Lecture 12: Further relational algebra, further SQL

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Lecture 12: Further relational algebra, further SQL. www.cl.cam.ac.uk/Teaching/current/Databases/. Today’s lecture. Where does SQL differ from relational model? What are some other features of SQL? How can we extend the relational algebra to match more closely SQL?. Duplicate rows. - PowerPoint PPT Presentation

Transcript of Lecture 12: Further relational algebra, further SQL

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Lecture 12:Further relational algebra,

further SQL

www.cl.cam.ac.uk/Teaching/current/Databases/

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Today’s lecture• Where does SQL differ from relational

model?• What are some other features of SQL?• How can we extend the relational algebra

to match more closely SQL?

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Duplicate rows• Consider our relation instances from lecture

6, Reserves, Sailors and Boats• Consider SELECT rating,age FROM Sailors;• We get a relation that doesn’t satisfy our

definition of a relation!• RECALL: We have the keyword DISTINCT to

remove duplicates

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Multiset semantics• A relation in SQL is really a multiset or

bag, rather than a set as in the relational model– A multiset has no order (unlike a list), but

allows duplicates– E.g. {1,2,1,3} is a bag– select, project and join work for bags as well

as sets• Just work on a tuple-by-tuple basis

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Bag operations• Bag union:

– Sum the number of times that an element appears in the two bags, e.g.

• {1,2,1}{1,2,3} = {1,1,1,2,2,3}

• Bag intersection:– Take the minimum of the number of occurrences in

each bag, e.g.• {1,2,1}{1,2,3,3} = {1,2}

• Bag difference:– Proper-subtract the number of occurrences in the two

bags, e.g.• {1,2,1}-{1,2,3,3} = {1}

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Laws for bags• Note that whilst some of the familiar (set-

theoretic) laws continue to hold, some of them do not

• Example: R(ST) = (RS)(RT) ??

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Extended relational algebraAdd features needed for SQL

1. Bag semantics2. Duplicate elimination operator, 3. Sorting operator, 4. Grouping and aggregation operator, 5. Outerjoin operators, oV, Vo, oVo

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Duplicate-elimination operator

(R) = relation R with any duplicated tuples removed

• R= (R)=

• This is used to model the DISTINCT feature of SQL

A B1 23 41 2

A B1 23 4

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Sorting L1,… Ln

(R) returns a list of tuples of R, ordered according to the attributes L1, …, Ln

• Note: does not return a relation• R= B(R)= [(5,2),(1,3),(3,4)]

• ORDER BY in SQL, e.g. SELECT * FROM Sailors WHERE rating>7 ORDER BY age, sname;

A B

1 3

3 4

5 2

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Extended projection• SQL allows us to use arithmetic operators SELECT age*5 FROM Sailors;• We extend the projection operator to allow the

columns in the projection to be functions of one or more columns in the argument relation, e.g.

• R= A+B,A,A(R)=

A B1 23 4

A+B A.1 A.23 1 17 3 3

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Arithmetic• Arithmetic (and other expressions) can not

be used at the top level– i.e. 2+2 is not a valid SQL query

• How would you get SQL to compute 2+2?

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Aggregation• SQL provides us with operations to summarise a

column in some way, e.g. SELECT COUNT(rating) FROM Sailors;

SELECT COUNT(DISTINCT rating) FROM Sailors;

SELECT COUNT(*) FROM Sailors WHERE rating>7;• We also have SUM, AVG, MIN and MAX

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Grouping• These aggregation operators have been

applied to all qualifying tuples. Sometimes we want to apply them to each of several groups of tuples, e.g.– For each rating, find the average age of the

sailors– For each rating, find the age of the youngest

sailor

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GROUP BY in SQL SELECT [DISTINCT] target-list FROM relation-list WHERE qualification GROUP BY grouping-list;• The target-list contains

1. List of column names2. Aggregate terms– NOTE: The variables in target-list must be

contained in grouping-list

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GROUP BY cont.For each rating, find the average age of the sailors SELECT rating,AVG(age) FROM Sailors GROUP BY rating;

For each rating find the age of the youngest sailor SELECT rating,MIN(age) FROM Sailors GROUP BY rating;

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Grouping and aggregationL(R) where L is a list of elements that are

either– Individual column names (“Grouping

attributes”), or– Of the form (A), where is an aggregation

operator (MIN, SUM, …) and A is the column it is applied to

• For example,rating,AVG(age)(Sailors)

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Semantics• Group R according to the grouping

attributes• Within each group, compute (A)• Result is the relation consisting of one

tuple for each group. The components of that tuple are the values associated with each element of L for that group

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Example• Let R=

• Compute beer,AVG(price)(R)

bar beer priceAnchor 6X 2.50Anchor Adnam’s 2.40Mill 6X 2.60Mill Fosters 2.80Eagle Fosters 2.90

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Example cont.1. Group according to the grouping attribute,

beer:

2. Compute average of price within groups:

bar beer priceAnchor 6X 2.50Mill 6X 2.60Anchor Adnam’s 2.40Mill Fosters 2.80Eagle Fosters 2.90

beer price6X 2.55Adnam’s 2.40Fosters 2.85

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NULL values• Sometimes field values are unknown (e.g. rating

not known yet), or inapplicable (e.g. no spouse name)

• SQL provides a special value, NULL, for both these situations

• This complicates several issues– Special operators needed to check for NULL– Is NULL>8? Is (NULL OR TRUE)=TRUE?– We need a three-valued logic– Need to carefully re-define semantics

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NULL values• Consider INSERT INTO Sailors (sid,sname) VALUES (101,”Julia”);

SELECT * FROM Sailors;

SELECT rating FROM Sailors;

SELECT sname FROM Sailors WHERE rating>0;

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Entity integrity constraint• An entity integrity constraint states that

no primary key value can be NULL

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Outer join• Note that with the usual join, a tuple that

doesn’t ‘join’ with any from the other relation is removed from the resulting relation

• Instead, we can ‘pad out’ the columns with NULLs

• This operator is called an full outer join, written oVo

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Example of full outer join• Let R= Let S=

• Then RVS =

• But RoVoS =

A B1 23 4

B C4 56 7

A B C3 4 5

A B C1 2 NULL

3 4 5NULL 6 7

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Outer joins in SQL• SQL/92 has three variants:

– LEFT OUTER JOIN (algebra: oV)– RIGHT OUTER JOIN (algebra: Vo)– FULL OUTER JOIN (algebra: oVo)

• For example: SELECT * FROM Reserves r LEFT OUTER JOIN Sailors s ON r.sid=s.sid;

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Views• A view is a query with a name that can be used in

further SELECT statements, e.g. CREATE VIEW ExpertSailors(sid,sname,age) AS SELECT sid,sname,age FROM Sailors WHERE rating>9;

• Note that ExpertSailors is not a stored relation• (WARNING: mysql does not support views )

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Querying views• So an example query SELECT sname FROM ExpertSailors WHERE age>27;• is translated by the system to the following: SELECT sname FROM Sailors WHERE rating>9 AND age>27;

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SummaryYou should now understand:• Multi-set semantics• Conditions• Aggregation• GROUP BY• NULLs, entity ICs and outer joins• Views• Extensions to the core relational algebra