‫‪SQL‬‬
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‫פרופ' יהושע שגיב‪ ,‬האוניברסיטה העברית‬
Basic SQL Query
SELECT [Distinct] target-list
FROM from-list
WHERE condition;
•from-list: A list of relation names (possibly with a rangevariable after each name)
•target-list: A list of fields of relations in relation-list
•condition: A boolean condition
•DISTINCT: Optional keyword to delete duplicates
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Basic SQL Query (continues)
SELECT [Distinct] target-list
FROM from-list
WHERE condition;
The result evaluation:
1.
2.
3.
4.
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.
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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))
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Example Tables Used
Sailors
sid
sname rating
age
22
31
58
Dustin 7
Lubber 8
Rusty 10
45.0
55.5
35.0
bid
101 Nancy
103 Gloria
sid
Reserves
bid
Day
22
58
101
103
Key
Boats
bname color
10/10/02
11/12/02
red
green
Key
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Boat Names and Reservation
Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
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Boat Names and Reservation
Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
sid
Reserves
bid
day
Boats
bname color
bid
22
22
101
101
10/10/02
10/10/02
101 Nancy
103 Gloria
red
green
58
58
103
103
11/12/02
11/12/02
101 Nancy
103 Gloria
red
green
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Boat Names and Reservation
Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
Reserves
Boats
sid
bid
day
bid
bname
color
22
101
10/10/02
101
Nancy
red
58
103
11/12/02
103
Gloria
green
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Boat Names and Reservation
Dates
SELECT bname,day
FROM Boats,Reserves
WHERE Boats.bid = Reserves.bid
Reserves
Boats
day
bname
10/10/02
Nancy
11/12/02
Gloria
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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))
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Sailors Who Reserved Boat 103
Sailors x Reserves
Sailors
Reserves
sid
sname
rating age
sid
bid
day
22
Dustin
7
45.0
22
101
10/10/02
22
31
31
58
58
Dustin
Lubber
Lubber
Rusty
Rusty
7
8
8
10
10
45.0
55.5
55.5
35.0
35.0
58
22
58
22
58
103
101
103
101
103
11/12/02
10/10/02
11/12/02
10/10/02
11/12/02
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Sailors Who Reserved Boat 103
Sailors.sid = Reserves.sid  bid = 103
Sailors
Reserves
sid
sname
rating age
sid
bid
day
22
Dustin
7
45.0
22
101
10/10/02
22
31
31
58
58
Dustin
Lubber
Lubber
Rusty
Rusty
7
8
8
10
10
45.0
55.5
55.5
35.0
35.0
58
22
58
22
58
103
101
103
101
103
11/12/02
10/10/02
11/12/02
10/10/02
11/12/02
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Sailors Who Reserved Boat 103
sname
Sailors
Reserves
sid
sname
rating age
sid
bid
day
22
Dustin
7
45.0
22
101
10/10/02
22
31
31
58
58
Dustin
Lubber
Lubber
Rusty
Rusty
7
8
8
10
10
45.0
55.5
55.5
35.0
35.0
58
22
58
22
58
103
101
103
101
103
11/12/02
10/10/02
11/12/02
10/10/02
11/12/02
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Range Variables
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
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A Few SELECT Options

Select all columns:
SELECT *
FROM Sailors S;

Rename selected columns:
SELECT S.sname AS Sailors_Name
FROM Sailors S;
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A Few SELECT Options

Applying functions:
 Mathematical manipulations:
SELECT (age-5)*2
FROM Sailors S;

Aggregate functions:
SELECT COUNT(*)
FROM Sailors S;
SELECT MAX(age)
FROM Sailors S;
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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 ;
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The WHERE Clause (continues)

The LIKE operator:



A pattern matching operator
Basic format: colname LIKE pattern
Example:
SELECT sid
FROM Sailors
WHERE sname LIKE ‘B_%g’;

_ is a single character

% is 0 or more characters
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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?
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Exercise
Formulate a query that finds the names of
sailors who reserved a yellow boat.
sid
22
31
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Sailors
sname rating
Dustin 7
Lubber 8
Rusty 10
Boats
age
45.0
55.5
35.0
bid
bname
color
101
103
Nancy
Gloria
red
green
sid
Reserves
bid
Day
22
58
101
103
10/10/02
11/12/02
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A Harder Exercise
1. Formulate a query that finds the bid
of boats that are reserved in at
least two different days
2. Fix the query you formulated to find
the names of these boats
Hint: A relation may appear more than once
in the FROM list…
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Union, Intersect and Except


Sqlplus supports the union, intersection and
difference operators
The syntax:
 query1 UNION query2;
 query1 INTERSECT query2;
 query1 MINUS query2;
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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 ?
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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
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid = R.sid and R.bid = B.bid
and B.color = ‘green’;
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The Second Version in
Relational Algebra
sname(color = ‘red’ (Sailors ⋈ Reserves ⋈ Boats)) 
sname(color = ‘green’ (Sailors ⋈ Reserves ⋈ Boats))
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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
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The IN Operator
Names of sailors who’ve reserved boat 103:
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT
FROM
WHERE
What would happen if
we wrote NOT IN?
R.sid
Reserves R
R.bid = 103);
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The EXISTS Operator
Names of sailors who’ve reserved boat 103:
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT
FROM
WHERE
Notice the correlation
between the examined row
and the inner query
What would happen if
we wrote NOT EXISTS?
*
Reserves R
R.bid = 103 and
S.sid = R.sid);
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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 <>)
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Another Exercise

What does the following query compute?
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
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid = R.sid and R.bid = B.bid
and B.color = ‘green’;

Using IN and INTERSECT operators, formulate a
query that finds the names of the sailors who
reserved both green and red boats.
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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)
)
;
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For The Brave Ones…
1.
2.
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!
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