SQL 2 – The Sequel
CS 186, Spring 2007, Lecture 8
R&G, Chapter 5
Mary Roth
The important thing is not to
stop questioning.
Albert Einstein
Life is just a bowl of queries.
-Anon
(not Forrest Gump)

Administrivia
•
Homework 1 due Tuesday, Feb 13 10 p.m.
•
Coming up:
– Homework 2 handed out Feb 13
– Midterm 1: in class February 22
•
Questions?
Review: SQL DML
• DML includes 4 main statements:
SELECT (query), INSERT, UPDATE and DELETE
PROJECT
• e.g:
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid = E.sid AND
JOIN
S.age=19
SELECT
Example: Find sailors who have reserved a red
and a green boat
SELECT R.sid
FROM Boats B,Reserves R
WHERE B.color = ‘red’
AND R.bid=B.bid
INTERSECT
SELECT R.sid
FROM Boats B,Reserves R
WHERE B.color = ‘green’
AND R.bid=B.bid
sid
1
2

sid
1
=
Now let’s do this with
a self-join
Boats
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria red
105
Titanic
green
Reserves
sid
bid
day
sid
1
101
9/12
1
2
103
9/13
1
105
9/13
Find sids of sailors who’ve reserved a red and a green
boat
SELECT R1.sid
FROM Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE B1.color = ‘red’
AND B1.bid=R1.bid
Find red reserved boats
Find green reserved boats
AND B2.color = ‘green’
AND B2.bid=R2.bid
Find matching green and red
reserved boats
AND R1.sid=R2.sid
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria
red
105
Titanic
green
sid
bid
day
1
101
9/12
2
103
9/13
1
105
9/13
sid
1
2
sid
1
sid
=
1
Nested Queries
• The WHERE clause can itself contain an SQL query!
(Actually, so can FROM and HAVING clauses can too)
e.g. Find the names of sailors who’ve reserved boat #103:
Reserves
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
sname
WHERE R.bid=103)
Sailors
sid
sid
bid
day
1
101
9/12
2
103
9/13
1
105
9/13
Bilbo
sname
rating
age
S 1
Frodo
7
22
S 2
Bilbo
2
39
S 3
Sam
8
27
First compute the set of all sailors that have
reserved boat 103…
…and then check each the sid of
each tuple in Sailors to see if it is
in S1
sid
2
Nested Queries
• This nested query was uncorrelated because the subquery
does not refer to anything in the enclosing query
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.bid=103)
It can evaluated once and then checked for
each tuple in enclosing query
Nested Queries with correlation
• Nested queries can also be correlated; the subquery
refers to the enclosing query
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND R.sid=S.sid)
The subquery must be re-evaluated for each tuple in
enclosing query
 EXISTS is a set operator that is true if result of set
expression has at least one tuple
Nested Queries
e.g. Find the names of sailors who’ve reserved boat #103:
Notice that this query
computes the same
answer as the previous
query!
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=103 AND R.sid=S.sid)
1
2
3
Sailors
sid
Reserves
sname
rating
age
S 1
Frodo
7
22
S 2
Bilbo
2
39
S 3
Sam
8
27
sid
bid
day
sid
1
101
9/12
2
2
103
9/13
1
105
9/13
bid
day
103 9/13
Set-Comparison Operators
• <tuple expression> IN <set expression>
•
•
– True if <tuple> is a member of <set>
– Also, NOT IN
EXISTS <set expression>
– True if <set expression> evaluates to a set with at least one member
– Also, NOT EXISTS
UNIQUE <set expression>
– True if <set expression> evaluates to a set with no duplicates; each
row can appear exactly once
– Also, NOT UNIQUE
• <tuple expression> comparison op ANY <set expression>
– True if <set expression> contains at least one member that makes the
comparison true
– Also, op ALL
Use NOT Exists for Division
Recall: X/Y means only give me X tuples that have a match in Y.
Find sailors who’ve reserved all boats.
X = set of sailors and Y = set of all boats with reservations.
SELECT S.sname
Find Sailors S such that ...
FROM Sailors S
WHERE NOT EXISTS
there is no boat B...
(SELECT B.bid
FROM Boats B
WHERE NOT EXISTS
(SELECT R.bid
without a reservation by
FROM Reserves R
Sailor S
WHERE R.bid=B.bid
AND R.sid=S.sid))
Division
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
(SELECT B.bid
FROM Boats B
WHERE NOT EXISTS
(SELECT R.bid
FROM Reserves R
101
103
WHERE R.bid=B.bid
1
3
AND R.sid=S.sid))
2
Reserves
Sailors
sid
sid
bid
day
R
1
103
9/12
R
R
R
R
2
103
9/13
3
103
9/14
3
101
9/12
1
103
9/13
Boats
sname
rating
age
S 1
Frodo
7
22
S 2
Bilbo
2
39
S 3
Sam
8
27
bid
bname
color
B
101
Nina
red
B
103
Pinta
blue
UNIQUE
Find the names of sailors who’ve reserved boat #103
exactly once
SELECT S.sname
FROM Sailors S
WHERE UNIQUE (SELECT sid, bid
FROM Reserves R
WHERE R.bid=103 AND S.sid=R.sid)
1
2
3
Sailors
sid
Reserves
sname
rating
age
S 1
Frodo
7
22
S 2
Bilbo
2
39
S 3
Sam
8
27
sid
bid
day
sid
1
103
9/12
2
2
103
9/13
1
103
9/13
bid
day
103 9/13
ANY
Find sailors whose rating is greater than that of some sailor
called Bilbo:
Correlated or
uncorrelated?
SELECT *
FROM Sailors S
WHERE S.rating > ANY(SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘Bilbo’)
S1
Uncorrelated!
S2
sid
sname
rating
age
sid
sname
rating
age
1
Frodo
7
22
2
1
Bilbo
Frodo
2
7
39
22
2
Bilbo
2
39
2
Bilbo
2
39
3
Sam
8
27
3
Sam
8
27
Intermission
Aggregate Operators
• Very powerful; enables computations over sets of tuples
• COUNT: returns a count of tuples in
the set
• AVG: returns average of column
values in the set
• SUM: returns sum of column values in
the set
• MIN, MAX: returns min (max) value of
column values in a set.
• DISTINCT can be added to COUNT,
AVG, SUM to perform computation
only over distinct values.
SELECT COUNT (*)
FROM Sailors S
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
SELECT AVG(DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
Aggregate Operators
Find name and age of the oldest sailor(s)
SELECT S.sname, MAX (S.age)
FROM Sailors S
X
What will the result be?
sname
age
Frodo
39
Bilbo
39
Sam
39
Sailors
sid
sname
rating
age
1
Frodo
7
22
2
Bilbo
2
39
3
Sam
8
27
Not legal syntax; no other columns
allowed in SELECT clause without a
GROUP BY clause
(we’ll learn about those next)
Aggregate Operators
Find name and age of the oldest sailor(s)
Instead:
SELECT S.sname, S.age
And then find the
FROM Sailors S
sailors(s) of that age…
WHERE S.age =
Find the maximum age…
(SELECT MAX (S2.age)
FROM Sailors S2)
GROUP BY and HAVING
• So far, we’ve seen aggregate operators applied to all
tuples.
– What if we want to apply ops to each of several groups of tuples?
• Consider: Find the age of the youngest sailor for each
rating level.
– In general, we don’t know how many rating levels exist, and what
the rating values for these levels are!
– Suppose we know that rating values go from 1 to 10; we can
write 10 queries that look like this (!):
For i = 1, 2, ... , 10:
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
Queries With GROUP BY
• To generate values for a column based on groups of rows,
use aggregate functions in SELECT statements with the
GROUP BY clause
Returning 1 row per group
SELECT [DISTINCT] target-list
FROM relation-list
[WHERE qualification]
GROUP BY grouping-list
And finally compute aggregate
function over each group…
First select these rows…
Then group them by the values
in these columns…
target-list contains:
• list of column names from grouping–list
• terms with aggregate operations (e.g., MIN (S.age)).
Group By Example
For each rating, find the age of the youngest
sailor with age  18
Sailors
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
7
22
2
21
8
27
Group 1
Group 2
Group 3
sid
sname
rating
age
1
Frodo
7
22
2
Bilbo
2
39
3
Sam
8
27
4
Pippin
2
21
5
Merry
8
17
1
Frodo
7
22
2
Bilbo
2
39
4
Pippin
2
21
3
Sam
8
27
Find the number of reservations for each
red boat.
SELECT B.bid, COUNT(*)AS tot_res
FROM Boats B, Reserves R
Boats
WHERE R.bid=B.bid AND
bid
bname
B.color=‘red’
101
Nina
GROUP BY B.bid
Pinta
Santa Maria red
1
103
101
2
Reserves
103
9/13
3
101
9/14
4
101
9/14
red
102
103
2
color
blue
sid
bid
day
1
102
9/12
2
103
9/13
3
101
9/14
4
101
9/14
Queries With GROUP BY and HAVING
SELECT
[DISTINCT] target-list
FROM
relation-list
WHERE
qualification
GROUP BY grouping-list
HAVING
group-qualification
• Use the HAVING clause with the GROUP BY clause to restrict
which group-rows are returned in the result set
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 S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1
rating
1
7
7
2
8
10
age
33.0
45.0
35.0
55.5
35.0
rating
1
7
38
10
m-age
33.0
35.0
55.0
35.0
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
count
1
2
1
1
rating
7
35.0
Answer
Sailors who have reserved
all boats
Sailors
sid sname rating age
1
Frodo
7
22
2
Bilbo
2
39
3
Sam
8
27
SELECT S.name
FROM Sailors S, reserves R
WHERE S.sid = R.sid
GROUP BY S.name, S.sid
HAVING COUNT(DISTINCT R.bid) =
( Select COUNT (*) FROM Boats)
Boats
sname
sid
Frodo
bid
bname
color
101
Nina
red
bid
102
Pinta
blue
1
102
103
Santa Maria
red
Bilbo
2
101
Bilbo
2
102
Frodo
1
102
Bilbo
2
103
Reserves
sname
sid
count
Frodo
1
1
count
sid
bid
day
Bilbo
2
3
3
1
102
9/12
2
102
9/12
sname
sid
bid
2
101
9/14
Frodo
1
102,102
Bilbo
2
101, 102, 103
1
102
9/10
2
103
9/13
More about Joins
SELECT (column_list)
FROM table_name
[INNER | {LEFT |RIGHT | FULL } OUTER] JOIN table_name
ON qualification_list
WHERE …
Explicit join semantics needed unless it is an INNER join
(INNER is default)
Default semantics: Inner Join
Only rows that match search conditions are returned.
SELECT s.sid, s.name, r.bid
FROM Sailors s INNER JOIN Reserves r
ON s.sid = r.sid
Returns only those sailors who have reserved boats
SQL-92 also allows:
SELECT s.sid, s.name, r.bid
FROM Sailors s NATURAL JOIN Reserves r
“NATURAL” means equi-join for each pair of attributes
with the same name
SELECT s.sid, s.name, r.bid
FROM Sailors s INNER JOIN Reserves r
ON s.sid = r.sid
sid
22
31
95
sname rating age
Dustin
7
45.0
Lubber 8
55.5
Bob
3
63.5
sid bid
day
22 101 10/10/96
95 103 11/12/96
s.sid s.name r.bid
22 Dustin
101
95 Bob
103
Left Outer Join
Left Outer Join returns all matched rows,
plus all unmatched rows from the table on the
left of the join clause
(use nulls in fields of non-matching tuples)
SELECT s.sid, s.name, r.bid
FROM Sailors s LEFT OUTER JOIN Reserves r
ON s.sid = r.sid
Returns all sailors & information on whether they
have reserved boats
SELECT s.sid, s.name, r.bid
FROM Sailors s LEFT OUTER JOIN Reserves r
ON s.sid = r.sid
sid
22
31
95
sname rating age
Dustin
7
45.0
Lubber 8
55.5
Bob
3
63.5
s.sid
22
95
31
sid bid
day
22 101 10/10/96
95 103 11/12/96
s.name r.bid
Dustin
101
Bob
103
Lubber
Right Outer Join
Right Outer Join returns all matched rows, plus
all unmatched rows from the table on the right
of the join clause
SELECT r.sid, b.bid, b.name
FROM Reserves r RIGHT OUTER JOIN Boats b
ON r.bid = b.bid
Returns all boats & information on which ones
are reserved.
SELECT r.sid, b.bid, b.name
FROM Reserves r RIGHT OUTER JOIN Boats b
ON r.bid = b.bid
sid bid
day
22 101 10/10/96
95 103 11/12/96
r.sid
bid
101
102
103
104
b.bid
22
95
101
102
103
104
bname
Interlake
Interlake
Clipper
Marine
b.name
Interlake
Interlake
Clipper
Marine
color
blue
red
green
red
Full Outer Join
Full Outer Join returns all (matched or
unmatched) rows from the tables on both
sides of the join clause
SELECT r.sid, b.bid, b.name
FROM Reserves r FULL OUTER JOIN Boats b
ON r.bid = b.bid
Returns all boats & all information on
reservations
SELECT r.sid, b.bid, b.name
FROM Reserves r FULL OUTER JOIN Boats b
ON r.bid = b.bid
sid bid
day
22 101 10/10/96
95 103 11/12/96
r.sid
bid
101
102
103
104
b.bid
22
95
101
102
103
104
bname
Interlake
Interlake
Clipper
Marine
color
blue
red
green
red
b.name
Interlake
Interlake
Clipper
Marine
Note: in this case it is the same as the ROJ because
bid is a foreign key in reserves, so all reservations must
have a corresponding tuple in boats.
INSERT
INSERT [INTO] table_name [(column_list)]
VALUES ( value_list)
INSERT [INTO] table_name [(column_list)]
<select statement>
INSERT INTO Boats VALUES ( 105, ‘Clipper’, ‘purple’)
INSERT INTO Boats (bid, color) VALUES (99, ‘yellow’)
“bulk insert” from one table to another (must be type compatible):
INSERT INTO TEMP(bid)
SELECT r.bid FROM Reserves R WHERE r.sid = 22;
“bulk insert” from files (in Postgres)
Copy
DELETE & UPDATE
DELETE [FROM] table_name
[WHERE
qualification]
DELETE FROM Boats WHERE color = ‘red’
DELETE FROM Boats b
WHERE b. bid =
(SELECT r.bid FROM Reserves R WHERE r.sid = 22)
Can also modify tuples using UPDATE statement.
UPDATE Boats
SET Color = “green”
WHERE bid = 103;
Null Values
• Values are sometimes
– unknown (e.g., a rating has not been assigned or
– inapplicable (e.g., no spouse’s name).
– SQL provides a special value null for such situations.
• The presence of null complicates many issues. E.g.:
– Special operators needed to check if value is/is not null.
– “rating>8” - true or false when rating is null? What about
AND, OR and NOT connectives?
– Need a 3-valued logic (true, false and unknown).
– Meaning of constructs must be defined carefully. (e.g.,
WHERE clause eliminates rows that don’t evaluate to true.)
– New operators (in particular, outer joins) possible/needed.
Null Values – 3 Valued Logic
(null > 0)
is null
(null + 1)
is null
(null = 0)
is null
null AND true
is null
AND
T
F
Null
OR
T
F
Null
T
T
F
Null
T
T
T
T
F
F
F
F
F
T
F
Null
NULL
Null
F
Null
NULL
T
Null
Null