Download SQL: Queries, Programming, Triggers

SQL: Queries, Programming,
Triggers
Chpt 5
Jianping Fan
UNC-Charlotte
Database Management Systems
Raghu Ramakrishnan
1
E-R Model for Example
sname
sid
rating
Age
bid
Sailors
bname
bcolor
Boats
Reserve
day
Database Management Systems
2
Example Instances
R1
bid
101
B1
bname
interlake
102
103
interlake red
clipper green
marine
Database Management Systems
We will use these instances of the
Sailors and Reserves relations in
our examples.
S1
sid bid
day
22 101 10/10/96
58 103 11/12/96
104

color S2
blue
red
sid
22
31
58
sname rating age
dustin
7
45.0
lubber
8
55.5
rusty
10 35.0
sid
28
31
44
58
sname rating age
yuppy
9
35.0
lubber
8
55.5
guppy
5
35.0
rusty
10 35.0
Raghu Ramakrishnan
3
SELECT
FROM
WHERE




[DISTINCT] target-list
relation-list
qualification
What you want

Where you can find A B
Conditions for candidates

relation-list A list of relation names (possibly with a
range-variable after each name).
target-list A list of attributes of relations in relationlist
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. Default is that
duplicates are not eliminated!
Database Management Systems
Raghu Ramakrishnan
4
Conceptual Evaluation Strategy

Semantics of an SQL query defined in terms of the
following conceptual evaluation strategy:
– Compute the cross-product of relation-list. A B
– Discard resulting tuples if they fail
qualifications.
– Delete attributes that are not in target-list.
– If DISTINCT is specified, eliminate duplicate
rows.


This strategy is probably the least efficient way to
compute a query! An optimizer will find more
efficient strategies to compute the same answers.
Database Management Systems
Raghu Ramakrishnan
5
Example of Conceptual Evaluation
Find the name of the sailors who reserve boat 103?
SELECT S.sname
FROM Sailors S, Reserves R, Boat B
WHERE S.sid=R.sid AND R.bid=103
(sid) sname rating age
(sid) bid day
22 dustin
7
45.0
22
101 10/10/96
22 dustin
7
45.0
58
103 11/12/96
31 lubber
8
55.5
22
101 10/10/96
31 lubber
8
55.5
58
103 11/12/96
58 rusty
10
35.0
22
101 10/10/96
58 rusty
10
35.0
58
103 11/12/96
Database Management Systems
Raghu Ramakrishnan
6
Example of Conceptual Evaluation
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
 S .sname( R.bid 103 (S  R))
We can also use “join”
 S .sname(S  ( R.bid 103R))
Database Management Systems
Raghu Ramakrishnan
7
A Note on Range Variables

Really needed only if the same relation
appears twice in the FROM clause. The
previous query can also be written as:
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid AND R.bid=103
OR
SELECT S.sname
FROM Sailors, Reserves
WHERE Sailors.sid=Reserves.sid
AND Reserves.bid=103
Database Management Systems
Raghu Ramakrishnan
It is good style,
however, to use
range variables
always!
8
Testing Example 1
1. Find courses’ name John Smith gets `A` at Spring 2017
2. Find students’ name who get `A` in ITCS6160 at Spring 2017
3. Find students’ name who get `A` from Database course
at Spring 2017
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
High-Level Query
Enrolled_In
SQL Interpretation
Relational Algebra
Grade
Database Management Systems
semester
9
Testing Example 2
1. Find Drug’s name which John Smith at ``9201 Uni. City BLVD” is taking
2. Find Drug’s expiration day which John Smith at ``9201 Uni. City BLVD” is taking
3. Find patient’s phone number who is taking ``Lipton” expired at 3/9/2017
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
High-Level Query
SQL Interpretation
Prescribed
Relational Algebra
Database Management Systems
Dosage
#days
10
Testing Example 3
High-Level Query
from
name
ssn
Employees
to
SQL Interpretation
dname
lot
did
Works_In2
budget
Departments
Relational Algebra
1. Find Dept’s ID which John Smith worked from 2001 to 2015
2. Find Dept’s name which John Smith worked from 2001 to 2015
3. Find Dept’s budget which John Smith worked from 2001 to 2015
Database Management Systems
11
Find the name of sailors who’ve reserved
a red 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’

Query contains a join of three tables, followed by a
selection on the color of boats
 S .sname( B.colorred ( B)  R  S )
Database Management Systems
Raghu Ramakrishnan
12
Find the sid of sailors who’ve reserved a
red boat
SELECT R.sid
FROM Boats B, Reserves R
WHERE R.bid=B.bid AND B.color = ‘red’


Query contains a join of two tables (cross product,
selection, projection), followed by a selection on the
color of boats
If we wanted the name of the sailors, we must include
Why?
the Sailors relation as well
 R.sid ( B.colorred ( R  B))
Database Management Systems
Raghu Ramakrishnan
13
Find the rating of sailors who’ve reserved
a green boat
SELECT S.rating
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color = ‘green’

Query contains a join of three tables, followed by a
selection on the color of boats
 S .rating (( B.color green B)  R  S )
Database Management Systems
Raghu Ramakrishnan
14
Find the sid of sailors who’ve reserved a
red boat and a green boat
SELECT S.sid
FROM Sailors AS S, Boats AS B1, Reserves AS R1,
Boats AS B2, Reserves AS R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color=‘red’ AND B2.color=‘green’)
OR
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color=‘red’ AND B2.color=‘green’)
Database Management Systems
Raghu Ramakrishnan
15
Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
High-Level Query
SQL Interpretation
Enrolled_In
Relational Algebra
Grade
semester
1. Find students’ name who are enrolled in both database
and visual database at Spring 2017
2. Find students’ ID who are enrolled in both database
and visual database at Spring 2017
Database Management Systems
16
Testing Example 2
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
High-Level Query
SQL Interpretation
Enrolled_In
Relational Algebra
Grade
semester
1. Find students’ name who are enrolled in database
or visual database at Spring 2017
2. Find students’ name who are enrolled in database
but not algorithm at Spring 2017
Database Management Systems
17
Testing Example 3
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
High-Level Query
SQL Interpretation
Prescribed
Relational Algebra
Dosage
#days
Find Drug’s name and manuf which both John Smith and Bob
Johnson are taking 2 pieces per day
Find Drug’s Exp. day which both John Smith and Bob
Johnson are taking 2 pieces per day
Database Management Systems
18
Testing Example 4
from
name
ssn
Employees
to
dname
lot
did
Works_In2
budget
Departments
Find Dept’s name which both John Smith and Bob Johnson
work from 2001 to 2015
Find Dept’s name which John Smith or Bob Johnson
work from 2001 to 2015
Find Dept’s DID which both John Smith and Bob Johnson
work from 2001 to 2015
Find Dept’s DID which John Smith or Bob Johnson
work from 2001 to 2015
Database Management Systems
19
Testing Example 5
from
name
ssn
Employees
to
dname
lot
did
Works_In2
budget
Departments
Find Dept’s name which John Smith but not Bob Johnson
work from 2001 to 2015
Find Dept’s name which Bob Johnson but not John Smith
work from 2001 to 2015
Find Dept’s DID which John Smith but not Bob Johnson
work from 2001 to 2015
Find Dept’s DID which Bob Johnson but not John Smith
work from 2001 to 2015
Database Management Systems
20
Find sailors who’ve reserved at least one boat
SELECT S.sid
FROM Sailors S, Reserves R
WHERE S.sid=R.sid
SELECT DISTINCT S.sid
FROM Sailors S, Reserves R
WHERE S.sid=R.sid

Query contains a join of two tables

Would adding DISTINCT to this query make a
difference? (yes, why?)
Database Management Systems
Raghu Ramakrishnan
21
Find sailors who’ve reserved at
least one boat
SELECT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid
SELECT DISTINCT S.sname
FROM Sailors S, Reserves R
WHERE S.sid=R.sid
What
is the effect of replacing S.sid by S.sname in
the SELECT clause? Would adding DISTINCT
to this variant of the query make a difference?
Database Management Systems
Raghu Ramakrishnan
22
Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
High-Level Query
SQL Interpretation
Enrolled_In
Relational Algebra
Grade
semester
Find students’ names who has enrolled in at least one course
Find students’ ID who has enrolled in at least one course
Database Management Systems
23
Testing Example 2
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
High-Level Query
SQL Interpretation
Prescribed
Relational Algebra
Dosage
#days
Find patient’s name who are taking at least one drug
called ``Lipton”
Database Management Systems
24
Testing Example 3
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
High-Level Query
SQL Interpretation
Prescribed
Relational Algebra
Dosage
#days
Find patient’s name who are taking at least one drug
expired at 3/9/2017
Database Management Systems
25
Testing Example 4
from
name
ssn
Employees
High-Level Query
to
dname
lot
did
Works_In2
budget
Departments
SQL Interpretation
Relational Algebra
Find Department’s name which has at least one employee
working from 2001 to 2015
Database Management Systems
26
Expressions and Strings
SELECT S.age, S.age-5 AS age1
FROM Sailors S
WHERE S.sname LIKE ‘B_%’;



Illustrates use of arithmetic expressions and string
pattern matching: Find triples (of ages of sailors and
two fields defined by expressions) for sailors whose
names begin with B and contain at least two
characters.
AS is way to name fields in result.
LIKE is used for string matching. `_’ stands for any
one character and `%’ stands for 0 or more arbitrary
B_%B
B%B
B_%%B
characters.
Database Management Systems
Raghu Ramakrishnan
What’s the difference?27
Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find students’ names who get ``A” from ITCS6160 and name
starting from `A’ and ending `S’ with at least 4 characters
Database Management Systems
28
Testing Example 2
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
Prescribed
Dosage
#days
Find Drug’s name which John Smith is taking now and name
Starting from `L” and ending as `R’ and having at least 5 characters
Database Management Systems
29
Find sid’s of sailors who’ve reserved a red or a green boat



UNION: Can be used to
compute the union of any
two union-compatible sets of
tuples (which are
themselves the result of
SQL queries).
If we replace OR by AND in
the first version, what do
we get? (intersection)
Also available: EXCEPT
(What do we get if we
replace UNION by EXCEPT?)
Database Management Systems
SELECT S.sid
?
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’)
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid AND B.color=‘red’
UNION
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid AND B.color=‘green’
Raghu Ramakrishnan
30
Find sid’s of sailors who’ve reserved a
red but not a green boat

EXCEPT
How to transform to
Relation Algebra?
Database Management Systems
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.color=‘red’ NOT B.color=‘green’)
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid AND B.color=‘red’
EXCEPT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid AND B.color=‘green’
Raghu Ramakrishnan
31
Find snames of sailors who’ve reserved a boat named
‘yellowboat or a boat named ‘purpleboat’

Another Union example
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.bname=‘yellowboat’
OR B.bname=‘purpleboat’)
How to transform to
Relation Algebra?
Database Management Systems
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid
AND B.bname=‘yellowboat’
UNION
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid
AND B.bname=‘purpleboat’
Raghu Ramakrishnan
32
Find sid’s of sailors who’ve reserved a red and a green boa



INTERSECT: Can be used to
compute the intersection
of any two unioncompatible sets of tuples.
Included in the SQL/92
standard, but some
systems don’t support it.
Contrast symmetry of the
UNION and INTERSECT
queries with how much
the other versions differ.
Database Management Systems
SELECT S.sid
FROM Sailors S, Boats B1, Reserves R1,
Boats B2, Reserves R2
WHERE S.sid=R1.sid AND R1.bid=B1.bid
AND S.sid=R2.sid AND R2.bid=B2.bid
AND (B1.color=‘red’ AND B2.color=‘green’)
Key field!
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid
AND B.color=‘red’
INTERSECT
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
R.bid=B.bid
Raghu Ramakrishnan
AND B.color=‘green’
33
Find snames of sailors who’ve reserved a boat named
‘yellow boat and a boat named ‘purpleboat’

Another Intersect example
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid
AND (B.bname=‘yellowboat’
AND B2.bname=‘purpleboat’)
R.bid=B.bid
AND B.bname=‘ yellowboat’
INTERSECT
SELECT S.sname
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND
How to transform to
Relation Algebra?
R.bid=B.bid
AND B.bname=‘purpleboat’
Database Management Systems
Raghu Ramakrishnan
34
Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find students’ names who are enrolled in ITCS6160 or ITCS6167
at Spring 2015
Find students’ names who are enrolled both database and visual
database at Spring 2015
Database Management Systems
35
Testing Example 2
Name
Addr
Patients
Phone
Name
Age
Manuf
Exp
Drug
Prescribed
Dosage
#days
Find Drug’s name which both John Smith and Bob Johnson are
taking 2 pieces per day
Find Drug’s name which John Smith but not Bob Johnson is
taking 2 pieces per day
36
Database Management Systems
Testing Example 3
from
name
ssn
Employees
to
dname
lot
did
Works_In2
budget
Departments
Find Department’s name which John Smith but not Bob Johnson
work from 2001 to 2015
Database Management Systems
37
Nested Queries
Find 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)
A very powerful feature of SQL: a WHERE clause can
itself contain an SQL query! (Actually, so can FROM
and HAVING clauses.)
 To find sailors who’ve not reserved #103, use NOT IN.
 To understand semantics of nested queries, think of a
nested loops evaluation: For each Sailors tuple, check the
qualification by computing the subquery.

Database Management Systems
Raghu Ramakrishnan
38
Nested Queries (continued)

Another example:
– Perform both select on sname and select on r.sid in Oracle
Find names of sailors who’ve reserved a boat
on October 21, 1999
How to transform to
Relation Algebra?
SELECT S.sname
FROM Sailors S
WHERE S.sid IN (SELECT R.sid
FROM Reserves R
WHERE R.day=‘21-oct-99’)
Database Management Systems
Raghu Ramakrishnan
How DBMS achieve this?
39
Nested Queries with Correlation
Find names of sailors who’ve reserved boat #2:
SELECT S.sname
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=2 AND S.sid=R.sid)

EXISTS is another set comparison operator, like IN.

Allows test whether a set is nonempty
Database Management Systems
Raghu Ramakrishnan
40
Nested Queries with Correlation
Find names of sailors who’ve reserved boat #2:
SELECT S.sname
FROM Sailors S
WHERE S.sid EXISTS (SELECT S.sid
FROM Reserves R
WHERE R.bid=2 AND S.sid=R.sid)
any difference on query result?
Database Management Systems
Raghu Ramakrishnan
41
More on Set-Comparison Operators
We’ve already seen IN, EXISTS. Can also use NOT IN,
NOT EXISTS. (will go over unique later)
,, ,,, 
 Also available: op ANY, op ALL
 Find sailors whose rating is greater than that of
some sailors called lubber:

SELECT *
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘lubber’)
Database Management Systems
Raghu Ramakrishnan
42
any difference on query result?
SELECT S.sid
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘lubber’)
SELECT *
FROM Sailors S
WHERE S.rating > ANY (SELECT S2.rating
FROM Sailors S2
WHERE S2.sname=‘lubber’)
Database Management Systems
Raghu Ramakrishnan
43
More on Set-Comparison Operators
(continued)

Another example: find name of the sailors with the
highest rating
SELECT s.sname
FROM Sailors S
WHERE S.rating >= ALL (SELECT S2.rating
FROM Sailors S2)
Database Management Systems
Raghu Ramakrishnan
44
More on Set-Comparison Operators
(continued)

Another example: find the sailors with the lowest
rating
SELECT s.sname
FROM Sailors S
WHERE S.rating <= ALL (SELECT S2.rating
FROM Sailors S2)
Database Management Systems
Raghu Ramakrishnan
45
Rewriting INTERSECT Queries Using IN
Find sid’s of sailors who’ve reserved both a red and a green boat:
SELECT S.sid
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
AND S.sid IN (SELECT S2.sid
FROM Sailors S2, Boats B2, Reserves R2
WHERE S2.sid=R2.sid AND R2.bid=B2.bid
AND B2.color=‘green’)
Similarly, EXCEPT queries re-written using NOT IN.
 To find names (not sid’s) of Sailors who’ve reserved
both red and green boats, just replace S.sid by S.sname
in SELECT clause.

Database Management Systems
Raghu Ramakrishnan
46
(1)
Division in SQL
Find sailors who’ve reserved all

Let’s do it the hard
way, without EXCEPT:
SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS
((SELECT B.bid
FROM Boats B)
EXCEPT
boats.
(SELECT R.bid
FROM Reserves R
WHERE R.sid=S.sid))
(2) SELECT S.sname
FROM Sailors S
WHERE NOT EXISTS (SELECT B.bid
FROM Boats B
WHERE NOT EXISTS (SELECT R.bid
Sailors S such that ...
FROM Reserves R
WHERE R.bid=B.bid
there is no boat B without ...
AND R.sid=S.sid))
a Reserves tuple showing S reserved B
Database Management Systems
Raghu Ramakrishnan
47
Aggregate Operators

Significant extension of
relational algebra.
SELECT COUNT (*)
FROM Sailors S
SELECT AVG (S.age)
FROM Sailors S
WHERE S.rating=10
single column
SELECT S.sname
FROM Sailors S
WHERE S.rating= (SELECT MAX(S2.rating)
FROM Sailors S2)
SELECT COUNT (DISTINCT S.rating)
FROM Sailors S
WHERE S.sname=‘Bob’
Database Management Systems
COUNT (*)
COUNT ( [DISTINCT] A)
SUM ( [DISTINCT] A)
AVG ( [DISTINCT] A)
MAX (A)
MIN (A)
SELECT AVG ( DISTINCT S.age)
FROM Sailors S
WHERE S.rating=10
Raghu Ramakrishnan
48
Aggregate Operators (continued)

Sum example: find the sum of all the ages of
sailors, and the count, who have reserved boat #2
SELECT SUM (S.age), Count (s.sname)
FROM Sailors S
WHERE EXISTS (SELECT *
FROM Reserves R
WHERE R.bid=2 AND S.sid=R.sid)

Show select * from reserves in Oracle as
confirmation
Database Management Systems
Raghu Ramakrishnan
49
Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find students from ITCS6160 at Spring 2016 whose grade is
greater than that of some students called John Smith
Database Management Systems
50
Testing Example 2
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find students from ITCS6160 at Spring 2016 with the highest
grade
Database Management Systems
51
Testing Example 3
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find students from ITCS6160 at Spring 2015 with the lowest
grade
Database Management Systems
52
Testing Example 4
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the ID of the students who are enrolled in both ITCS6160
and ITCS6157 at Spring 2016
Database Management Systems
53
Testing Example 5
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the sum of the ages of the students and the total number
of students who are enrolled in ITCS6160 at Spring 2016
Database Management Systems
54
Testing Example 6
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the names and the age for the oldest students in ITCS6160
at Spring 2016
Database Management Systems
55
Testing Example 7
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the names and the age for the youngest students in
ITCS6160 at Spring 2016
Database Management Systems
56
Find name and age of the oldest sailor(s)
The first query is illegal!
(We’ll look into the
reason a bit later, when
we discuss GROUP BY.)
 The third query is
equivalent to the second
query, and is allowed in
the SQL/92 standard,
but is not supported in
some systems.

Database Management Systems
SELECT S.sname, MAX (S.age)
FROM Sailors S
SELECT S.sname, S.age
FROM Sailors S
WHERE S.age =
(SELECT MAX (S2.age)
FROM Sailors S2)
SELECT S.sname, S.age
FROM Sailors S
WHERE (SELECT MAX (S2.age)
FROM Sailors S2)
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GROUP BY and HAVING
So far, we’ve applied aggregate operators to all
(qualifying) tuples. Sometimes, we want to apply
them 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)
For i = 1, 2, ... , 10:
FROM Sailors S
WHERE S.rating = i
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Queries With GROUP BY and HAVING
SELECT
FROM
WHERE
GROUP BY
HAVING

[DISTINCT] target-list
relation-list
qualification
grouping-list
group-qualification
The target-list contains (i) attribute names (ii) terms
with aggregate operations (e.g., MIN (S.age)).
– The attribute list (i) must be a subset of grouping-list.
Intuitively, each answer tuple corresponds to a group, and
these attributes must have a single value per group. (A
group is a set of tuples that have the same value for all
attributes in grouping-list.)
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GROUP BY and HAVING

Find the age of the youngest sailor for each rating
level.
SELECT MIN (S.age)
For i = 1, 2, ... , 10:
FROM Sailors S
WHERE S.rating = i
SELECT S.rating, MIN (S.age)
FROM Sailors S
GROUP BY S.rating
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Conceptual Evaluation
The cross-product of relation-list is computed, tuples
that fail qualification are discarded, `unnecessary’ fields
are deleted, and the remaining tuples are partitioned
into groups by the value of attributes in grouping-list.
 The group-qualification is then applied to eliminate
some groups. Expressions in group-qualification must
have a single value per group!

– In effect, an attribute in group-qualification that is not an
argument of an aggregate op also appears in grouping-list.
(SQL does not exploit primary key semantics here!)

One answer tuple is generated per qualifying group.
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Find the age of the youngest sailor with age  18,
for each rating with at least 2 such sailors
sid sname rating age
SELECT S.rating, MIN (S.age)
22 dustin
7
45.0
FROM Sailors S
31 lubber
8
55.5
WHERE S.age >= 18
71 zorba
10 16.0
GROUP BY S.rating
64 horatio
7
35.0
HAVING COUNT (*) > 1
29 brutus
1
33.0
 Only S.rating and S.age are
58 rusty
10 35.0
mentioned in the SELECT,
rating age
GROUP BY or HAVING clauses;
1
33.0
other attributes `unnecessary’.
7
45.0
rating
 2nd column of result is
7
35.0
7
35.0
unnamed. (Use AS to name it.)
8
55.5
Answer relation
10 35.0
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Ramakrishnan
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Database Management Systems
For each red boat, find the number of
reservations for this boat
SELECT B.bid, COUNT (*) AS scount
FROM Sailors S, Boats B, Reserves R
WHERE S.sid=R.sid AND R.bid=B.bid AND B.color=‘red’
GROUP BY B.bid
Grouping over a join of three relations.
 What do we get if we remove B.color=‘red’
from the WHERE clause and add a HAVING
clause with this condition?
 What if we drop Sailors and the condition
involving S.sid?

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Find the age of the youngest sailor with age > 18,
for each rating with at least 2 sailors (of any age)
SELECT S.rating, MIN (S.age)
FROM Sailors S
WHERE S.age > 18
GROUP BY S.rating
HAVING 1 < (SELECT COUNT (*)
FROM Sailors S2
WHERE S.rating=S2.rating)
Shows HAVING clause can also contain a subquery.
 Compare this with the query where we considered
only ratings with 2 sailors over 18!
 What if HAVING clause is replaced by:

– HAVING COUNT(*) >1
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Find those ratings for which the average
age is the minimum over all ratings

Aggregate operations cannot be nested! WRONG:
SELECT S.rating
FROM Sailors S
WHERE S.age = (SELECT MIN (AVG (S2.age)) FROM Sailors S2)

Correct solution (in SQL/92):
SELECT Temp.rating, Temp.avgage
FROM (SELECT S.rating, AVG (S.age) AS avgage
FROM Sailors S
GROUP BY S.rating) AS Temp
WHERE Temp.avgage = (SELECT MIN (Temp.avgage)
FROM Temp)
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Testing Example 1
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find name and age of the oldest students
Find name and age of the youngest students
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Testing Example 2
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the age of the youngest student for
each grade level at ITCS6160.
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Testing Example 3
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
find the sum of all the ages of students, and
the count, who are enrolled in ITCS61060
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Testing Example 4
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the age of the youngest student with
age>18, for each grade level in ITCS6160
with at least 2 such students
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Testing Example 5
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the age of the youngest student with
age>18, for each grade level in Database at
Spring 2016 with at least 2 such students
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Testing Example 6
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
For each course, find the number of enrolled
students for this course
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Testing Example 7
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the age of the youngest student with
age > 18, for each grade level with at least 2
students (of any age)
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Testing Example 8
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find the age of the youngest student in Database
at Spring 2016 with age > 18, for each grade
level with at least 2 students (of any age)
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Testing Example 9
Name
Sid
Login
Students
Age
Cid
GPA
Name
Credit
Courses
Enrolled_In
Grade
semester
Find those grade levels for which the average
age is the minimum over all grade levels
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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.
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Summary (Contd.)
NULL for unknown field values brings many
complications
 Embedded SQL allows execution within a host
language; cursor mechanism allows retrieval of
one record at a time
 APIs such as ODBC and ODBC introduce a layer
of abstraction between application and DBMS
 SQL allows specification of rich integrity
constraints
 Triggers respond to changes in the database

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Homework 4
Transform the following high-level queries
into SQL and obtain their relational algebra
(follow the first three homeworks)

1. Find the name of the chair for toy department
2. Find the budget for employee Dr. Li’s working department
3. Find the name for employee Dr. Li’s wife, Dr. Li is working
toy department since 2002.
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since
name
dname
ssn
did
lot
Employees
Manages
budget
Departments
Works_In
since
pname
Policy
age
Dependents
cost
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