SQL REVIEW
Example Schema
We will use these table definitions in our examples.
Sailors(sid: integer, sname: string, rating: integer, age: real)
Boats(bid: integer, bname: string, color: string)
Reserves(sid: integer, bid: integer, day: date)
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 an optional keyword indicating that the answer
should not contain duplicates.
Querying Relations (1)
What does the following query compute?
Enrolled
sid
53831
53831
53650
53666
cid
grade
Carnatic101
C
Reggae203
B
Topology112
A
History105
B
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.sid AND E.grade=“A”
A student with
“sid” has an
entry in Enrolled
Students
sid
name
login
age
gpa
53831
Zhang
zhang@ee
19
3.2
53650
Smith
smith@cs
21
3.9
53666
Jones
jones@cs
20
3.5
The Enrolled
entry has a
grade of “A”
Retrieve names of students
and the courses they
received an “A” grade
Querying Relations (2)
Enrolled
sid
53831
53831
53650
53666
cid
grade
Carnatic101
C
Reggae203
B
Topology112
A
History105
B
Students
sid
name
login
age
gpa
53831
Zhang
zhang@ee
19
3.2
53650
Smith
smith@cs
21
3.9
53666
Jones
jones@cs
20
3.5
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid=E.sid AND E.grade=“A”
S.name
E.cid
Smith
Topology112
Semantics of SQL
QUERY PROCESSOR
SELECT
FROM
WHERE
[DISTINCT] target-list
relation-list
qualification
This strategy is probably the least efficient way
to compute a query! An optimizer will find
more efficient strategies to compute the same
answers.
Query Result
R1 × R2 × R3 × · · ·
Define search space
˂, ˃, ≤, ≥, =, ≠
Select rows
Projection
Select columns
Example of Conceptual Evaluation
Range
SELECT S.sname
variable
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
Reservation sid bid
22
58
101
103
Sailors sid sname rating age
22
31
58
dustin
lubber
rusty
7
8
10
45.0
55.5
35.0
day
10/10/96
11/12/96
Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
S×R
(sid)
22
sname
dustin
rating
7
age
45.0
(sid)
22
bid
day
101 10/10/96
22
31
31
dustin
lubber
lubber
7
8
8
45.0
55.5
55.5
58
22
58
103 11/12/96
101 10/10/96
103 11/12/96
58
58
rusty
rusty
10
10
35.0
35.0
22
58
101 10/10/96
103 11/12/96
Answer
A Note on Range Variables
Really needed only if the same relation appears twice in the
FROM clause. The previous query can be written in two ways:
Range variable
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.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!
Aggregate Operators
Significant extension of relational algebra
COUNT (*)
The number of rows in the relation
The number of (unique) values in the A
COUNT ([DISTINCT] A)
column
The sum of all (unique) values in the A
SUM ([DISTINCT] A)
column
The average of all (unique) values in the A
AVG ([DISTINCT] A)
column
MAX (A)
The maximum value in the A column
MIN (A)
The minimum value in the A column
Aggregate Operators
SELECT COUNT (*)
FROM Sailors S
SELECT AVG (S.age)
FROM
Sailors S
WHERE S.rating=10
Count the Find the name of
number of sailors with the
sailors
highest rating
Compute
maximum
rating
SELECT S.sname
FROM
S age
Find the Sailors
average
WHERE
S.rating=
(S2.rating)
Find the MAX
average
of the
of sailors
with a(SELECT
Count the number FROM
ofdistinct ages
Sailors
ofS2)
sailors
rating of 10
distinct ratings of with a rating of 10
sailors called “Bob”
SELECT COUNT (DISTINCT S.rating)
FROM
Sailors S
WHERE S.sname=‘Bob’
SELECT AVG (DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
GROUP BY and HAVING (1)
• So far, we’ve applied aggregate operators to all
(qualifying) tuples.
32
Aggregator
Qualifier
Relation
Aggregator
SELECT AVG (S.age)
FROM
Sailors S
WHERE S.rating=10
Qualifier
Find the average age
of sailors with a
rating of 10
GROUP BY and HAVING (1)
• So far, we’ve applied aggregate operators to all
(qualifying) tuples.
Aggregator
32
Qualifier
Relation
• Sometimes, we want to apply them to each of several
groups of tuples.
12
Aggregator
Group 1
9
Aggregator
Group 2
11
Aggregator
Group 3
Relation
Queries With GROUP BY and HAVING
SELECT
FROM
WHERE
GROUP BY
HAVING
[DISTINCT] target-list
relation-list
qualification
MIN(Attribute)
grouping-list
group-qualification
HAVING
12
Aggregator
9
Aggregator
Qualifier
selecting
groups
GROUP BY
Group 1
Group 2
Group 3
Output
a table
SELECT
FROM
WHERE
Find the age of the youngest sailor with age ≥
18, for each rating with at least 2 such sailors
SELECT
FROM
WHERE
GROUP BY
HAVING
S.rating, MIN (S.age)
Sailors S
S.age >= 18
S.rating
COUNT (*) > 1
rating age
7
35.0
Only one
4 rating
group
groups
satisfies
HAVING
Answer
rating age
Disqualify
1
7
7
8
10
33.0
45.0
35.0
55.5
35.0
Input relation
Sailors
sid
22
31
71
64
29
58
sname rating age
dustin
7
45.0
lubber
8
55.5
zorba
10 16.0
horatio
7
35.0
brutus
1
33.0
rusty
10 35.0
Only S.rating and S.age are
mentioned in SELECT
Summary
• SQL was an important factor in the early acceptance of
the relational model; more natural than earlier,
procedural query languages.
• Relationally complete; in fact, significantly more
expressive power than relational algebra.
• Even queries that can be expressed in RA can often be
expressed more naturally in SQL.
• Many alternative ways to write a query; optimizer
should look for most efficient evaluation plan.

In practice, users need to be aware of how queries are
optimized and evaluated for best results.