1. © פרופ' יהושע שגיב, האוניברסיטה העברית
SQL
© פרופ' יהושע שגיב, האוניברסיטה העברית

2. Basic SQL Query SELECT [Distinct] target-list FROM from-list
WHERE condition;
from-list: A list of relation names (possibly with a range-variable after each name)
target-list: A list of fields of relations in relation-list
condition: A boolean condition
DISTINCT: Optional keyword to delete duplicates

3. Basic SQL Query (continues)
SELECT [Distinct] target-list
FROM from-list
WHERE condition;
The result evaluation:
Compute the cross product of the tables in from-list.
Delete all rows that do not satisfy condition.
Delete all columns that do not appear in target-list.
If Distinct is specified eliminate duplicate rows.

4. Basic SQL Query (continues)
SELECT Distinct A1,…,An
FROM R1,…,Rm
WHERE C;
This translates to the expression in relational algebra:
A1,…,An (C(R1 x…x Rm))

5. Example Tables Used Sailors sid sname rating age 22 31 58 Dustin
Lubber
Rusty 7 8 10
45.0
55.5
35.0
Boats
bid
bname
color
101
103
Nancy
Gloria
red
green
Reserves
sid
bid
Day 22 58
101
103
10/10/02
11/12/02
Key
Key

6. Boat Names and Reservation Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid

7. Boat Names and Reservation Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
Reserves
sid
bid
day 22
101
10/10/02 58
103
11/12/02
Boats
bid
bname
color
101
103
Nancy
Gloria
red
green

8. Boat Names and Reservation Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
Reserves
sid
bid
day 22
101
10/10/02 58
103
11/12/02
Boats
bid
bname
color
101
Nancy
red
103
Gloria
green

9. Boat Names and Reservation Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
Reserves
day
10/10/02
11/12/02
Boats
bname
Nancy
Gloria

10. Sailors Who Reserved Boat 103
SELECT sname
FROM Sailors, Reserves
WHERE Sailors.sid = Reserves.sid and
bid = 103;
sname(Sailors.sid = Reserves.sid  bid = 103 (Sailors x Reserves))

11. Sailors Who Reserved Boat 103
Sailors x Reserves
Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/02 58
103
11/12/02 31
Lubber 8
55.5
Rusty 10
35.0

12. Sailors Who Reserved Boat 103
Sailors.sid = Reserves.sid  bid = 103
Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/02 58
103
11/12/02 31
Lubber 8
55.5
Rusty 10
35.0

13. Sailors Who Reserved Boat 103
sname
Sailors
Reserves
sid
sname
rating
age
bid
day 22
Dustin 7
45.0
101
10/10/02 58
103
11/12/02 31
Lubber 8
55.5
Rusty 10
35.0

14. Range Variables Range variables are good style
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid = R.sid and
R.bid = 103;
Range variables are good style
They are necessary if the same relation appears twice in the FROM clause

15. A Few SELECT Options Rename selected columns: Select all columns:
FROM Sailors S;
Rename selected columns:
SELECT S.sname AS Sailors_Name

16. A Few SELECT Options Applying functions: Mathematical manipulations:
SELECT (age-5)*2
FROM Sailors S;
Aggregate functions:
SELECT COUNT(*)
SELECT MAX(age)

17. The WHERE Clause Numerical and string comparison:
!=,<>,=, <, >, >=, <=, between(between val1 AND val2)
String comparisson is according to the alphabetical order!
Logical components: AND, OR
Null verification: IS NULL, IS NOT NULL
Example:
SELECT sname
FROM Sailors
WHERE age>=40 AND rating IS NOT NULL ;

18. The WHERE Clause (continues)
The LIKE operator:
A pattern matching operator
Basic format: colname LIKE pattern
Example:
_ is a single character
% is 0 or more characters
SELECT sid
FROM Sailors
WHERE sname LIKE ‘B_%g’;

19. Sailors Who’ve Reserved a Boat
SELECT sname
FROM Sailors S, Reserves R
WHERE S.sid = R.sid;
Would adding DISTINCT give a different result?

20. Exercise
Formulate a query that finds the names of sailors who reserved a yellow boat.
Sailors
sid
sname
rating
age 22 31 58
Dustin
Lubber
Rusty 7 8 10
45.0
55.5
35.0
Boats
bid
bname
color
101
103
Nancy
Gloria
red
green
Reserves
sid
bid
Day 22 58
101
103
10/10/02
11/12/02

21. A Harder Exercise
Formulate a query that finds the bid of boats that are reserved in at least two different days
Fix the query you formulated to find the names of these boats
Hint: A relation may appear more than once in the FROM list…

22. Union, Intersect and Except
Sqlplus supports the union, intersection and difference operators
The syntax:
query1 UNION query2;
query1 INTERSECT query2;
query1 MINUS query2;

23. Sailors who’ve reserved a red or 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’ or
B.color = ‘green’);
sname(color = ‘red’  color = ‘green’ (Sailors ⋈ Reserves ⋈ Boats))
What would happen if we replaced or by and ?

24. Sailors who’ve reserved red or 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’
UNION
WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘green’;

25. The Second Version in Relational Algebra
sname(color = ‘red’ (Sailors ⋈ Reserves ⋈ Boats)) 
sname(color = ‘green’ (Sailors ⋈ Reserves ⋈ Boats))

26. Nested Queries
A query may be nested within another through the following operators:
(NOT) IN
(NOT) EXISTS
ANY
ALL
A query with nested queries is computed using nested loops

27. The IN Operator 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);
What would happen if we wrote NOT IN?

28. Notice the correlation between the examined row and the inner query
The EXISTS Operator
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);
What would happen if we wrote NOT EXISTS?
Notice the correlation between the examined row and the inner query

29. Set-Comparison Queries: ANY,ALL
Sailors who are not the youngest:
SELECT *
FROM Sailors S1
WHERE S1.age > ANY (SELECT S2.age
FROM Sailors S2);
We can also use op ALL (op is >, <, =, >=, <=, or <>)

30. Another Exercise What does the following query compute?
Using IN and INTERSECT operators, formulate a query that finds the names of the sailors who reserved both green and red boats.
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘red’
INTERSECT
WHERE S.sid = R.sid and R.bid = B.bid and B.color = ‘green’;

31. Another Exercise (continues)
What does the following query compute?
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (
(SELECT B.bid FROM Boats B)
MINUS
(SELECT R.bid FROM Reserves R
WHERE R.sid=S.sid) ) ;

32. For The Brave Ones…
Formulate a query that computes the sid of all sailors who have not reserved a green boat.
Prove formally that in the Sailors-Reserves-Boats database, the query of question 1 cannot be as simple as a query of the form:
SELECT Vj.sid
FROM R1 V1, R2 V2…Rn Vn
WHERE C
Where C is a simple condition (contains only comparison clauses, AND and OR, and does not contain nested queries)
Notice that the records in each table are arbitrary!