BKM_DATS: Databázové systémy
2. SQL
Vlastislav Dohnal
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 2
Contents
History of the SQL Query Language
Basic Query Structure
Additional Basic Operations
Set Operations
Null Values
Aggregate Functions
Nested Subqueries
Join Expressions
Views
Modification of the Database
Data Definition Language
SQL Data Types and Schemas
Integrity Constraints
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 3
History
IBM Sequel language developed as part of System R project at the
IBM San Jose Research Laboratory
Renamed to Structured Query Language (SQL)
ANSI and ISO standard SQL:
SQL-86; SQL-89
SQL-92
SQL:1999 (language name became Y2K compliant!)
SQL:2003
SQL:2006 (adds XML support)
SQL:2008
SQL:2011 (adds support for temporal databases)
Commercial systems offer most, if not all, SQL-99 features
plus varying feature sets from later standards and
special proprietary features
Not all examples here may work on your particular system.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 4
Basic Query Structure
A typical SQL query has the form:
select A1, A2, ..., An
from r1, r2, ..., rm
where C
Ai represents an attribute
Ri represents a relation
C is a condition.
The result of an SQL query is a relation.
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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:
Relation
instructor (id, name, dept_name, salary)
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.
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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 names of all departments of instructors, and remove duplicates
select distinct dept_name
from instructor
The keyword all specifies that duplicates not to be removed.
select all dept_name
from instructor
It is also an implicit behavior when the keyword all is omitted.
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The select Clause (Cont.)
Relation
instructor (id, name, dept_name, salary)
An asterisk in the select clause denotes “all attributes”
select *
from instructor
The select clause can contain arithmetic expressions
Involving the operations: +, –, , and /,
Operating on constants or attributes of tuples.
Also, function can be used (nullif(), upper(), to_char(), …)
The query:
select id, name, dept_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.
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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.' department with salary > 80000
select name
from instructor
where dept_name = 'Comp. Sci.' and salary > 80000
Comparison results can be combined using the logical connectives
and, or, not
Comparisons can be applied to results of arithmetic expressions.
select name
from instructor
where salary / 12 > 6000
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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  teaches
select 
from instructor, teaches
Generates every possible instructor-teaches pair, with all attributes
from both relations.
Cartesian product not very useful directly,
but useful when combined with a where-clause condition (selection
operation in relational algebra).
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Cartesian Product
instructor teaches
instructor  teaches
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Joins
Relations:
instructor (id, name, dept_name, salary)
course (course_id, title, dept_name)
section (sec_id, semestr, year)
teaches (id, course_id, sec_id)
For all instructors who teach courses, find their names and the course id
of the courses they teach.
select name, course_id
from instructor, teaches
where instructor.id = teaches.id
Find the course id, title, semester and year of each course offered by the
“Comp. Sci.” department
select course.course_id, title, semester, year
from course, teaches, section
where course.course_id = teaches.course_id and
teaches.sec_id = section.sec_id and
dept_name = 'Comp. Sci.'
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Natural Join
Natural join matches tuples with the same values for all common
attributes, and retains only one copy of each common column
For relations:
instructor (id, name, dept_name, salary)
teaches (id, course_id, sec_id, semestr, year)
select * from instructor natural join teaches;
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Natural Join (Cont.)
Danger in natural join:
beware of unrelated attributes with same name which get equated incorrectly
Relations:
instructor (id, name, dept_name, salary)
course (course_id, title, dept_name)
section (sec_id, semester, year)
teaches (id, course_id, sec_id)
List the names of instructors along with the titles of courses that they teach.
Incorrect version (equates course.dept_name with instructor.dept_name)
select name, title
from (instructor natural join teaches) natural join course;
Correct version
select name, title
from (instructor natural join teaches), course
where teaches.course_id= course.course_id;
Another correct version
select name, title
from (instructor natural join teaches) join course using(course_id);
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The Rename Operation
The SQL allows renaming relations and attributes using the as clause:
old-name as new-name
E.g.,
select id, name, salary/12 as monthly_salary
from instructor
Find the names of all instructors who have a salary higher 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 in renaming relations
instructor as T ≡ instructor T
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Ordering the Display of Tuples
List in alphabetic order the names of all instructors
select 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
or … order by dept_name desc, name asc
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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
select name, course_id
from instructor, teaches
where (instructor.ID, dept_name) = (teaches.ID, ’Biology’);
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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 containing “100 %”
… like ‘%100 \%%' escape '\'
SQL supports a variety of string operations such as
concatenation (using “||”)
converting from upper to lower case (and vice versa)
functions upper() and lower()
finding string length, extracting substrings, etc.
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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
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Null Values and Three-valued Logic
Any comparison with null returns null
Example: 5 < null or null <> null or null = null
Three-valued logic using the truth value null:
OR: (null or true) = true
(null or false) = null
(null or null) = null
AND: (true and null) = null
(false and null) = false
(null and null) = null
NOT: (not null) = null
Result of where clause predicate is treated as false if it evaluates to
null
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Set Operations (union, intersect, except)
Relation:
teaches (id, course_id, sec_id, semester, year)
Find courses that ran in Fall 2009 or in Spring 2010
Find courses that ran in Fall 2009 and in Spring 2010
Find courses that ran in Fall 2009 but not in Spring 2010
(select course_id from teaches where semester = ‘Fall’ and year = 2009)
union
(select course_id from teaches where semester = ‘Spring’ and year = 2010)
(select course_id from teaches where semester = ‘Fall’ and year = 2009)
intersect
(select course_id from teaches where semester = ‘Spring’ and year = 2010)
(select course_id from teaches where semester = ‘Fall’ and year = 2009)
except
(select course_id from teaches where semester = ‘Spring’ and year = 2010)
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Set Operations
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
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Praktické cvičení
Cvičení SQL, první část
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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
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Aggregate Functions (Cont.)
Relations:
instructor (id, name, dept_name, salary)
teaches (id, course_id, sec_id, semestr, year)
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;
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Aggregate Functions – Group By
Find the average salary of instructors in each department
select dept_name, avg (salary)
from instructor
group by dept_name;
avg
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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;
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Aggregate Functions – Having Clause
Relations:
instructor (id, name, dept_name, salary)
Find the names and average salaries of all departments whose
average salary is greater than 42,000
Note: predicates in the having clause are applied after the
formation of groups whereas predicates in the where
clause are applied before forming groups.
Note2: so aggregate functions cannot be used in where clause.
select dept_name, avg (salary)
from instructor
group by dept_name
having avg (salary) > 42000;
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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
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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.
A common use of subqueries is to perform tests for
set membership,
set comparisons, and
set cardinality.
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Example Query: set membership
Operators: IN NOT IN
Relations:
teaches (id, course_id, sec_id, semester, year)
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);
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Example Query: set membership (cont.)
Relations:
instructor (id, name, dept_name, salary)
teaches (id, course_id, sec_id, semester, year)
takes (id, course_id, sec_id, semester, year)
student (id, name)
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 course_id, sec_id, semester, year
from teaches
where teaches.ID = 10101);
select count (distinct ID)
from takes
where (course_id, sec_id, semester, year) in
Attribute ID is a
reference to
student, here.
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Set Comparison
Relations:
instructor (id, name, dept_name, salary)
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
from instructor as T, instructor as S
where T.salary > S.salary and S.dept name = ’Biology’;
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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
any is an
equivalent
of some
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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
Set comparison and NULL values
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 35
0
5
null
(5 in ) = true
0
5
6
(null in ) = false
0
5
null
(2 not in ) = false!!!
0
5
null
(null in ) = false!!!
0
5
null
(2 in ) = false
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Example Query
Relations:
instructor (id, name, dept_name, salary)
Find the names of 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’);
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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
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Correlation Variables
Relations:
section (sec_id, semestr, year)
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);
Correlated subquery
Correlation name or correlation variable
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Not Exists
Relations:
student (id, name)
takes (id, course_id, sec_id, semester, year)
course (course_id, title, dept_name)
Find students who have taken all courses offered in the Biology department.
Remark that X – Y = Ø  X  Y
Note: Cannot write this query using = all and its variants
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));
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 40
Derived Relations
SQL allows a subquery expression to be used in the from clause
Find the departments where the average salary is greater than $42,000.
Print the average salary too.
select dept_name, avg_salary
from (select dept_name, avg (salary) as avg_salary
from instructor
group by dept_name) as dept_avg
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;
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Scalar Subquery
Relations:
instructor (id, name, dept_name, salary)
department (dept_name, building, budget)
select dept_name,
(select count(*)
from instructor
where department.dept_name = instructor.dept_name)
as num_instructors
from department;
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Joined Relations
Join operations take two relations and return as a result another
relation.
A join operation is a Cartesian product which requires that tuples in
the two relations match (under some condition).
It also specifies the attributes that are present in the result of the
join.
The join operations are typically used as sub-query expressions in the
from clause.
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Outer Join
An extension of the join operation that avoids loss of information.
Computes the join and then adds tuples form one relation that does
not match tuples in the other relation to the result of the join.
Uses null values.
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Left Outer Join
course prereq
course natural left outer join prereq
prereq_id
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Right Outer Join
course prereq
course natural right outer join prereq
prereq_id
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Full Outer Join
course prereq
course natural full outer join prereq
prereq_id
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Joined Relations
Join operations take two relations and return as a result another
relation.
These additional operations are typically used as subquery
expressions in the from clause
Join condition – defines which tuples in the two relations match, and
what attributes are present in the result of the join.
Join type – defines how tuples in each relation that do not match any
tuple in the other relation (based on the join condition) are treated.
Join type
inner join
left outer join
right outer join
full outer join
Join condition Usage
natural r1 natural <join_type> r2
on <predicate> r1 <join_type> r2 on <predicate>
using (A1,A2,…An) r1 <join_type> r2 using (A1,A2,…,An)
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Joined Relations – Examples
course inner join prereq on course.course_id = prereq.course_id
course left outer join prereq on course.course_id = prereq.course_id
prereq_id
prereq_idcourse_id
course_id
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Joined Relations – Examples
course natural right outer join prereq
course full outer join prereq using (course_id)
prereq_id
prereq_id
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Praktické cvičení
Cvičení SQL, druhá část
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 51
Views
In some cases, it is not desirable for all users to see the entire
logical model (that is, all the actual relations stored in the
database.)
Consider a person who needs to know an instructor’s name and
department, but not the salary. This person should see a relation
described, in SQL, by
select id, name, dept_name
from instructor
A view provides a mechanism to hide certain data from the view
of certain users.
Any relation that is not of the conceptual model but is made
visible to a user as a “virtual relation” is called a view.
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View Definition
A view is defined using the create view statement which has the
form
create view v as < query expression >
where <query expression> is any legal SQL expression. The view
name is represented by v.
Once a view is defined, the view name can be used to refer to the
virtual relation that the view generates.
View definition is not the same as creating a new relation by
evaluating the query expression
Rather, a view definition causes the saving of an expression;
the expression is substituted into queries using the view.
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Example Views
A view of instructors without their salary
create view faculty as
select ID, name, dept_name
from instructor
Find all instructors in the Biology department
select name
from faculty
where dept_name = ‘Biology’
Create a view of department salary totals
create view departments_total_salary(dept_name, total_salary) as
select dept_name, sum (salary)
from instructor
group by dept_name;
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 54
Views Defined Using Other Views
create view physics_fall_2009 as
select course.course_id, sec_id, building, room_number
from course, section
where course.course_id = section.course_id
and course.dept_name = ’Physics’
and section.semester = ’Fall’
and section.year = ’2009’;
create view physics_fall_2009_watson as
select course_id, room_number
from physics_fall_2009
where building = ’Watson’;
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View Expansion
Expand use of a view in a query/another view
create view physics_fall_2009_watson as
select course_id, room_number
from (select course.course_id, building, room_number
from course, section
where course.course_id = section.course_id
and course.dept_name = ’Physics’
and section.semester = ’Fall’
and section.year = ’2009’)
where building = ’Watson’;
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Views Defined Using Other Views
One view may be used in the expression defining another view,
A view relation v1 is said to depend directly on a view relation v2
if v2 is used in the expression defining v1
A view relation v1 is said to depend on view relation v2 if either
v1 depends directly to v2 or there is a path of dependencies
from v1 to v2
A view relation v is said to be recursive if it depends on itself.
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57
View Expansion
A way to define the meaning of views defined in terms of other views.
Let view v1 be defined by an expression e1 that may itself contain uses
of view relations.
View expansion of an expression repeats the following replacement
step:
repeat
Find any view relation vi in e1
Replace the view relation vi by the expression defining vi
until no more view relations are present in e1
As long as the view definitions are not recursive, this loop will
terminate.
Recursive views/queries are typically limited to the construct:
WITH RECURSIVE myquery (A, B, …) AS (
SELECT A, B, … FROM table WHERE …
UNION
SELECT A, B, … FROM myquery, table, …
) SELECT * FROM myquery
Non-recursive
part of query
Recursive
part
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Modification of the Database – Deletion
Relations:
instructor (id, name, dept_name, salary)
department (dept_name, building, budget)
Delete all instructors
delete from instructor ;
Delete all instructors from the Finance department
delete from instructor
where dept_name= ’Finance’;
Delete all tuples in the instructor relation for those instructors
associated with a department located in the Watson building.
delete from instructor
where dept name in (select dept name
from department
where building = ’Watson’);
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Example Query
Relations:
instructor (id, name, dept_name, salary)
Delete all instructors whose salary is less than the average salary of
instructors
Problem: as we delete tuples from instructor, the average salary
changes
Solution used in SQL:
First, compute avg salary and find all tuples to delete
Next, delete all tuples found above (without recomputing avg or
retesting the tuples)
delete from instructor
where salary < (select avg (salary) from instructor);
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Modification of the Database – Insertion
Relations:
course (course_id, title, dept_name, credits)
Add a new tuple to course
insert into course
values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4);
or equivalently (this is a recommended variant!)
insert into course (course_id, title, dept_name, credits)
values (’CS-437’, ’Database Systems’, ’Comp. Sci.’, 4);
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 61
Modification of the Database – Insertion
Relations:
student (id, name, dept_name, tot_credits)
Add a new tuple to student with tot_credits set to null
insert into student
values (’3003’, ’Green’, ’Finance’, null);
or equivalently
insert into student (id, name, dept_name)
values (’3003’, ’Green’, ’Finance’);
The value for the unspecified attribute is automatically set to null
or the default value assigned to the attribute is used instead.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 62
Modification of the Database – Insertion
Add all instructors to the student relation with tot_credits 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 this would cause problems
insert into table1 select * from table1
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 63
Modification of the Database – Updates
Increase salaries of instructors whose salary is over $100,000 by 3%,
and all others receive a 5% raise
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)
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 64
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
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 65
Updates with Scalar Subqueries
Re-compute and update tot_credits value for all students
update student
set tot_credits = ( select sum(credits)
from takes natural join course
where student.ID= takes.ID and
takes.grade <> ’F’ and
takes.grade is not null );
Sets tot_credits to null for students who have not taken any course
So, instead of sum(credits), use:
case
when sum(credits) is not null then sum(credits)
else 0
end
Or, use the function COALESCE
… (select coalesce( sum(credits), 0 ) from …
Modification of the Database – Views
Modifications of views must be translated to modifications of the actual
relations in the database.
Consider the view faculty where instructors’ salary is hidden:
create view faculty as
select ID, name, dept_name
from instructor
Since we allow a view name to appear wherever a relation name is
allowed, the user may write:
insert into faculty
values (’3003’, ’Green’, ’Finance’);
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 66
Recall: instructor (id, name, dept_name, salary)
Modification of the Database – Views (cont.)
The previous insertion must be represented by an insertion into the
actual relation instructor from which the view faculty is constructed.
An insertion into instructor requires a value for salary. The insertion
can be dealt with by either
rejecting the insertion and returning an error message to the user;
or
inserting the tuple
(’3003’, ’Green’, ’Finance’, null)
into the instructor relation.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 67
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 68
Praktické cvičení
Cvičení SQL, třetí část
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 69
Data Definition Language
Allows the specification of not only a set of relations but also
information about each relation, including:
The schema for each relation.
The domain of values associated with each attribute.
Integrity constraints
The set of indices to be maintained for each relation.
Security and authorization information for each relation.
The physical storage structure of each relation on disk.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 70
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),
primary key (id) )
insert into instructor values (‘10211’, ’Smith’, ’Biology’, 66000);
insert into instructor values (‘10211’, null, ’Biology’, 66000);
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 71
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.
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.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 72
Domain Types in SQL (cont.)
date: Dates, containing a (4 digit) year, month and date
Example: date ‘2005-07-27’
time: Time of day, in hours, minutes and seconds.
Example: time ‘09:00:30’ time ‘09:00:30.75’
timestamp: date plus time of day
Example: timestamp ‘2005-07-27 09:00:30.75’
interval: period of time
Example: interval ‘1’ day
Subtracting a date/time/timestamp value from another gives an
interval value
Interval values can be added to date/time/timestamp values
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 73
Integrity Constraints
Integrity constraints guard against accidental damage to the database,
by ensuring that authorized changes to the database do not result in a
loss of data consistency.
A checking account must have a balance greater than $10,000.00.
A salary of a bank employee must be at least $4.00 an hour.
A customer must have a (non-null) phone number.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 74
Not Null and Unique Constraints
not null
Declare name and budget to be not null
name varchar(20) not null
budget numeric(12,2) not null
primary key ( A1, A2, …, Am)
Attributes A1, A2, … Am forms the relation’s primary key.
Equals to unique and not null.
unique ( A1, A2, …, Am)
The unique specification states that the values in attributes A1, A2,
… Am cannot repeat within the relation.
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 75
The Check Constraint
check (P)
where P is a predicate
Example: Ensure that semester is one of fall or spring:
create table section (
course_id varchar (8),
sec_id varchar (8),
semester varchar (6),
year numeric (4,0),
building varchar (15),
room_number varchar (7),
time slot id varchar (4),
primary key (course_id, sec_id, semester, year),
check (semester in (’Fall’, ’Spring’))
);
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 76
Referential Integrity
Ensures that a value that appears in one relation for a given set of
attributes also appears for a certain set of attributes in another
relation.
Example: If “Biology” is a department name appearing in one of
the tuples in the instructor relation, then there exists a tuple in the
department relation for “Biology”.
Let A be a set of attributes. Let R and S be two relations that contain
attributes A and where A is the primary key of S.
E.g.: S(A,…) R(X, …, A, …)
A is said to be a foreign key of R if for any value of A appearing in R
it also appears in S.
Π 𝐴 𝑅 ⊆ Π 𝐴 𝑆
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 77
Referential Integrity in Create Table
foreign key (Am, ..., An ) references r
Example: Declare dept_name as the foreign key referencing
department relation
create table instructor (
ID char(5),
name varchar(20) not null,
dept_name varchar(20),
salary numeric(8,2),
primary key (ID),
foreign key (dept_name) references department
);
Notice: Schema of department is (dept_name, building).
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 78
Cascading Actions in Referential Integrity
create table course (
course_id char(5) primary key,
title varchar(20),
dept_name varchar(20) references department
)
create table course (
…
dept_name varchar(20),
foreign key (dept_name) references department
on delete cascade
on update cascade,
. . .
)
Alternative actions to cascade: set null, set default
E.g. … ON DELETE CASCADE SET NULL
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 79
Complex Check Clauses
Assume table section(course_id, sec_id, semester, year,
time_slot_id, building, room_number)
We define this constraint:
check (time_slot_id in (select time_slot_id from time_slot))
Why not use a foreign key here?
If time_slot_id is not the primary key in time_slot
Every section has at least one instructor teaching the section.
How to write this?
By a subquery…
Unfortunately: subquery in check clause not supported by pretty
much any database
Alternative: triggers
BKM_DATS, Vlastislav Dohnal, FI MUNI, 2022 80
Drop and Alter Table Constructs
drop table r
DROP TABLE instructor;
alter table r …
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 tuples in the relation are assigned null as the value for the
new attribute.
ALTER TABLE instructor ADD rating CHAR(1);
alter table r drop A
where A is the name of an attribute of relation r
Dropping of attributes not supported by many databases.
ALTER TABLE instructor DROP rating;