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SQL Queries

Chapter 5

Sample Database used in these

notes: sailors.sql(Posted on Ctools)

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SQL Query Language

Implements relational algebra Select, Project, Join, Set operators (we will see

these in the next lecture)

And so much more

Correlated subqueries

Ordering of results

Aggregate queries (e.g., SUM, MAX, AVG)

Three-valued logic for NULL values

Etc.

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Basic SQL Query

SELECT [DISTINCT] attr-list

FROM relation-list

WHERE qualification

Optional

List of relations

Attr1 op Attr2

OPS: , =, =,

Combine using AND, OR, NOT

Attributes frominput relations

(Conceptual)Evaluation:

1. Take cross-product of relation-list2. Select rows satisfying qualification

3. Project columns in attr-list

(eliminate duplicates only if DISTINCT)

Optimizerchoosesefficient

plan!!

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Example of Basic Query

Schema: Sailors (sid, sname, rating, age) Boats (bid, bname, color)

Reserves (sid, bid, rday)

Find the names of sailors who havereserved boat #103

SELECT S.snameFROM Sailors S, Reserves RWHERE S.sid = R.sid AND R.bid = 103;

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sid bid rday

22 101 10/10

58 103 11/12

Reserves

sid sname rating age

22 Dustin 7 45

58 Rusty 10 35

31 Lubber 8 55

Sailors

558Lubber3110/1010122

457Dustin2211/1210358

3510Rusty5811/1210358

58

22

22

sid

103

101

101

bid

11/12

10/10

10/10

rday

3510Rusty58

8

7

rating

55Lubber31

45Dustin22

agesnamesid

Reserves x Sailors

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SELECT DISTINCT snameFROM Sailors S, Reserves RWHERE S.sid = R.sid AND R.bid = 103

What

s the effect of adding DISTINCT?

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Another Example

Schema: Sailors (sid, sname, rating, age) Boats (bid, bname, color)

Reserves (sid, bid, day)

Find the colors of boats reserved by asailor named rusty

SELECT B.colorFROM Sailors S, Reserves R, Boats BWHERE S.sid = R.sid AND R.bid = B.bid AND

S.sname = 'Rusty';

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Note on Range Variables

Needed when same relation appears twicein FROM clause

SELECT S1.sname, S2.snameFROM Sailors S1, Sailors S2WHERE S1.age > S2.age;

What does thisQuery compute?

Good style to always use range variables anyway

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ORDER BY clause

Helps sort the result for presentation

SELECT S.sname, S.ageFROM Sailors SORDER BY S.age [ASC]

SELECT S.sname, S.ageFROM Sailors SORDER BY S.age DESC

Find the names and agesof all sailors, in increasingorder of age

Find the names and agesof all sailors, in decreasingorder of age

Attribute(s) in ORDER BY clause must be in SELECT list.

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ORDER BY clause

SELECT S.sname, S.age, S.ratingFROM Sailors SORDER BY S.age ASC, S.rating DESC

Find the names, ages, and rankings of all sailors.

Sort the result in increasing order of age.

If there is a tie, sort those tuples in decreasing order ofrating.

What does this query compute?

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Union

Find names of sailors who have reserved ared or a green boat

SELECT S.snameFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid AND R.bid = B.bidAND (B.color = 'red' OR B.color = 'green');


WHERE S.sid = R.sid AND R.bid = B.bid AND B.color = 'red'UNIONSELECT S.snameFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid and R.bid = B.bid AND B.color = 'green';

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Intersect

Find sids of sailors who have reserved ared and a green boat

What is wrong with the above query?


WHERE S.sid = R.sid AND R.bid = B.bidAND (B.color = 'red' AND B.color = 'green');

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Intersect

Find sids of sailors who have reserved ared and a green boat

SELECT S.sidFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid AND R.bid = B.bid AND B.color = 'red'INTERSECTSELECT S.sidFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid and R.bid = B.bid AND B.color = 'green';

What if we replace sid with sname (non-key)?

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Intersect on Non-Key

Find the names of sailors who havereserved a red and a green boat


WHERE S.sid = R.sid AND R.bid = B.bid AND B.color = 'red'INTERSECTSELECT S.snameFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid and R.bid = B.bid AND B.color = 'green'

What is wrong with the above query?

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Intersect on non-key

Find the names of sailors who havereserved a red and a green boat

CREATE VIEW RedGreenSailors ASSELECT S.sidFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid AND R.bid = B.bid AND B.color = 'red'INTERSECTSELECT S.sidFROM Sailors S, Reserves R, Boats BWHERE S.sid = R.sid and R.bid = B.bid AND B.color = 'green';

SELECT S.sname FROM

Sailors S, RedGreenSailors R WHERE

S.sid = R.sid;

DROP VIEW RedGreenSailors;

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Intersect with non-key

Find the names of sailors who havereserved a red and a green boat. Get rid ofthe VIEW.

SELECT S.sname FROM

Sailors S,

(SELECT S.sidFROM Sailors S, Reserves R, Boats BWHERE S.sid = R.sid AND R.bid = B.bid AND B.color = 'red'INTERSECTSELECT S.sidFROM Sailors S, Reserves R, Boats B

WHERE S.sid = R.sid and R.bid = B.bid AND B.color = 'green')RedGreenSailors WHERES.sid = RedGreenSailors.sid;

Another sol:See Q8 inCh. 5, p. 150

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Nested Queries

Query with another query embedded insideSELECT S.snameFROM Sailors SWHERE S.sid IN (SELECT R.sid

FROM Reserves RWHERE R.bid=103)

What does thisCompute?

Conceptual evaluation: For each row of Sailors,evaluate the subquery over reserves

To find sailors who have not reserved 103, use NOT IN

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More Set Operators

Also EXCEPT (set difference) (Painful) SQL oddities

For UNION, default is to eliminate duplicates!

To keep duplicates, must specify UNION ALL

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Over-Use of Nesting

Common error by novice SQLprogrammers.

Query optimizers not as good at optimizingqueries across nesting boundaries.

Try hard first to write non-nested.

SELECT DISTINCT S.sname

FROM Sailors S, Reserves RWHERE S.sid = R.sid AND R.bid = 103;

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More Set Comparison Operators

Weve already seen IN, EXISTS. Can also use NOT IN,NOT EXISTS.

Also available: opANY, opALL, op is , , =,

Find sailors whose rating is greater than that ofall

sailors called Horatio:SELECT *FROM Sailors SWHERE S.rating >ALL (SELECT S2.rating

FROM Sailors S2WHERE S2.sname='Horatio')

What if we replace ALL with ANY?

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Aggregate OperatorsCOUNT (*)COUNT ( [DISTINCT] A)SUM ( [DISTINCT] A)

AVG ( [DISTINCT] A)MAX (A) Can use DistinctMIN (A) Can use Distinct

SELECT AVG (S.age)FROM Sailors SWHERE S.rating=10

SELECT COUNT (*) FROM Sailors S

SELECT AVG ( DISTINCT S.age)FROM Sailors SWHERE S.rating=10

SELECT S.sname FROM Sailors SWHERE S.rating= (SELECT MAX(S2.rating) FROM Sailors S2)

single column

SELECT COUNT (DISTINCT S.name)FROM Sailors S

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Aggregate Query - Example

Find name and age of oldest sailor(s)SELECT S.sname, MAX (S.age)FROM Sailors S

SELECT S.sname, S.ageFROM Sailors SWHERE S.age =

(SELECT MAX (S2.age)

FROM Sailors S2)

How many tuplesin the result?

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GROUP BY

Conceptual evaluation Partition data into groups according to somecriterion

Evaluate the aggregate for each group

Example: For each rating level, find the ageof the youngest sailor

SELECT MIN (S.age), S.ratingFROM Sailors SGROUP BY S.rating

How many tuplesin the result?

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GROUP BY and HAVING

SELECT [DISTINCT] target-listFROM relation-listWHERE qualificationGROUP BY grouping-listHAVING group-qualification

Conceptual Evaluation:

1. Eliminate tuples that dont satisfy qualification

2. Partition remaining data into groups3. Eliminate groups according to group-qualification

4. Evaluate aggregate operation(s) for each group

Target-list contains:Attribute names (subsetof grouping-list)Aggregate operations(e.g., min(age))

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Find the age of the youngest sailor with age >= 18,for each rating with at least 2 such sailors

SELECT S.rating, MIN (S.age)FROM Sailors SWHERE S.age >= 18GROUP BY S.ratingHAVING COUNT (*) >= 2


22 dustin 7 45.031 lubber 8 55.5

71 zorba 10 16.0

64 horatio 7 35.0

29 brutus 1 33.058 rusty 10 35.0

rating age

1 33.0

7 45.0

7 35.0

8 55.5

10 35.0

rating

7 35.0

Answer relation

rating

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NULL Values in SQL NULL represents unknown

or inapplicable Query evaluation

complications Q: Is (rating > 10) true when

rating is NULL?

A: Condition evaluates tounknown (not T or F)

What about AND, ORconnectives?

Need 3-valued logic WHERE clause eliminates

rows that dont evaluate totrue

p q p AND q p OR q

T T T T

T F F T

T U U T

F T F T

F F F F

F U F U

U T U T

U F F U

U U U U

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NULL Values Example


22 dustin 7 45

58 rusty 10 NULL

31 lubber 8 55

sailorsWhat does this queryreturn?

SELECT snameFROM sailors

WHERE age > 45OR age

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NULL Values in Aggregates

NULL values generally ignored whencomputing aggregates

Modulo some special cases (see textbook)

SELECT AVG(age)FROM sailors sid sname rating age

22 dustin 7 45

58 rusty 10 NULL31 lubber 8 55

sailors

Returns 50!

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Outer Join*

sid bid day

22 101 10/10/99

Reservessid sname rating age

22 dustin 7 45.0

58 rusty 10 35.0

Sailors

Select S.sid, R.sidFrom Sailors S NATURAL LEFT OUTER JOIN Reserves R

sid bid

22 101

58 null

ResultSimilarly:

RIGHT OUTER JOIN FULL OUTER JOIN

For each red boat find the number of

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For each red boat, find the number ofreservations for this boat*

SELECT B.bid, COUNT (*) AS scountFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sidAND R.bid=B.bidAND B.color='red'

GROUP BY B.bid

SELECT B.bid, COUNT (*) AS scountFROM Sailors S, Boats B, Reserves RWHERE S.sid=R.sidAND R.bid=B.bidGROUP BY B.bid -- Would this work?HAVING B.color = 'red' --note: one color per bid

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Subtle Errors

Find the sid of sailors who have reservedexactly one boat.

SELECT S1.sid

FROM Sailors S1EXCEPT

SELECT R1.sid

FROM Reserves R1, Boats B1, Reserves R2, Boats B2 WHERE R1.sid=R2.sid ANDR1.bid=B1.bid

AND R2.bid=B2.bid AND R1.bidR2.bid;

There is a subtle error in the above.

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Error fixed


SELECT R3.sid

FROM Reserves R3MINUS

SELECT R1.sid

FROM Reserves R1, Boats B1, Reserves R2, Boats B2 WHERE R1.sid=R2.sid ANDR1.bid=B1.bid

AND R2.bid=B2.bid AND R1.bid R2.bid;

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Another Solution


SELECT S.sid FROM Sailors S, Boats B, Reserves R

WHERES.sid = R.sid AND B.bid = R.bid

GROUP BY S.sid

HAVING COUNT(*) = 1;

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Fi d h f h il i h 18

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Find the age of the youngest sailor with age>18,for each rating with at least 2 sailors (of any age)

Subquery in the HAVING clause

Compare this with the query where we consideredonly ratings with 2 sailors over 18!

SELECT S.rating, MIN (S.age) AS MINAGEFROM Sailors SWHERE S.age > 18GROUP BY S.rating

HAVING 1 < (SELECT COUNT (*) FROM Sailors S2WHERE S2.rating=S.rating)

i d i f hi h h i h

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Find ratings for which the average age is theminimum of the average age over all ratings*

Aggregate operations cannot be nested! WRONG:SELECT S.ratingFROM Sailors SWHERE AVG(S.age) =

(SELECT MIN (AVG (S2.age)) FROM Sailors S2)

SELECT T.rating, T.avgageFROM (SELECT S.rating,AVG (S.age) AS avgage FROM Sailors S

GROUP BY S.rating) TWHERE T.avgage = (SELECT MIN (T.avgage) FROM T);

Correct solution (in SQL/92 may fail in practice):

If above does not work, one solution is to define T as a view .

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Solution Using Views

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CREATE VIEW AVG_AGE_BY_RATING ASSELECT S.rating, AVG(S.age) AS avgageFROM Sailors S;

SELECT T.rating, T.avgage FROMAVG_AGE_BY_RATING TWHERE T.avgage= (SELECT MIN(A.avgage) FROM

AVG_AGE_BY_RATING A);

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Triggers, general ICs

Review the end of lecture 4Avoid triggers whenever possible

Why?

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Suggested Review

Exercises 5.1, 5.3, 5.7

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