LaRn, on AWS

7 Dan Suciu -- p544 Fall 2011 Due: Mondays’, by 11:59pm. Three late days per person

Take-‐home Final

•  Posted on December 8, at 11:59pm

•  Due on December 10, by 10:00pm

•  No late days/hours/minutes/seconds


Sohware Tools •  Postgres:

–  Preferred usage: download from download hSp://www.postgresql.org/download/ –  Other opRon: use postgres on lab machines

•  SQL Server 2008 –  Download client from hSp://msdnaa.cs.washington.edu –  Username is your full @cs.washington.edu email address –  Doesn’t work ? Email ms-‐sw-‐[email protected] –  Connect to IPROJSRV (may need tunneling) –  OK to use you own server, just import IMDB

•  Xquery: download one interpreter from –  Preferred: Saxon: hSp://saxon.sourceforge.net/ (from apache; very popular) –  Others:

•  Zorba: hSp://www.zorba-‐xquery.com/ (I used this one: ½ day installaRon) •  Galax: hSp://galax.sourceforge.net/ (great in the past, seems less well maintained)

•  Pig LaRn: –  We will run it on Amazon Web Services –  You may download from hSp://hadoop.apache.org/pig/, but you won’t need it


Accessing SQL Server •  SQL Server Management Studio •  Server Type = Database Engine •  Server Name = IPROJSRV •  AuthenRcaRon = SQL Server AuthenRcaRon

–  Login = your UW email address (not the CSE email) –  Password = [in class]

•  Must connect from within CSE, or must use tunneling •  AlternaRvely: install your own, get it from MSDNAA (see earlier slide)

•  Then play with IMDB, start working on HW 1


Rest of Today’s Lecture

• Overview of DBMS

• Overview of the course content

•  SQL Dan Suciu -- p544 Fall 2011 11

Database

What is a database ?

Give examples of databases


Database

What is a database ? •  A collecRon of files storing related data

Give examples of databases •  Accounts database; payroll database; UW’s students database; Amazon’s products database; airline reservaRon database


Database Management System

What is a DBMS ?

Give examples of DBMS


Database Management System

What is a DBMS ? •  A big C program wri;en by someone else that allows us to manage efficiently a large database and allows it to persist over long periods of Bme

Give examples of DBMS •  DB2 (IBM), SQL Server (MS), Oracle, Sybase •  MySQL, Postgres, …

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SQL for Nerds, Greenspun, hSp://philip.greenspun.com/sql/ (Chap 1) Dan Suciu -- p544 Fall 2011

Market Shares

From 2006 Gartner report: •  IBM: 21% market with $3.2BN in sales

•  Oracle: 47% market with $7.1BN in sales

•  Microsoh: 17% market with $2.6BN in sales


An Example

The Internet Movie Database hSp://www.imdb.com

•  EnRRes: Actors (800k), Movies (400k), Directors, …

•  RelaRonships: who played where, who directed what, …


Key concept 1: RelaRonal Data Model


Actor: Cast:

Movie:

id fName lName gender

195428 Tom Hanks M 645947 Amy Hanks F

. . .

id Name year

337166 Toy Story 1995

. . . . . . . ..

pid mid

195428 337166 . . .

Key concept 2: DeclaraRve Query Language

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SELECT * FROM Actor

Dan Suciu -- p544 Fall 2011

SELECT count(*) FROM Actor

SELECT * FROM Actor WHERE lName = ‘Hanks’ SQL

We write what we want, not how we want it.

Key concept 3: Data Independence

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SELECT * FROM Actor, Casts, Movie WHERE lname='Hanks' and Actor.id = Casts.pid and Casts.mid=Movie.id and Movie.year=1995

817k actors, 3.5M casts, 380k movies; How can it be so fast ?

Physical data independence: query is independent of physical storage

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How Can We Evaluate the Query ?

Actor: Cast: Movie: id fName lName gender

. . . Hanks

. . .

id Name year

. . . 1995

. . .

pid mid

. . .

. . .

Plan 1: . . . . [ in class ] Plan 2: . . . . [ in class ]



Actor Cast Movie

σlName=‘Hanks’ σyear=1995

Actor Cast Movie

σlName=‘Hanks’ σyear=1995

Indexes: on Actor.lName, on Movie.year

AlternaRve query plans:

Query opRmizaRon Database StaRsRcs histograms, synopses, etc

Key concept 4: TransacRons


X = Read(Account_1); X.amount = X.amount - 100; Write(Account_1, X); Y = Read(Account_2); Y.amount = Y.amount + 100; Write(Account_2, Y);

CRASH !

What is the problem ?

Recovery from systems failures: Transfer $100 from account 1 to account 2:


X = Read(Account); if (X.amount >= 100) { dispense_money( ); X.amount = X.amount – 100; } else error(“Insufficient funds”);

X = Read(Account); if (X.amount >= 100) { dispense_money( ); X.amount = X.amount – 100; } else error(“Insufficient funds”);

What can go wrong ?

Concurrency Control

Overdrahing an account:

User 1: User 2:

TransacRons

ACID = •  Atomicity ( = recovery) •  Consistency •  IsolaRon ( = concurrency control) •  Durability


Client/Server Database Architecture

•  Single server that stores the database •  Many clients running apps and connecRng to DBMS •  Performance boSlenecks:

–  Client/server communicaRon –  TransacRonal semanRcs

•  Other architectures: –  main memory database –  replicated databases


Two Types of Database Usage

•  OLTP (online-‐transacRon-‐processing) – Many updates – Many simple “point queries” – Few (or no) complex aggregate queries

•  Decision-‐Support – Many aggregate/group-‐by queries. – Few (or no) updates


Trends in Data Management

•  Large scale data analyRcs: Map/Reduce, Pig, … •  Cloud based database service: AWS, Azure, … •  NoSQL: sacrifice ACID for performance •  Data privacy •  Data provenance •  Complex data analyRcs: probabilisRc databases


Outline of Course Content 1.  SQL 2.  RelaRonal Calculus, Database Design 3.  Constraints, Views 4.  TransacRons: recovery 5.  TransacRons: concurrency control 6.  XML, XPath, XQuery 7.  Data storage, indexes, physical tuning 8.  Query execuRon 9.  Query opRmizaRon 10.  Big Data: Parallel databases, Map/Reduce, Pig LaRn 11.  Advanced topics: privacy, provenance, probabilisRc dbs


Announcement: Homework 1

•  Homework 1 is posted; •  Due on Monday, Oct. 10 •  Tools:

– Postgres: install on your computer (PREFERRED) or use the installaRon in the lab

– SQL Server, for tesRng only; connect to IPROJSRV: login: your UW email address; password: ……..

•  Tasks: create db, import data, create indices, write 11 SQL queries


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Outline for rest of today

•  Basics SQL (Chapters 5.2, 5.3) •  Aggregates (Chapter 5.5.) •  Nulls, Outer joins (Chapter 5.6) •  Subqueries (Chapters 5.4)

– This is tough ! Next lecture we will discuss RelaRonal Calculus (a.k.a. Tuple Calculus, Chapter 4.3). See supplementary text Three Query Language Formalisms


32

SQL

•  Data DefiniRon Language (DDL) – Create/alter/delete tables and their aSributes – Read from the book

•  Data ManipulaRon Language (DML) – Query tables, Insert/delete/modify – Discussed in class


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Tables in SQL

PName Price Category Manufacturer

Gizmo $19.99 Gadgets GizmoWorks

Powergizmo $29.99 Gadgets GizmoWorks

SingleTouch $149.99 Photography Canon

MultiTouch $203.99 Household Hitachi

Product

Attribute names Table name

Tuples or rows

Key


The RelaRonal Data Model Data is stored in tables , a.k.a. relaBons

Each relaRon has: 1.  A schema = name+aSributes

–  Product(PName, Price, Category, Manufacturer) –  Each relaRon has a key, which we underline

2.  An instance = set of rows

SQL departs from the pure relaRonal model in that it allows duplicate tuples •  Set semanBcs à bag semanBcs

{1, 2, 3} à {1, 1, 2, 3, 3, 3}


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Data Types in SQL

•  Atomic types: – Characters: CHAR(20), VARCHAR(50) – Numbers: INT, BIGINT, SMALLINT, FLOAT – Others: MONEY, DATETIME, …

•  Record (aka tuple) – Has atomic aSributes

•  Table (relaRon) – A set of tuples


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Simple SQL Query

PName Price Category Manufacturer Gizmo $19.99 Gadgets GizmoWorks

Powergizmo $29.99 Gadgets GizmoWorks SingleTouch $149.99 Photography Canon MultiTouch $203.99 Household Hitachi

SELECT * FROM Product WHERE category=‘Gadgets’

Product


Powergizmo $29.99 Gadgets GizmoWorks “selection” Dan Suciu -- p544 Fall 2011

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Simple SQL Query PName Price Category Manufacturer Gizmo $19.99 Gadgets GizmoWorks


SELECT PName, Price, Manufacturer FROM Product WHERE Price > ‘$100’

Product

PName Price Manufacturer SingleTouch $149.99 Canon MultiTouch $203.99 Hitachi

“selection” and “projection”


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Details •  Case insensiRve:

SELECT = Select = select Product = product BUT: ‘SeaSle’ ≠ ‘seaSle’

•  Constants: ‘abc’ -‐ yes “abc” -‐ no


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EliminaRng Duplicates

SELECT DISTINCT category FROM Product

Compare to:

SELECT category FROM Product

Category Gadgets Gadgets

Photography Household

Category Gadgets

Photography Household


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Ordering the Results

SELECT pname, price, manufacturer FROM Product WHERE category=‘Gadgets’ AND price > ‘$10’ ORDER BY price, pname

Ties are broken by the second attribute on the ORDER BY list. Ordering is ascending, unless you specify the DESC keyword.


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SELECT Category FROM Product ORDER BY PName



? SELECT DISTINCT category FROM Product ORDER BY category

SELECT DISTINCT category FROM Product ORDER BY PName

? ?


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Keys and Foreign Keys



Product

Company

CName Country

GizmoWorks USA

Canon Japan

Hitachi Japan

Key

Foreign key


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Joins

Product (PName, Price, Category, Manufacturer) Company (CName,, Country) Find all products under $200 manufactured in Japan; return their names and prices.

SELECT PName, Price FROM Product, Company WHERE Manufacturer=CName AND Country=‘Japan’ AND Price <= ‘$200’

Join between Product

and Company


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Joins






Product Company

Cname Country

GizmoWorks USA

Canon Japan

Hitachi Japan

PName Price

SingleTouch $149.99

SELECT PName, Price FROM Product, Company WHERE Manufacturer=CName AND Country=‘Japan’ AND Price <= ‘$200’


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Tuple Variables

SELECT DISTINCT name, country FROM Person, Company WHERE worksfor = cname

Which country ?

Product (PName, Price, Category, Manufacturer) Company (CName,, Country) Person(name, Country, Worksfor)

SELECT DISTINCT Person.name, Company.country FROM Person, Company WHERE Person.worksfor = Company.cname


SELECT DISTINCT x.name, y.country FROM Person AS x, Company AS y WHERE x.worksfor = y.cname

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In Class

Product (pname, price, category, manufacturer) Company (cname, country) Find all Chinese companies that manufacture products in the ‘toy’ category

SELECT cname FROM WHERE


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In Class

Product (pname, price, category, manufacturer) Company (cname, country) Find all Chinese companies that manufacture products both in the ‘electronic’ and ‘toy’ categories

SELECT cname FROM WHERE


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The Nested Loop SemanRcs of SQL Queries

SELECT a1, a2, …, ak FROM R1 AS x1, R2 AS x2, …, Rn AS xn WHERE CondiRons


Answer = {} for x1 in R1 do for x2 in R2 do ….. for xn in Rn do if Conditions then Answer = Answer ∪ {(a1,…,ak)} return Answer

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SELECT DISTINCT R.A FROM R, S, T WHERE R.A=S.A OR R.A=T.A

Using the Formal SemanRcs

If S ≠ ∅ and T ≠ ∅ then returns R ∩ (S ∪ T) else returns ∅

What do these queries compute ?

SELECT DISTINCT R.A FROM R, S WHERE R.A=S.A

Returns R ∩ S


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AggregaRon

SELECT count(*) FROM Product

Except count, all aggregations apply to a single attribute

SELECT sum(price) FROM Product WHERE manufacturer=‘GizmoWorks’

SQL supports several aggregation operations: sum, count, min, max, avg


Product (pname, price, category, manufacturer) Company (cname, country)

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COUNT applies to duplicates, unless otherwise stated:

SELECT count(category) FROM Product WHERE price > ‘$20’

If category has no nulls, then count(category)=count(*)

We probably want:

SELECT count(DISTINCT category) FROM Product WHERE price > ‘$20’

AggregaRon: Count



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Grouping and AggregaRon

SELECT manufacturer, count(*) AS total FROM Product WHERE price < ‘$200’ GROUP BY manufacturer

Let’s see what this means…

For each manufacturer, find total number of its products under $200.



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Grouping and AggregaRon

1. Compute the FROM and WHERE clauses. 2. Group by the attributes in the GROUPBY 3. Compute the SELECT clause, including aggregates.


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1&2. FROM-‐WHERE-‐GROUPBY








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3. SELECT








count(*) Manufacturer

2 GizmoWorks

1 Canon

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HAVING Clause

SELECT manufacturer, count(*) AS total FROM Product WHERE price < ‘$200’ GROUP BY manufacturer HAVING min(price) >’$20’

Same query, except that we return only those manufacturers that make only products with price > $20

HAVING clause contains conditions on aggregates. Dan Suciu -- p544 Fall 2011


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General form of Grouping and AggregaRon

SELECT S FROM R1,…,Rn WHERE C1 GROUP BY a1,…,ak HAVING C2 S = may contain aSributes a1,…,ak and/or any aggregates but NO

OTHER ATTRIBUTES C1 = is any condiRon on the aSributes in R1,…,Rn C2 = is any condiRon on aggregate expressions

Why ?


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General form of Grouping and AggregaRon

EvaluaRon steps: 1.  Evaluate FROM-‐WHERE, apply condiRon C1 2.  Group by the aSributes a1,…,ak 3.  Apply condiRon C2 to each group (may have aggregates) 4.  Compute aggregates in S and return the result

SELECT S FROM R1,…,Rn WHERE C1 GROUP BY a1,…,ak HAVING C2


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NULLS in SQL

•  Whenever we don’t have a value, we can put a NULL •  Can mean many things:

–  Value does not exists –  Value exists but is unknown –  Value not applicable –  Etc.

•  The schema specifies for each aSribute if can be null (nullable aSribute) or not

•  How does SQL cope with tables that have NULLs ?


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Null Values

•  If x= NULL then 4*(3-‐x)/7 is sRll NULL

•  If x= NULL then x=‘Joe’ is UNKNOWN •  In SQL there are three boolean values:

FALSE = 0 UNKNOWN = 0.5 TRUE = 1


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Null Values

•  C1 AND C2 = min(C1, C2) •  C1 OR C2 = max(C1, C2) •  NOT C1 = 1 – C1 Rule in SQL: include only tuples that yield TRUE

SELECT * FROM Person WHERE (age < 25) AND (height > 6 OR weight > 190)

E.g. age=20 heigth=NULL weight=200


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Null Values

Unexpected behavior: Some Persons are not included !

SELECT * FROM Person WHERE age < 25 OR age >= 25


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Null Values

Can test for NULL explicitly: –  x IS NULL –  x IS NOT NULL

Now it includes all Persons

SELECT * FROM Person WHERE age < 25 OR age >= 25 OR age IS NULL


Outerjoins

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SELECT x.country, y.pname FROM Company x JOIN Product y ON x.cname = y.manufacturer

SELECT x.country, y.pname FROM Company x, Product y WHERE x.cname = y.manufacturer

Same as:

But countries that don’t manufacture will not be listed !

Product (pname, price, category, manufacturer) Company (cname, country) Normally, joins are “inner joins”:


Outerjoins

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SELECT x.country, y.pname FROM Company x LEFT OUTER JOIN Product y ON x.cname = y.manufacturer

If we want to see the companies that don’t produce anything, then we use an outer join:



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Product Company







Cname Country

GizmoWorks USA

Canon Japan

Hitachi Japan

MuseumPass Vatican

Cname PName

USA GizmoWorks

USA GizmoWorks

Japan Canon

Japan Hitachi

Vatican NULL

ApplicaRon


SELECT x.country, count(*) FROM Company x, Product y WHERE x.cname = y.manufacturer GROUP BY x.country

What’s wrong ?


Compute the total number of products made by each country

ApplicaRon


SELECT x.country, count(y.pname) FROM Company x LEFT OUTER JOIN Product y ON x.cname = y.manufacturer GROUP BY x.country

Now we also get the products who sold in 0 quantity


Compute the total number of products made by each country

Note: we don’t use count(*)

WHY ?

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Outer Joins

•  Leh outer join: –  Include the leh tuple even if there’s no match

•  Right outer join: –  Include the right tuple even if there’s no match

•  Full outer join: –  Include the both leh and right tuples even if there’s no match


Subqueries

•  A subquery is another SQL query nested inside a larger query

•  Such inner-‐outer queries are called nested queries •  A subquery may occur in:

1.  A SELECT clause 2.  A FROM clause 3.  A WHERE clause


Rule of thumb: avoid wriRng nested queries when possible; someRmes it’s impossible

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1. Subqueries in SELECT


For each product return the country that manufactures it

SELECT X.pname, (SELECT Y.country FROM Company Y WHERE Y.cname=X.manufacturer) FROM Product X

What happens if a subquery returns more than one country ?


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Whenever possible, don’t use a nested queries:

= We have “unnested” the query


SELECT X.pname, (SELECT Y.country FROM Company Y WHERE Y.cname=X.manufacturer) FROM Product X

SELECT pname, country FROM Product, Company WHERE cname=manufacturer


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Compute the number of products made by each country

SELECT DISTINCT x.country, (SELECT count(*) FROM Company y, Product WHERE y.cname=manufacturer and y.country = x.country) FROM Company x

Better: we can unnest by using a GROUP BY



SELECT x.country, count(*) FROM Company x, Product z WHERE x.cname = z.manufacturer GROUP BY x.country

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GROUP BY v.s. Nested Quereis

SELECT manufacturer, count(*) AS total FROM Product WHERE price < '$200’ GROUP BY manufacturer

SELECT DISTINCT x.manufacturer, (SELECT count(*) FROM Product y WHERE x.manufacturer = y.manufacturer AND price < '$200’) AS total FROM Product x WHERE price < '$200’

Why twice ? Dan Suciu -- p544 Fall 2011

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2. Subqueries in FROM

Find all products whose prices is > 20 and < 30

SELECT * FROM (SELECT * FROM Product AS Y WHERE Y.price > ‘$20’) AS x WHERE x.price < ‘$30’

Unnest this query !



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3. Subqueries in WHERE

Find all countries that make some products with price < 100

SELECT DISTINCT x.country FROM Company x WHERE EXISTS (SELECT * FROM Product y WHERE y.manufacturer = x.cname and y.price < ‘$100’)

Existential quantifiers

Using EXISTS:



Correlated subqery: uses x from outer query

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Predicate Calculus (a.k.a. First Order Logic)


{ y |∃x.Company(x,y)∧(∃z.∃p.∃c.Product(z,p,c,x)∧p<100)}

Existential quantifiers Product (pname, price, category, manufacturer) Company (cname, country)

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SELECT DISTINCT country FROM Company WHERE cname IN (SELECT Product.manufacturer FROM Product WHERE Product.price < ‘$100’)

Using IN



De-‐correlated subqery

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SELECT DISTINCT Company.country FROM Company WHERE ‘$100’ > ANY (SELECT price FROM Product WHERE manufacturer = cname)

Using ANY:



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SELECT DISTINCT x.country FROM Company x, Product y WHERE x.cname = y.manufacturer and y.price < ‘$100’

Existential quantifiers are easy ! J

Now let’s unnest it:



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Universal quantifiers are hard ! L

Find the countries of all companies that make only products with price < 100


Universal quantifiers Product (pname, price, category, manufacturer) Company (cname, country)

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Predicate Calculus (a.k.a. First Order Logic)


{ y | ∃x.Company(x,y)∧(∀z.∀p.∀c.Product(z,p,c,x)èp<100) }



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{ y |∃x. Company(x,y)∧(∀z.∀p.∀c.Product(z,p,c,x)èp<100) }

De Morgan’s Laws: ¬(A ∧ B) = ¬A ∨ ¬B ¬(A ∨ B) = ¬A ∧ ¬B ¬∀x. P(x) = ∃x. ¬ P(x) ¬∃x. P(x) = ∀x. ¬ P(x)

{ y|∃x.Company(x,y)∧¬(∃z∃p.∃p.Product(z,p,c,x)∧p≥100) }

{ y | ∃x. Company(x,y)) } − { y | ∃x. Company(x,y) ∧(∃z∃p.∃c.Product(z,p,c,x)∧p≥100) }

¬(A è B) = A ∧ ¬B

=

=

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2. Find all companies s.t. all their products have price < 100

1. Find the other companies: i.e. s.t. some product ≥ 100


SELECT DISTINCT country FROM Company WHERE cname IN (SELECT manufacturer FROM Product WHERE price >= ‘$100’)

SELECT DISTINCT country FROM Company WHERE cname NOT IN (SELECT manufacturer FROM Product WHERE price >= ‘$100’)

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Universal quantifiers

Using EXISTS:


SELECT DISTINCT x.country FROM Company x WHERE NOT EXISTS (SELECT * FROM Product y WHERE y.manufacturer = x.cname and y.price >= ‘$100’)


86


SELECT DISTINCT Company.country FROM Company WHERE ‘$100’ > ALL (SELECT price FROM Product WHERE manufacturer = cname)

Using ALL:




87

QuesRon for Database Fans and their Friends

•  Can we unnest this query ?



88

Monotone Queries •  A query Q is monotone if:

– Whenever we add tuples to one or more of the tables… –  … the answer to the query cannot contain fewer tuples

•  Fact: all unnested queries are monotone –  Proof: using the “nested for loops” semanRcs

•  Fact: A query a universal quanRfier is not monotone

•  Consequence: we cannot unnest a query with a universal quanRfier Dan Suciu -- p544 Fall 2011

Queries that must be nested


Rule of Thumb: Non-‐monotone queries cannot be unnested. In parRcular, queries with a universal quanRfier cannot be unnested

More SQL

Read the following commands in the book

•  CREATE TABLE •  INSERT •  DELETE •  UPDATE

They are easy; but we need/use them all the Rme in class, and in the homework assignments


91

Advanced SQLizing

1.  UnnesRng Aggregates

2.  Finding witnesses


UnnesRng Aggregates

For each category, find the maximum price

SELECT DISTINCT X.category, (SELECT max(Y.price) FROM Product Y WHERE X.category = Y.category) FROM Product X

SELECT category, max(price) FROM Product GROUP BY category

Equivalent queries

Note: no need for DISTINCT (DISTINCT is the same as GROUP BY)


UnnesRng Aggregates

Find the number of products made in each country SELECT DISTINCT X.country, (SELECT count(*) FROM Company Y, Product Z WHERE Y.cname=Z.manufacturer AND Y.country = X.country) FROM Company X

SELECT X.country, count(*) FROM Company X, Product Y WHERE X.cname=Y.manufacturer GROUP BY X.country

They are NOT equivalent !

(WHY?)


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More UnnesRng

•  Find authors who wrote ≥ 10 documents: •  ASempt 1: with nested queries

SELECT DISTINCT Author.name FROM Author WHERE count(SELECT Wrote.url FROM Wrote WHERE Author.login=Wrote.login) > 10

This is SQL by a novice

Author(login,name) Wrote(login,url)


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More UnnesRng

•  Find all authors who wrote at least 10 documents:

•  ASempt 2: SQL style (with GROUP BY)

SELECT DISTINCT Author.name FROM Author, Wrote WHERE Author.login=Wrote.login GROUP BY Author.name HAVING count(wrote.url) > 10

This is SQL by

an expert


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Finding Witnesses

For each country, find its most expensive products



Finding Witnesses

SELECT x.country, max(y.price) FROM Company x, Product y WHERE x.cname = y.manufacturer GROUP BY x.country

Finding the maximum price is easy…

But we need the witnesses, i.e. the products with max price

For each country, find its most expensive products


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Finding Witnesses

SELECT u.country, v.pname, v.price FROM Company u, Product v, (SELECT x.country, max(y.price) as mprice FROM Company x, Product y WHERE x.cname = y.manufacturer GROUP BY x.country) AS p WHERE u.country = p.country and v.price = p.mprice

To find the witnesses, compute the maximum price in a subquery


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Finding Witnesses

There is a more concise solution here:

SELECT x.country, y.pname, y.price FROM Company x, Product y WHERE x.cname = y.manufacturer and y.price >= ALL (SELECT z.price FROM Product z WHERE x.cname = z.manufacturer)



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