Lecture8-SQL-PartI-Jan30-2018 test Lecture8-SQL-PartI-Jan30-2018 test

Database Applications (15-415)
SQL-Part I
Lecture 8, January 30, 2018
Mohammad Hammoud

Today…
 Last Session:
 Relational Calculus
 Today’s Session:
 Standard Query Language (SQL)- Part I
 Announcements:
 PS2 is due on Sunday, Feb 04 by midnight
 Quiz I will be on Sunday, Feb 11
 P1 is due on Thursday, Feb 15 by midnight
 In this week’s recitation, we will practice on SQL

Outline
SQL Major Aspects
Basic SQL Queries
Set Operations
Aggregate Functions & Group By, Having and
Order By Clauses


SQL Major Aspects
 A major strength of the relational model is that it supports
simple and powerful querying of data
 Structured Query Language (SQL) is the most widely used
commercial relational database language
 SQL has several aspects to it:
1. Data Manipulation Language (DML)
 It allows users to pose queries and insert, delete
and modify rows
2. Data Definition Language (DDL)
 It allows users to create, delete, and modify tables and views

SQL Major Aspects
3. Triggers and Advanced Integrity Constraints
 It supports “triggers”, which are actions executed by the
DBMS whenever changes to the database meet conditions
specified in triggers
4. Embedded and Dynamic Language
 Embedded SQL allows SQL code to be called from a host
language (e.g., Java)
 Dynamic SQL allows SQL queries to be constructed and
executed at run-time

SQL Major Aspects
5. Remote Database Access
 It allows connecting client programs to remote
database servers
6. Transaction Management
 It allows users to explicitly control aspects of how a
transaction is to be executed (later in the semester)
7. Security
 It provides mechanisms to control users’ accesses to data
objects (e.g., tables and views)
And others…

Basic SQL Queries
 The basic form of an SQL query is as follows:
select a1, a2, … an
from r1, r2, … rm
where P
The Column-List
The Relation-List
Qualification (Optional)

Equivalence to Relational Algebra
 The basic form of an SQL query is as follows:
select a1, a2, … an
from r1, r2, … rm
where P
))
...
2
1
(
(
,...
2
,
1 rm
r
r
P
an
a
a 




join

Reminder: Our Mini-U DB
STUDENT
Ssn Name Address
123 smith main str
234 jones QF ave
CLASS
c-id c-name units
15-413 s.e. 2
15-412 o.s. 2
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B

The WHERE Clause
 Find the ssn(s) of everybody called “smith”
select ssn
from student
where name=‘smith’
STUDENT
Ssn Name Address
123 smith main str
234 jones QF ave

The WHERE Clause
 Find ssn(s) of all “smith”s on “main”
select ssn
from student
where address=‘main’ and
name = ‘smith’
STUDENT
Ssn Name Address
123 smith main str
234 jones QF ave

The WHERE Clause
 Boolean operators (and, or, not)
 Comparison operators (<, ≤, >, ≥, =, ≠)
 And more…

What About Strings?
 Find student ssn(s) who live on “main” (st or
str or street – i.e., “main st” or “main str” or
“main street”)
select ssn
from student
where address like ‘main%’
%: Variable-length do not care (i.e., stands for 0 or more arbitrary characters)
_: Single-character do not care (i.e., stands for any 1 character)

Another Example on Pattern Matching
 Find the ages of sailors whose names begin and end
with B and have at least 3 characters
Sailors
Sid Sname Rating age
22 Dustin 7 45.0
29 Brutus 1 33.0
select S.age
from Sailors S
where S.sname like ‘B_%B’

The FROM Clause
 Find the names of students taking 15-415
STUDENT
Ssn Name Address
123 smith main str
234 jones QF ave
CLASS
c-id c-name units
15-413 s.e. 2
15-412 o.s. 2
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B
2-way Join!

The FROM Clause
select Name
from STUDENT, TAKES
where ???

The FROM Clause
select Name
from STUDENT, TAKES
where STUDENT.ssn = TAKES.ssn
and TAKES.c-id = ‘15-415’

Renaming: Tuple Variables
select Name
from STUDENT as S, TAKES as T
where S.ssn = T.ssn
and T.c-id = “15-415”
Optional!

Renaming: Self-Joins
 Find Tom’s grandparent(s)
PC
p-id c-id
Mary Tom
Peter Mary
John Tom
PC
p-id c-id
Mary Tom
Peter Mary
John Tom
select gp.p-id
from PC as gp, PC
where gp.c-id= PC.p-id
and PC.c-id = ‘Tom’

More on Self-Joins
 Find names and increments for the ratings of persons
who have sailed two different boats on the same day
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
Reserves
Sid Bid Day
22 101 10/10/2013
22 102 10/10/2013

More on Self-Joins
 Find names and increments for the ratings of persons
who have sailed two different boats on the same day
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
Reserves
Sid Bid Day
22 101 10/10/2013
22 102 10/10/2013
select S.sname, S.rating+1 as rating
from Sailors S, Reserves R1, Reserves R2
where S.sid = R1.sid and S.sid = R2.sid
and R1.day = R2.day and R1.bid != R2.bid

Renaming: Theta Joins
 Find course names with more units than 15-415
select c1.c-name
from class as c1, class as c2
where c1.units > c2.units
and c2.c-id = ‘15-415’
CLASS
c-id c-name units
15-413 s.e. 2
15-412 o.s. 2

Set Operations
 Find ssn(s) of students taking both 15-415 and 15-413
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B
select ssn
from takes
where c-id=‘15-415’ and
c-id=‘15-413’

Set Operations
 Find ssn(s) of students taking both 15-415 and 15-413
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B
(select ssn from takes where c-id=“15-415” )
intersect
Other operations: union , except

Set Operations
 Find ssn(s) of students taking 15-415 or 15-413
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B
union

Set Operations
 Find ssn(s) of students taking 15-415 but not 15-413
TAKES
SSN c-id grade
123 15-413 A
234 15-413 B
except

Another Example on Set Operations
 Find the names of sailors who have reserved both a
red and a green boat
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
Reserves
Sid Bid Day
22 101 10/10/2013
22 102 10/11/2013
Boats
Bid Bname Color
101 Interlake Red
102 Clipper Green

Another Example on Set Operations
 Find the names of sailors who have reserved both a
red and a green boat
(select S.sname from Sailors S, Reserves R, Boats B
where S.sid = R.sid and R.bid = B.bid and B.color = ‘green’)
intersect
(select S2.sname from Sailors S2, Reserves R2, Boats B2
where S2.sid = R2.sid and R2.bid = B2.bid and B2.color = ‘red’)
The query contains a “subtle bug” which arises because we are using sname to
identify Sailors, and “sname” is not a key for Sailors!
We can compute the names of such Sailors using a NESTED query (which we
cover next lecture!)

Outline
SQL Major Aspects
Basic SQL Queries
Set Operations
Aggregate Functions & Group By, Having and
Order By Clauses 

Aggregate Functions
 Find average grade, across all students
SSN c-id grade
123 15-413 4
234 15-413 3
select ??
from takes

Aggregate Functions
 Find average grade, across all students
SSN c-id grade
123 15-413 4
234 15-413 3
select avg(grade)
from takes
Other functions: Count ([Distinct] A), Sum ([Distinct] A), Max (A), Min (A),
assuming column A

Aggregate Functions
 Find total number of enrollments
SSN c-id grade
123 15-413 4
234 15-413 3
select count(*)
from takes

Aggregate Functions
 Find total number of students in 15-415
SSN c-id grade
123 15-413 4
234 15-413 3
select count(*)
from takes
where c-id=‘15-415’

Aggregate Functions
 Find the name and age of the oldest sailor
select S.sname, max (S.age)
from Sailors S
This query is illegal in SQL- If the “select” clause uses an aggregate function, it
must use ONLY aggregate function unless the query contains a “group by” clause!
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0

The GROUP BY and HAVING Clauses
 Find the age of the youngest sailor for each rating level
 In general, we do not 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 (!):
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
SELECT MIN (S.age)
FROM Sailors S
WHERE S.rating = i
For i = 1, 2, ... , 10:

 Find the age of the youngest sailor for each rating level
 Using the GROUP BY clause, we can write this query as follows:
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
select S.rating, min (S.age)
from Sailors S
group by S.rating
“Every” column that appears in the Column-List
“must” also appear in the Grouping-List
The Grouping-List

 Find age of the youngest sailor with age ≥ 18, for each
rating level with at least 2 sailors
Sailors
22 Dustin 7 45.0
29 Brutus 1 33.0
SELECT S.rating, MIN (S.age) AS minage
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1

rating level with at least 2 sailors
rating age
7 45.0
1 33.0
8 55.5
8 25.5
10 35.0
7 35.0
10 16.0
9 35.0
3 25.5
3 63.5
3 25.5
rating minage
3 25.5
7 35.0
8 25.5
rating age
1 33.0
3 25.5
3 63.5
3 25.5
7 45.0
7 35.0
8 55.5
8 25.5
9 35.0
10 35.0

rating level with at least 2 sailors, and with every sailor
under 60
FROM Sailors S
WHERE S.age >= 18
GROUP BY S.rating
HAVING COUNT (*) > 1 AND EVERY (S.age <=60)

rating level with at least 2 sailors, and with every sailor
under 60
rating age
7 45.0
1 33.0
8 55.5
8 25.5
10 35.0
7 35.0
10 16.0
9 35.0
3 25.5
3 63.5
3 25.5
rating age
1 33.0
3 25.5
3 63.5
3 25.5
7 45.0
7 35.0
8 55.5
8 25.5
9 35.0
10 35.0
rating minage
7 35.0
8 25.5
What would be the result if
we change EVERY to ANY in
“HAVING COUNT (*) > 1
AND EVERY (S.age <=60)”?

rating level with at least 2 sailors between 18 and 60
FROM Sailors S
WHERE S.age >= 18 AND S.age <= 60
GROUP BY S.rating
HAVING COUNT (*) > 1
Will this give the same result as the previous query which uses the
EVERY clause?
Will this give the same result as the previous query which uses the
ANY clause?

select *
from student
where ??
The ORDER BY Clause
 Find student records, sorted in name order

select *
from student
order by name asc
The ORDER BY Clause
 Find student records, sorted in name order
asc is the default

select *
from student
order by name, ssn desc
The ORDER BY Clause
 Find student records, sorted in name order;
break ties by reverse ssn

More Examples
 Find the total number of students in each course
SSN c-id grade
123 15-413 4
234 15-413 3
select count(*)
from takes
where ???

More Examples
 Find the total number of students in each course
SSN c-id grade
123 15-413 4
234 15-413 3
select c-id, count(*)
from takes
group by c-id
c-id count
15-413 2

More Examples
 Find total number of students in each course,
and sort by count, in decreasing order
SSN c-id grade
123 15-413 4
234 15-413 3
select c-id, count(*) as pop
from takes
group by c-id
order by pop desc
c-id pop
15-413 2

Concluding Remarks
 SQL was an important factor in the early acceptance of
the relational model
 It is more natural than earlier procedural
query languages
 SQL is relationally complete; in fact, significantly more
expressive power than relational algebra
 Even queries that can be expressed in relational
algebra can often be expressed more naturally in SQL

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