SQL.ppt

Chapter 3: Introduction to
SQL

Outline
• Overview of The SQL Query Language
• Data Definition
• Basic Query Structure
• Additional Basic Operations
• Set Operations
• Null Values
• Aggregate Functions
• Nested Subqueries
• Modification of the Database

History
IBM Sequel language developed as part of System R project at the
IBM San Jose Research Laboratory
Renamed Structured Query Language (SQL)
ANSI and ISO standard SQL:
 SQL-86
 SQL-89
 SQL-92
 SQL:1999 (language name became Y2K compliant!)
 SQL:2003
Commercial systems offer most, if not all, SQL-92 features, plus
varying feature sets from later standards and special proprietary
features.
 Not all examples here may work on your particular system.

Data Definition Language
The schema for each relation.
The domain of values associated with each attribute.
Integrity constraints
And as we will see later, also other information such as
 The set of indices to be maintained for each relations.
 Security and authorization information for each relation.
 The physical storage structure of each relation on disk.
The SQL data-definition language (DDL) allows the specification
of information about relations, including:

Domain Types in SQL
char(n). Fixed length character string, with user-specified length n.
varchar(n). Variable length character strings, with user-specified
maximum length n.
int. Integer (a finite subset of the integers that is machine-
dependent).
smallint. Small integer (a machine-dependent subset of the integer
domain type).
numeric(p,d). Fixed point number, with user-specified precision of
p digits, with d digits to the right of decimal point. (ex.,
numeric(3,1), allows 44.5 to be stores exactly, but not 444.5 or 0.32)
real, double precision. Floating point and double-precision floating
point numbers, with machine-dependent precision.
float(n). Floating point number, with user-specified precision of at
least n digits.
More are covered in Chapter 4.

Create Table Construct
• An SQL relation is defined using the create table command:
create table r (A1 D1, A2 D2, ..., An Dn,
(integrity-constraint1),
...,
(integrity-constraintk))
• r is the name of the relation
• each Ai is an attribute name in the schema of relation r
• Di is the data type of values in the domain of attribute Ai
• Example:
create table instructor (
ID char(5),
name varchar(20),
dept_name varchar(20),
salary numeric(8,2))

Integrity Constraints in Create Table
• not null
• primary key (A1, ..., An )
• foreign key (Am, ..., An ) references r
Example:
create table instructor (
ID char(5),
name varchar(20) not null,
salary numeric(8,2),
primary key (ID),
foreign key (dept_name) references department);
primary key declaration on an attribute automatically ensures not null

And a Few More Relation Definitions
create table student (
ID varchar(5),
name varchar(20) not null,
tot_cred numeric(3,0),
primary key (ID),
create table takes (
ID varchar(5),
course_id varchar(8),
sec_id varchar(8),
semester varchar(6),
year numeric(4,0),
grade varchar(2),
primary key (ID, course_id, sec_id, semester, year) ,
foreign key (ID) references student,
foreign key (course_id, sec_id, semester, year) references section);

And more still
• create table course (
course_id varchar(8),
title varchar(50),
credits numeric(2,0),
primary key (course_id),

Updates to tables
Insert
 insert into instructor values (‘10211’, ’Smith’, ’Biology’, 66000);
Delete
 Remove all tuples from the student relation
delete from student
Drop Table
 drop table r
Alter
 alter table r add A D
 where A is the name of the attribute to be added to relation r and D is the
domain of A.
All exiting tuples in the relation are assigned null as the value for the new
attribute.
 alter table r drop A
where A is the name of an attribute of relation r
Dropping of attributes not supported by many databases.

Basic Query Structure
• A typical SQL query has the form:
select A1, A2, ..., An
from r1, r2, ..., rm
where P
• Ai represents an attribute
• Ri represents a relation
• P is a predicate.
• The result of an SQL query is a relation.

The select Clause
• The select clause lists the attributes desired in the result of a query
• corresponds to the projection operation of the relational algebra
• Example: find the names of all instructors:
select name
from instructor
• NOTE: SQL names are case insensitive (i.e., you may use upper- or
lower-case letters.)
• E.g., Name ≡ NAME ≡ name
• Some people use upper case wherever we use bold font.

The select Clause (Cont.)
• SQL allows duplicates in relations as well as in query results.
• To force the elimination of duplicates, insert the keyword distinct
after select.
• Find the department names of all instructors, and remove
duplicates
select distinct dept_name
from instructor
• The keyword all specifies that duplicates should not be removed.
select all dept_name
from instructor

• An asterisk in the select clause denotes “all attributes”
select *
from instructor
• An attribute can be a literal with no from clause
select ‘437’
• Results is a table with one column and a single row with value “437”
• Can give the column a name using:
select ‘437’ as FOO
• An attribute can be a literal with from clause
select ‘A’
from instructor
• Result is a table with one column and N rows (number of tuples in the
instructors table), each row with value “A”

• The select clause can contain arithmetic expressions involving the
operation, +, –, , and /, and operating on constants or attributes of
tuples.
• The query:
select ID, name, salary/12
from instructor
would return a relation that is the same as the instructor relation, except
that the value of the attribute salary is divided by 12.
• Can rename “salary/12” using the as clause:
select ID, name, salary/12 as monthly_salary

The where Clause
The where clause specifies conditions that the result must satisfy
 Corresponds to the selection predicate of the relational algebra.
To find all instructors in Comp. Sci. dept
select name
from instructor
where dept_name = ‘Comp. Sci.'
Comparison results can be combined using the logical
connectives and, or, and not
 To find all instructors in Comp. Sci. dept with salary > 80000
select name
from instructor
where dept_name = ‘Comp. Sci.' and salary > 80000
Comparisons can be applied to results of arithmetic expressions.

The from Clause
• The from clause lists the relations involved in the query
• Corresponds to the Cartesian product operation of the relational algebra.
• Find the Cartesian product instructor X teaches
select 
from instructor, teaches
• generates every possible instructor – teaches pair, with all attributes from
both relations.
• For common attributes (e.g., ID), the attributes in the resulting table are
renamed using the relation name (e.g., instructor.ID)
• Cartesian product not very useful directly, but useful combined with
where-clause condition (selection operation in relational algebra).

Cartesian Product
instructor teaches

Examples
• Find the names of all instructors who have taught some course and
the course_id
• select name, course_id
from instructor , teaches
where instructor.ID = teaches.ID
• Find the names of all instructors in the Art department who have
taught some course and the course_id
from instructor , teaches
where instructor.ID = teaches.ID and instructor. dept_name = ‘Art’

The Rename Operation
• The SQL allows renaming relations and attributes using the as clause:
old-name as new-name
• Find the names of all instructors who have a higher salary than
some instructor in ‘Comp. Sci’.
• select distinct T.name
from instructor as T, instructor as S
where T.salary > S.salary and S.dept_name = ‘Comp. Sci.’
• Keyword as is optional and may be omitted
instructor as T ≡ instructor T

Self Join Example
 Relation emp-super
 Find the supervisor of “Bob”
 Find the supervisor of the supervisor of “Bob”
 Find ALL the supervisors (direct and indirect) of “Bob
person supervisor
Bob Alice
Mary Susan
Alice David
David Mary

String Operations
• SQL includes a string-matching operator for comparisons on
character strings. The operator like uses patterns that are described
using two special characters:
• percent ( % ). The % character matches any substring.
• underscore ( _ ). The _ character matches any character.
• Find the names of all instructors whose name includes the substring
“dar”.
select name
from instructor
where name like '%dar%'
• Match the string “100%”
like ‘100 %' escape ''
in that above we use backslash () as the escape character.

String Operations (Cont.)
• Patterns are case sensitive.
• Pattern matching examples:
• ‘Intro%’ matches any string beginning with “Intro”.
• ‘%Comp%’ matches any string containing “Comp” as a substring.
• ‘_ _ _’ matches any string of exactly three characters.
• ‘_ _ _ %’ matches any string of at least three characters.
• SQL supports a variety of string operations such as
• concatenation (using “||”)
• converting from upper to lower case (and vice versa)
• finding string length, extracting substrings, etc.

Ordering the Display of Tuples
• List in alphabetic order the names of all instructors
select distinct name
from instructor
order by name
• We may specify desc for descending order or asc for ascending
order, for each attribute; ascending order is the default.
• Example: order by name desc
• Can sort on multiple attributes
• Example: order by dept_name, name

Where Clause Predicates
• SQL includes a between comparison operator
• Example: Find the names of all instructors with salary between
$90,000 and $100,000 (that is,  $90,000 and  $100,000)
• select name
from instructor
where salary between 90000 and 100000
• Tuple comparison
from instructor, teaches
where (instructor.ID, dept_name) = (teaches.ID, ’Biology’);

Duplicates
• In relations with duplicates, SQL can define how many copies of
tuples appear in the result.
• Multiset versions of some of the relational algebra operators –
given multiset relations r1 and r2:
1.  (r1): If there are c1 copies of tuple t1 in r1, and t1 satisfies selections
,, then there are c1 copies of t1 in  (r1).
2. A (r ): For each copy of tuple t1 in r1, there is a copy of tuple A (t1) in
A (r1) where A (t1) denotes the projection of the single tuple t1.
3. r1 x r2: If there are c1 copies of tuple t1 in r1 and c2 copies of tuple t2 in
r2, there are c1 x c2 copies of the tuple t1. t2 in r1 x r2

Duplicates (Cont.)
• Example: Suppose multiset relations r1 (A, B) and r2 (C) are as
follows:
r1 = {(1, a) (2,a)} r2 = {(2), (3), (3)}
• Then B(r1) would be {(a), (a)}, while B(r1) x r2 would be
{(a,2), (a,2), (a,3), (a,3), (a,3), (a,3)}
• SQL duplicate semantics:
select A1,, A2, ..., An
where P
is equivalent to the multiset version of the expression:
))
(
( 2
1
,
,
, 2
1 m
P
A
A
A r
r
r
n



 
 

Set Operations
• Find courses that ran in Fall 2009 or in Spring 2010
 Find courses that ran in Fall 2009 but not in Spring 2010
(select course_id from section where sem = ‘Fall’ and year = 2009)
union
(select course_id from section where sem = ‘Spring’ and year = 2010)
 Find courses that ran in Fall 2009 and in Spring 2010
intersect
except

Set Operations (Cont.)
• Find the salaries of all instructors that are less than the largest salary.
• select distinct T.salary
where T.salary < S.salary
• Find all the salaries of all instructors
• select distinct salary
from instructor
• Find the largest salary of all instructors.
• (select “second query” )
except
(select “first query”)

Set Operations (Cont.)
• Set operations union, intersect, and except
• Each of the above operations automatically eliminates duplicates
To retain all duplicates use the corresponding multiset versions
union all, intersect all and except all.
Suppose a tuple occurs m times in r and n times in s, then, it
occurs:
• m + n times in r union all s
• min(m,n) times in r intersect all s
• max(0, m – n) times in r except all s

Null Values
• It is possible for tuples to have a null value, denoted by null,
for some of their attributes
• null signifies an unknown value or that a value does not exist.
• The result of any arithmetic expression involving null is null
• Example: 5 + null returns null
• The predicate is null can be used to check for null values.
• Example: Find all instructors whose salary is null.
select name
from instructor
where salary is null

Null Values and Three Valued Logic
Three values – true, false, unknown
Any comparison with null returns unknown
 Example: 5 < null or null <> null or null = null
Three-valued logic using the value unknown:
 OR: (unknown or true) = true,
(unknown or false) = unknown
(unknown or unknown) = unknown
 AND: (true and unknown) = unknown,
(false and unknown) = false,
(unknown and unknown) = unknown
 NOT: (not unknown) = unknown
 “P is unknown” evaluates to true if predicate P evaluates to unknown
Result of where clause predicate is treated as false if it evaluates
to unknown

Aggregate Functions
• These functions operate on the multiset of values of a
column of a relation, and return a value
avg: average value
min: minimum value
max: maximum value
sum: sum of values
count: number of values

Aggregate Functions (Cont.)
• Find the average salary of instructors in the Computer Science
department
• select avg (salary)
from instructor
where dept_name= ’Comp. Sci.’;
• Find the total number of instructors who teach a course in the Spring
2010 semester
• select count (distinct ID)
from teaches
where semester = ’Spring’ and year = 2010;
• Find the number of tuples in the course relation
• select count (*)
from course;

Aggregate Functions – Group By
• Find the average salary of instructors in each department
• select dept_name, avg (salary) as avg_salary
from instructor
group by dept_name;
avg_salary

Aggregation (Cont.)
• Attributes in select clause outside of aggregate functions must
appear in group by list
• /* erroneous query */
select dept_name, ID, avg (salary)
from instructor
group by dept_name;

Aggregate Functions – Having Clause
• Find the names and average salaries of all departments whose
average salary is greater than 42000
Note: predicates in the having clause are applied after the
formation of groups whereas predicates in the where
clause are applied before forming groups
select dept_name, avg (salary)
from instructor
group by dept_name
having avg (salary) > 42000;

Null Values and Aggregates
• Total all salaries
select sum (salary )
from instructor
• Above statement ignores null amounts
• Result is null if there is no non-null amount
• All aggregate operations except count(*) ignore tuples with null
values on the aggregated attributes
• What if collection has only null values?
• count returns 0
• all other aggregates return null

Nested Subqueries
SQL provides a mechanism for the nesting of subqueries. A
subquery is a select-from-where expression that is nested within
another query.
The nesting can be done in the following SQL query
select A1, A2, ..., An
where P
as follows:
 Ai can be replaced be a subquery that generates a single value.
 ri can be replaced by any valid subquery
 P can be replaced with an expression of the form:
B <operation> (subquery)
Where B is an attribute and <operation> to be defined later.

Subqueries in the Where Clause

Subqueries in the Where Clause
• A common use of subqueries is to perform tests:
• For set membership
• For set comparisons
• For set cardinality.

Set Membership
• Find courses offered in Fall 2009 and in Spring 2010
 Find courses offered in Fall 2009 but not in Spring 2010
select distinct course_id
from section
where semester = ’Fall’ and year= 2009 and
course_id in (select course_id
from section
where semester = ’Spring’ and year= 2010);
select distinct course_id
from section
where semester = ’Fall’ and year= 2009 and
course_id not in (select course_id
from section
where semester = ’Spring’ and year= 2010);

Set Membership (Cont.)
• Find the total number of (distinct) students who have taken course
sections taught by the instructor with ID 10101
 Note: Above query can be written in a much simpler manner.
The formulation above is simply to illustrate SQL features.
select count (distinct ID)
from takes
where (course_id, sec_id, semester, year) in
(select course_id, sec_id, semester, year
from teaches
where teaches.ID= 10101);

Set Comparison – “some” Clause
• Find names of instructors with salary greater than that of some (at
least one) instructor in the Biology department.
 Same query using > some clause
select name
from instructor
where salary > some (select salary
from instructor
where dept name = ’Biology’);
select distinct T.name
where T.salary > S.salary and S.dept name = ’Biology’;

Definition of “some” Clause
• F <comp> some r t  r such that (F <comp> t )
Where <comp> can be:     
0
5
6
(5 < some ) = true
0
5
0
) = false
5
0
5
(5  some ) = true (since 0  5)
(read: 5 < some tuple in the relation)
(5 < some
) = true
(5 = some
(= some)  in
However, ( some)  not in

Set Comparison – “all” Clause
• Find the names of all instructors whose salary is greater than the
salary of all instructors in the Biology department.
select name
from instructor
where salary > all (select salary
from instructor
where dept name = ’Biology’);

Definition of “all” Clause
• F <comp> all r t  r (F <comp> t)
0
5
6
(5 < all ) = false
6
10
4
) = true
5
4
6
(5  all ) = true (since 5  4 and 5  6)
(5 < all
) = false
(5 = all
( all)  not in
However, (= all)  in

Test for Empty Relations
• The exists construct returns the value true if the
argument subquery is nonempty.
• exists r  r  Ø
• not exists r  r = Ø

Use of “exists” Clause
Yet another way of specifying the query “Find all courses taught in
both the Fall 2009 semester and in the Spring 2010 semester”
select course_id
from section as S
where semester = ’Fall’ and year = 2009 and
exists (select *
from section as T
where semester = ’Spring’ and year= 2010
and S.course_id = T.course_id);
Correlation name – variable S in the outer query
Correlated subquery – the inner query

Use of “not exists” Clause
• Find all students who have taken all courses offered in the Biology
department.
select distinct S.ID, S.name
from student as S
where not exists ( (select course_id
from course
where dept_name = ’Biology’)
except
(select T.course_id
from takes as T
where S.ID = T.ID));
• First nested query lists all courses offered in Biology
• Second nested query lists all courses a particular student took
 Note that X – Y = Ø  X Y
 Note: Cannot write this query using = all and its variants

Test for Absence of Duplicate Tuples
• The unique construct tests whether a subquery has any
duplicate tuples in its result.
• The unique construct evaluates to “true” if a given subquery
contains no duplicates .
• Find all courses that were offered at most once in 2009
select T.course_id
from course as T
where unique (select R.course_id
from section as R
where T.course_id= R.course_id
and R.year = 2009);

Subqueries in the Form Clause
• SQL allows a subquery expression to be used in the from clause
• Find the average instructors’ salaries of those departments where
the average salary is greater than $42,000.”
select dept_name, avg_salary
from (select dept_name, avg (salary) as avg_salary
from instructor
group by dept_name)
where avg_salary > 42000;
• Note that we do not need to use the having clause
• Another way to write above query
select dept_name, avg_salary
from (select dept_name, avg (salary)
from instructor
group by dept_name) as dept_avg (dept_name, avg_salary)
where avg_salary > 42000;

With Clause
• The with clause provides a way of defining a temporary relation
whose definition is available only to the query in which the with
clause occurs.
• Find all departments with the maximum budget
with max_budget (value) as
(select max(budget)
from department)
select department.name
from department, max_budget
where department.budget = max_budget.value;

Complex Queries using With Clause
• Find all departments where the total salary is greater than the average
of the total salary at all departments
with dept _total (dept_name, value) as
(select dept_name, sum(salary)
from instructor
group by dept_name),
dept_total_avg(value) as
(select avg(value)
from dept_total)
select dept_name
from dept_total, dept_total_avg
where dept_total.value > dept_total_avg.value;

Subqueries in the Select Clause

Scalar Subquery
• Scalar subquery is one which is used where a single value is expected
• List all departments along with the number of instructors in each
department
select dept_name,
(select count(*)
from instructor
where department.dept_name = instructor.dept_name)
as num_instructors
from department;
• Runtime error if subquery returns more than one result tuple

Modification of the Database
• Deletion of tuples from a given relation.
• Insertion of new tuples into a given relation
• Updating of values in some tuples in a given relation

Deletion
• Delete all instructors
delete from instructor
• Delete all instructors from the Finance department
where dept_name= ’Finance’;
• Delete all tuples in the instructor relation for those instructors
associated with a department located in the Watson building.
where dept name in (select dept name
from department
where building = ’Watson’);

Deletion (Cont.)
• Delete all instructors whose salary is less than the average salary of
instructors
where salary < (select avg (salary)
from instructor);
 Problem: as we delete tuples from deposit, the average salary
changes
 Solution used in SQL:
1. First, compute avg (salary) and find all tuples to delete
2. Next, delete all tuples found above (without
recomputing avg or retesting the tuples)

Insertion
• Add a new tuple to course
insert into course
values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4);
• or equivalently
insert into course (course_id, title, dept_name, credits)
values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4);
• Add a new tuple to student with tot_creds set to null
insert into student
values (’3003’, ’Green’, ’Finance’, null);

Insertion (Cont.)
• Add all instructors to the student relation with tot_creds set to 0
insert into student
select ID, name, dept_name, 0
from instructor
• The select from where statement is evaluated fully before any of its
results are inserted into the relation.
Otherwise queries like
insert into table1 select * from table1
would cause problem

Updates
• Increase salaries of instructors whose salary is over $100,000
by 3%, and all others by a 5%
• Write two update statements:
update instructor
set salary = salary * 1.03
where salary > 100000;
update instructor
set salary = salary * 1.05
where salary <= 100000;
• The order is important
• Can be done better using the case statement (next slide)

Case Statement for Conditional Updates
• Same query as before but with case statement
update instructor
set salary = case
when salary <= 100000 then salary * 1.05
else salary * 1.03
end

Updates with Scalar Subqueries
Recompute and update tot_creds value for all students
update student S
set tot_cred = (select sum(credits)
from takes, course
where takes.course_id = course.course_id and
S.ID= takes.ID.and
takes.grade <> ’F’ and
takes.grade is not null);
Sets tot_creds to null for students who have not taken any course
Instead of sum(credits), use:
case
when sum(credits) is not null then sum(credits)
else 0
end

SQL.ppt

More Related Content

Similar to SQL.ppt (20)

Recently uploaded (20)

SQL.ppt