SQL: Structured Query Language
(‘Sequel’)
Chapter 5
1
Running Example

Instances of the Sailors
and Reserves relations
in our examples.
R1
sid
22
58
bid
101
103
day
10/10/96
11/12/96
S1
sid
22
31
58
sname
dustin
lubber
rusty
rating
7
8
10
S2
sid
28
31
44
58
sname
yuppy
lubber
guppy
rusty
rating
9
8
5
10
age
45.0
55.5
35.0
age
35.0
55.5
35.0
35.0
3
Basic SQL Query
SELECT
FROM
WHERE



[DISTINCT] target-list
relation-list
qualification
relation-list A list of relation names
target-list A list of attributes of relations in relation-list
qualification
• Comparisons (Attr op const or Attr1 op Attr2, where op is one of
, ,  , , ,  ) combined using AND, OR and NOT.

DISTINCT is optional keyword indicating that answer
should not contain duplicates. Default is that
duplicates are not eliminated!
4
Example Query
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
5
Comparisons in Oracle
6
Conceptual Evaluation Strategy
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103

Semantics of SQL query defined in terms of
conceptual evaluation strategy:
Compute cross-product of relation-list.
• Discard resulting tuples if they fail qualifications.
• Delete attributes that are not in target-list.
• If DISTINCT is specified, eliminate duplicate rows.
•
Probably the least efficient way!
 Optimizer must find more efficient strategies to
compute same answers.

7
R
sid
22
58
bid
101
103
day
10/10/96
11/12/96
S
sid
22
31
58
sname
dustin
lubber
rusty
rating
7
8
10
age
45.0
55.5
35.0
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
(sid) sname rating age
(sid) bid day
22 dustin
7
45.0
22
101 10/10/96
22 dustin
7
45.0
58
103 11/12/96
31 lubber
8
55.5
22
101 10/10/96
31 lubber
8
55.5
58
103 11/12/96
58 rusty
10
35.0
22
101 10/10/96
58 rusty
10
35.0
58
103 11/12/96
8
A Note on Range Variables


Needed only if same relation appears twice in FROM clause.
The previous query can also be written as:
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND bid=103
OR
SELECT sname
FROM Sailors, Reserves
WHERE Sailors.sid=Reserves.sid
AND bid=103
It is good style, however, to use range variables always!
10
Find sailors who’ve reserved at least one boat
SELECT S.sid
FROM Sailors S, Reserves R
WHERE S.sid=R.sid

Question: What is the output of this query ?
11
Expressions and Strings
SELECT S.age, age1=S.age-5, 2*S.age AS age2
FROM Sailors S
WHERE S.sname LIKE ‘B_%B’

AS and = are two ways to name fields in result.

LIKE is used for string matching. `_’ stands for any one
character and `%’ stands for 0 or more arbitrary characters.


MEANING of Query:
Find sailors (age of sailor and two fields defined by expressions)
whose names begin and end with B and contain at least three
characters.
13
SQL Queries
Using Set Operations
14
Find sid’s of sailors who’ve reserved a red
or a green boat
SELECT S.sid
FROM Sailors S, Reserves R, Boats B
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.color=‘red’ OR B.color=‘green’)

UNION
compute union
of any two
union-compatible
sets of tuples
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
UNION
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid AND B.color=‘green’
15
Find sid’s of sailors who’ve reserved
a red and a green boat.

If we replace OR by AND as done below,
what are the semantics?
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.color=‘red’ AND B.color=‘green’)
17
Find sid’s of sailors who’ve reserved a red and a
green boat


INTERSECT: compute
intersection of two
union-compatible sets of
tuples.
What happens if
change S.sid as
S.sname ?
Key field!
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
INTERSECT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘green’
18
Find sid’s of sailors who’ve reserved a red and a
green boat
SELECT S.sid

Could query be
written without
INTERSECT?
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
INTERSECT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘green’
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE (S.sid=R1.sid AND R1.bid=B1.bid AND B1.color=‘red’ )
AND (S.sid=R2.sid AND R2.bid=B2.bid AND B2.color=‘green’)

Do we need Sailors relation in the FROM clauses?
20
Find sid’s of sailors who’ve reserved a red and not a green boat

What if we want to compute
the DIFFERENCE instead?
SQL/92
standard,
but some systems don’t
support EXCEPT.
What
then?
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
EXCEPT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘green’
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color='red' AND NOT(B2.color = 'green'))
22
Find sid’s of sailors who’ve reserved a red and not a green boat

What if we want to compute
the DIFFERENCE instead?
Except
is SQL/92 standard,
but some systems don’t
. support EXCEPT.
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘red’
EXCEPT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND B.color=‘green’
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color='red' AND NOT(B2.color = 'green'))

Is the above query correct, or not correct ?
23
About duplication

Default: Eliminate duplicate

Unless we use special keyword “ALL” :
# of copies of a row in result
UNION ALL
INTERSECT ALL
EXCEPT ALL
(m+n)
min(m,n)
m-n
24
Nested Queries
25
Nested Queries
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)

WHERE, FROM, HAVING clauses can itself contain
another SQL query, called a subquery.
 powerful feature of SQL.
26
Nested Queries Evaluation
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)

Nested loops evaluation:
 For each Sailors tuple,
• Evaluate the WHERE clause qualification
• Which here means re-compute subquery.
27
Nested Queries
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)

Change query to find sailors who’ve not reserved #103:
SELECT S.sname
FROM Sailors S
WHERE S.sid NOT IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)
28
Nested Queries
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)

Could this query be rewritten without nesting?
29
Nested Queries with Correlation
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)

EXISTS is another set comparison operator, like IN.
30
Nested Queries with Correlation
Find names of sailors who’ve reserved boat #103:
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)

Why, in general, this subquery must be recomputed for each Sailors tuple ?

Could this query be rewritten into a flat
SQL query?
31
More on Set-Comparison Operators
op = {
, , , ,, 

op ANY and op ALL where
}

Find sailors whose rating is greater than that of some sailor
called Horatio:
SELECT *
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘Horatio’)

IN is equal to =ANY

NOT IN is equal to <>ANY
33
More on Set-Comparison Operators
Find sailors whose rating is greater than that
of all sailors called Horatio:

SELECT *
FROM Sailors S
WHERE S.rating > ALL (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘Horatio’)
What if there is no other sailor with that name?
 “> ALL” would return TRUE if set is empty.

34
Rewriting INTERSECT Queries Using IN
Find sid’s of sailors who’ve reserved both a red and a green boat:
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
AND S.sid IN (SELECT S2.sid
FROM Sailors S2, Boats B2, Reserves R2
WHERE S2.sid=R2.sid AND R2.bid=B2.bid
AND B2.color=‘green’)
35
Rewriting INTERSECT Queries Using IN
Find sname’s of sailors who’ve reserved both a red and a green boat:
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
AND S.sid IN (SELECT S2.sid
FROM Sailors S2, Boats B2, Reserves R2
WHERE S2.sid=R2.sid AND R2.bid=B2.bid
AND B2.color=‘green’)

INTERSECT queries re-written using IN.

Note: This query can return “sname” (and not just “sid) of
Sailors who’ve reserved both red and green boats
36
Division in SQL
Find sailors who’ve reserved all boats.
37
Division in SQL
Find sailors who’ve reserved all boats.
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
((SELECT B.bid
FROM Boats B)
EXCEPT
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
38
(1)
Division in SQL
Find sailors
who’ve reserved all boats.

Can we do it without EXCEPT ?
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
((SELECT B.bid
FROM Boats B)
EXCEPT
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
39
(1)
Division in SQL
Find sailors
who’ve reserved all boats.

Let’s do it without EXCEPT:
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
((SELECT B.bid
FROM Boats B)
EXCEPT
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
(2) SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (SELECT B.bid
FROM Boats B
WHERE NOT EXISTS (SELECT R.bid
Sailors S such that ...
FROM Reserves R
there is no boat B without ...
WHERE R.bid=B.bid
AND R.sid=S.sid))
a Reserves tuple showing S reserved B
40
Aggregation and
Having Clauses
41
Aggregate Operators

Significant extension of relational algebra.
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
Why no Distinct?
42
Aggregate Operators
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
SELECT COUNT (*)
FROM Sailors S
Why no Distinct?
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
SELECT COUNT (DISTINCT S.rating)
FROM Sailors S
WHERE S.sname=‘Bob’
43
More Examples of Aggregate Operators
SELECT S.sname
FROM Sailors S
WHERE S.rating= (SELECT MAX(S2.rating)
FROM Sailors S2)
SELECT AVG ( DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
44
Find name and age of the oldest sailor(s)
SELECT S.sname, MAX (S.age)
FROM Sailors S
•Is this first query legal?
•No!
•Why not ?
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age =
(SELECT MAX (S2.age)
FROM Sailors S2)
Is this second query legal?
47
Motivation for Grouping


So far, aggregate operators to all (qualifying) tuples.
Question?
 What if want to apply aggregate to each group of tuples.

Example :
 Find the age of the youngest sailor for each rating level.

Example Procedure :
 Suppose rating values  {1,2,…,10}, 10 queries:
For i = 1, 2, ... , 10:
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
48
Motivation for Grouping
For i = 1, 2, ... , 10:

SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
What’s the problem with above ?
• We don’t know how many rating levels exist.
• Nor what the rating values for these levels are.
49
Add Group By to SQL
50
Queries With GROUP BY and HAVING
SELECT
FROM
WHERE
GROUP BY
HAVING


[DISTINCT] target-list
relation-list
qualification
grouping-list
group-qualification
A group is a set of tuples that each have the same
value for all attributes in grouping-list.
HAVING clause is a restriction on each group.
51
Queries With GROUP BY and HAVING
SELECT
FROM
WHERE
GROUP BY
HAVING

[DISTINCT] target-list
relation-list
qualification
grouping-list
group-qualification
target-list contains :
(i) attribute names
(ii) aggregate-op (column-name)
(e.g., MIN (S.age)).

A group is a set of tuples that each have the same value for all
attributes in grouping-list.

REQUIREMENT:
- target-list (i)  grouping-list.
- Why?
- Each answer tuple of a group must have single value.
52
Find age of the youngest sailor with age  18,
for each rating with at least 2 such sailors
SELECT S.rating, MIN (S.age) AS min-age
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
53
GroupBy --- Conceptual Evaluation
1.
2.
3.
4.
5.
Compute the cross-product of relation-list
Discard tuples that fail qualification
Delete `unnecessary’ fields
Partition the remaining tuples into groups by the value of
attributes in grouping-list. (GroupBy)
Eliminate groups using the group-qualification (Having)
We will have a single value per group
That is, one answer tuple is generated per qualifying group.
54
Find age of the youngest sailor with age  18,
for each rating with at least 2 such sailors
SELECT S.rating, MIN (S.age)
AS min-age
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
Sailors instance:
sid
22
29
31
32
58
64
71
74
85
95
96
sname rating age
dustin
7 45.0
brutus
1 33.0
lubber
8 55.5
andy
8 25.5
rusty
10 35.0
horatio
7 35.0
zorba
10 16.0
horatio
9 35.0
art
3 25.5
bob
3 63.5
frodo
3 25.5
55
Find age of the youngest sailor with age  18,
for each rating with at least 2 such sailors.
rating
7
1
8
8
10
7
10
9
3
3
3
age
45.0
33.0
55.5
25.5
35.0
35.0
16.0
35.0
25.5
63.5
25.5
rating
1
3
3
3
7
7
8
8
9
10
age
33.0
25.5
63.5
25.5
45.0
35.0
55.5
25.5
35.0
35.0
rating
3
7
8
minage
25.5
35.0
25.5
56
Find age of the youngest sailor with age  18, for
each rating with at least 2 sailors between 18 and 60.
SELECT S.rating, MIN (S.age)
AS min-age
FROM Sailors S
WHERE S.age >= 18 AND S.age <= 60
GROUP BY S.rating
HAVING COUNT (*) > 1
Answer relation:
rating
3
7
8
minage
25.5
35.0
25.5
Sailors instance:
sid
22
29
31
32
58
64
71
74
85
95
96
sname rating age
dustin
7 45.0
brutus
1 33.0
lubber
8 55.5
andy
8 25.5
rusty
10 35.0
horatio
7 35.0
zorba
10 16.0
horatio
9 35.0
art
3 25.5
bob
3 63.5
frodo
3 25.5
60
For each red boat, find the number of
reservations for this boat
SELECT B.bid, COUNT (*) AS s-count
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid


Grouping over Join of three relations.
Q: What do we get if we remove B.color=‘red’ from WHERE
clause and add HAVING clause with this condition?
• No difference. But Illegal.
• Only column in GroupBy can appear in Having,
unless in aggregate operator of Having
61
Find age of the youngest sailor with age > 18,
for each rating with at least 2 sailors (of any age)

HAVING clause can also contain a subquery.
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age > 18
GROUP BY S.rating
HAVING 1 < (SELECT COUNT (*)
FROM Sailors S2
WHERE S.rating=S2.rating)
Find age of the youngest sailor with age > 18,
for each rating with at least 2 such sailors (of age 18)
HAVING COUNT(*) >1
62
Find those ratings for which the average age is
the minimum over all ratings
SELECT S.rating
FROM Sailors S
WHERE S.age = (SELECT MIN (AVG (S2.age))
FROM Sailors S2)

WRONG : Aggregate operations cannot be nested!
63
Find those ratings for which the average age is
the minimum over all ratings

Correct solution (in SQL/92):
SELECT Temp.rating, Temp.avg-age
FROM (SELECT S.rating, AVG (S.age) AS avg-age
FROM Sailors S
GROUP BY S.rating) AS Temp
WHERE Temp.avg-age = (SELECT MIN (avg-age)
FROM (SELECT S.rating, AVG(S.age) as avg_age
FROM Sailors S
GROUP BY S.rating))
65
Null Values
66
Null Values

Field values in a tuple are sometimes :
 Unknown (e.g., a rating has not been assigned) or
 Inapplicable (e.g., no maiden-name when a male person)


SQL provides a special value null for such situations.
The presence of null complicates many issues.
67
Comparisons Using Null Values

(S.rating = 8) is TRUE or FALSE ??
What if rating has null value in tuple ?

We need a 3-valued logic : condition can be true, false or unknown.

SQL special operators IS NULL to check if value is/is not null.

Disallow NULL value :

rating INTEGER NOT NULL
68
Comparison with NULL values

Arithmetic operations on NULL return NULL.

Comparison operators on NULL return UNKNOWN.

We can explicitly check whether a value is null or not
by IS NULL and by IS NOT NULL.
69
Truth table with UNKNOWN
UNKNOWN AND TRUE = UNKNOWN
UNKNOWN OR TRUE = TRUE
UNKNOWN AND FALSE = FALSE
UNKNOWN OR FALSE = UNKNOWN
UNKNOWN AND UNKNOWN = UNKNOWN
UNKNOWN OR UNKNOWN = UNKNOWN
NOT UNKNOWN = UNKNOWN
WHERE clause is satisfied only when it evaluates to TRUE.
70
Outer Joins : Special Operators



Left Outer Join
Right Outer Join
Full Outer Join
SELECT S.sid, R.bid
FROM Sailors S LEFT OUTER JOIN Reserves R
ON S.sid = R.sid
Sailors rows (left) without a matching Reserves row
(right) appear in result, but not vice versa.
SELECT S.sid, R.bid
FROM Sailors S NATURAL LEFT OUTER JOIN Reserves R
71
Sorting: ORDER BY clause
SELECT *
FROM Student
WHERE sNumber >= 1
ORDER BY sNumber, sName
 (sNumber, sName) ( (sNumber >= 1) (Student))
72
SQL Queries - Summary
SELECT [DISTINCT] a1, a2, …, an
FROM R1, R2, …, Rm
[WHERE C1]
[GROUP BY g1, g2, …, gl [HAVING C2]]
[ORDER BY o1, o2, …, oj]
73
Summary

SQL was important factor in early acceptance of
relational model : easy-to-understand !

Relationally complete: in fact, significantly more
expressive power than relational algebra.

Many alternative ways to write a query. So optimizer
must look for most efficient evaluation plan.

In practice, users (still) need to be aware of how
queries are optimized and evaluated for best results.
74