1. CS 405G: Introduction to Database Systems
SQL II
Instructor: Jinze Liu

2. Review: SQL DML DML includes 4 main statements:
SELECT (query), INSERT, UPDATE and DELETE
e.g:
PROJECT
SELECT S.name, E.cid
FROM Students S, Enrolled E
WHERE S.sid = E.sid AND
S.age=19
JOIN
SELECT

3. Example: Find sailors who have reserved a red and a green boat
Now let’s do this with a self-join
SELECT R.sid
FROM Boats B,Reserves R
WHERE B.color = ‘red’
AND R.bid=B.bid
INTERSECT
WHERE B.color = ‘green’ AND R.bid=B.bid
Boats
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria
105
Titanic
green
Reserves
sid
bid
day 1
101
9/12 2
103
9/13
105
sid 1 2
sid 1
sid 1  =

4. 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
AND B2.color = ‘green’
AND B2.bid=R2.bid
AND R1.sid=R2.sid
Find red reserved boats
Find green reserved boats
Find matching green and red reserved boats
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria
105
Titanic
green
sid
bid
day 1
101
9/12 2
103
9/13
105
sid 1 2
sid 1 =
sid 1

5. 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
WHERE R.bid=103)
sid
bid
day 1
101
9/12 2
103
9/13
105
sname
Bilbo
Sailors
First compute the set of all sailors that have reserved boat 103…
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
…and then check each the sid of each tuple in Sailors to see if it is in S1 S
sid 2 S S

6. 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 be evaluated once and then checked for each tuple in enclosing query

7. 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

8. 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) 2 1 3
Sailors
Reserves
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
sid
bid
day
sid
bid
day 1
101
9/12 2
103
9/13
105 S
/13 S S

9. 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

10. 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
FROM Sailors S
WHERE NOT EXISTS
(SELECT B.bid
FROM Boats B
WHERE NOT EXISTS
(SELECT R.bid
FROM Reserves R
WHERE R.bid=B.bid
AND R.sid=S.sid))
Find Sailors S such that ...
there is no boat B...
without a reservation by Sailor S

11. 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
WHERE R.bid=B.bid
AND R.sid=S.sid))
101
103 1 3 2
Reserves
Sailors
sid
bid
day 1
103
9/12 2
9/13 3
9/14
101
Boats R
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
bid
bname
color
101
Nina
red
103
Pinta
blue S B R S R B R S R

12. 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) 3 2 1
Sailors
Reserves
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
sid
bid
day
sid
bid
day 1
103
9/12 2
9/13 S
/13 S S

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’)
Uncorrelated! S1 S2
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
sid
sname
rating
age 2
Bilbo 39
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27

14. Aggregate Operators
Very powerful; enables computations over sets of tuples
SELECT COUNT (*)
FROM Sailors S
COUNT: returns a count of tuples in the set
AVG: returns average of column values in the set
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
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 AVG(DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10

15. Find name and age of the oldest sailor(s)
Aggregate Operators
Find name and age of the oldest sailor(s) X
Sailors
SELECT S.sname, MAX (S.age)
FROM Sailors S
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27
What will the result be?
Not legal syntax; no other columns allowed in SELECT clause without a GROUP BY clause (we’ll learn about those next)
sname
age
Frodo 39
Bilbo
Sam

16. Find name and age of the oldest sailor(s)
Aggregate Operators
Find name and age of the oldest sailor(s)
Instead:
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age =
(SELECT MAX (S2.age)
FROM Sailors S2)
And then find the sailors(s) of that age…
Find the maximum age…

17. 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 (!):
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
For i = 1, 2, ... , 10:

18. 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
And finally compute aggregate function over each group…
SELECT [DISTINCT] target-list
FROM relation-list
[WHERE qualification]
GROUP BY grouping-list
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)).

19. For each rating, find the age of the youngest sailor with age  18
Group By Example
For each rating, find the age of the youngest
sailor with age  18
Sailors
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 39 3
Sam 8 27 4
Pippin 21 5
Merry 17
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
Group 1 1
Frodo 7 22 7 22 2 21 8 27 2
Bilbo 39 4
Pippin 21
Group 2
Group 3 3
Sam 8 27

20. Find the number of reservations for each red boat.
SELECT B.bid, COUNT(*)AS tot_res
FROM Boats B, Reserves R
WHERE R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid
Boats
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria
103 1
101 2
Reserves
sid
bid
day 1
102
9/12 2
103
9/13 3
101
9/14 4 2
103
9/13 3
101
9/14 4

21. 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

22. SELECT S.rating, MIN (S.age) FROM Sailors S WHERE S.age >= 18
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 2 3
Answer

23. Sailors who have reserved all boats
sid
sname
rating
age 1
Frodo 7 22 2
Bilbo 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
bid
bname
color
101
Nina
red
102
Pinta
blue
103
Santa Maria
sname
sid
bid
Frodo 1
102
Bilbo 2
101
103
Reserves
sname
sid
count
Frodo 1
Bilbo 2 3
count 3
sid
bid
day 1
102
9/12 2
101
9/14
9/10
103
9/13
sname
sid
bid
Frodo 1
102,102
Bilbo 2
101, 102, 103

24. 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)

25. 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:
FROM Sailors s NATURAL JOIN Reserves r
“NATURAL” means equi-join for each pair of attributes with the same name

26. SELECT s.sid, s.name, r.bid FROM Sailors s INNER JOIN Reserves r ON s.sid = r.sid

27. 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

28. SELECT s.sid, s.name, r.bid FROM Sailors s LEFT OUTER JOIN Reserves r ON s.sid = r.sid

29. 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.

30. SELECT r.sid, b.bid, b.name FROM Reserves r RIGHT OUTER JOIN Boats b ON r.bid = b.bid

31. 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

32. SELECT r.sid, b.bid, b.name FROM Reserves r FULL OUTER JOIN Boats b ON r.bid = b.bid
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.

33. INSERT INSERT [INTO] table_name [(column_list)] VALUES ( value_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

34. DELETE & UPDATE DELETE FROM Boats WHERE color = ‘red’
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;

35. 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.

36. Null Values – 3 Valued Logic
null AND true
is null
is null
is null
is null
AND T F
Null
NULL OR T F
Null
NULL T F
Null T T T F F F T F
Null
Null F
Null T
Null
Null