Database System Concepts - ULisboa · Database System Concepts Chapter 3: SQL Departamento de...

DatabaseSystem

Concepts

DataDefinitionLanguage

Querying theDatabase

Views

Modificationof theDatabase

JoinedRelations

Examples

Database System ConceptsChapter 3: SQL

Departamento de Engenharia InformaticaInstituto Superior Tecnico

1st Semester2007/2008

Slides (fortemente) baseados nos slides oficiais do livro“Database System Concepts”c©Silberschatz, Korth and Sudarshan.

DatabaseSystem

Concepts



Views


JoinedRelations

Examples

Outline

1 Data Definition Language

2 Querying the DatabaseBasic SQL QueriesAggregation and Set OperationsNull ValuesComplex Queries

3 Views

4 Modification of the Database

5 Joined Relations

6 Examples

DatabaseSystem

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Examples

Outline


2 Querying the Database

3 Views


5 Joined Relations

6 Examples

DatabaseSystem

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Views


JoinedRelations

Examples

Data Definition Language

Allows the specification of not only a set of relations but alsoinformation 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 relations.

Security and authorization information for each relation.

The physical storage structure of each relation on disk.

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Domain Types in SQL

char(n) - Fixed length character string, with user-specifiedlength n.varchar(n) - Variable length character strings, withuser-specified maximum length n.int - Integer (a finite subset of the integers that ismachine-dependent).smallint - Small integer (a machine-dependent subset ofthe integer domain type).numeric(p,d) - Fixed point number, with user-specifiedprecision of p digits, with n digits to the right of decimalpoint.real, double - Floating point and double-precision floatingpoint numbers, with machine-dependent precision.float(n) - Floating point number, with user-specifiedprecision of at least n digits.More are covered in the book...

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Creating Tables

An SQL relation is defined using the create table

command:

create table r ((A1 D1,A2 D2, . . . ,An Dn)constraint1, . . . , constraintk

)

r is the name of the relationeach 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 branch (branch name char(15),branch city char(30),assets integer )

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Integrity Constraints

Ensuring non-null values: not null

Defining the identifier: primary key (A1, . . . ,An)

Example:

create table branch (branch name char(15),branch city char(30),assets integer not null,primary key (branch name)

)

Note: primary key also ensures that the attribute is not null

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Deleting and Altering Tables

The drop table command deletes all information aboutthe dropped relation from the database.

drop table r

The alter table command is used to add/removeattributes to/from an existing relation:

alter table r add A D

alter table r drop A

where A is the name of the attribute to be added torelation r and D is the domain of A.

All tuples in the relation are assigned null as the value forthe new attribute.Dropping of attributes is not supported by many databases

DatabaseSystem

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Basic SQLQueries

Aggregation andSet Operations

Null Values

Complex Queries

Views


JoinedRelations

Examples

Outline


2 Querying the DatabaseBasic SQL QueriesAggregation and Set OperationsNull ValuesComplex Queries

3 Views


5 Joined Relations

6 Examples

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Basic SQLQueries


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Basic Query Structure

SQL is based on set and relational operations with certainmodifications and enhancements

A typical SQL query has the form:

select A1,A2, . . . ,An

from r1, r2, . . . , rmwhere P

Ai represents an attributeri represents a relationP is a predicate.

This query is equivalent to the relational algebra expression

πA1,A2,...,An

(

σP(r1 × r2 × · · · × rm))

The result of an SQL query is a relation

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The select Clause

The select clause list the attributes desired in the result ofa query

Corresponds to the projection operation of the relationalalgebra

Example: find the names of all branches in the loanrelation:

select branch name

from loan

In the relational algebra, the query would be:

πbranch name(loan)

Note : SQL names are case insensitive (i.e., you may use upper- orlower-case letters.)

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The select Clause (cont.)

SQL allows duplicates in relations as well as in queryresults

To force the elimination of duplicates, insert the keyworddistinct after select

Example: find the names of all branches in the loanrelations, and remove duplicates

select distinct branch name

from loan

The keyword all specifies that duplicates not be removed

select all branch name

from loan

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The select Clause (cont.)

An asterisk in the select clause denotes “all attributes”

select ∗from loan

The select clause can contain arithmetic expressionsinvolving the operation, +, −, ∗, and /, and operating onconstants or attributes of tuples

select loan number , branch name, amount ∗ 100from loan

would return a relation that is the same as the loan relation,except that the value of the attribute amount is multiplied by100.

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The where Clause

The where clause specifies conditions that the result mustsatisfy

Corresponds to the selection predicate of the relationalalgebra

Example: to find all loan number for loans made at thePerryridge branch with loan amounts greater than $1200

select loan number

from loan

where branch name = ’Perryridge’ and amount > 1200

Comparison results can be combined using the logicalconnectives and, or, and not

Comparisons can be applied to results of arithmeticexpressions

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The where Clause (cont.)

SQL includes a between comparison operator

Example: Find the loan number of those loans with loanamounts between $90 000 and $100 000 (that is,≥ $90 000 and ≤ $100 000)

select loan number

from loan

where amount between 90000 and 100000

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The from Clause

The from clause lists the relations involved in the query

Corresponds to the Cartesian product operation of therelational algebra

Example: find the Cartesian product borrower × loan

select ∗from borrower , loan

Example: find the name, loan number and loan amount ofall customers having a loan at the Perryridge branch

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The from Clause






loan(loan number , branch name, amount)borrower(customer name, loan number)

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The from Clause






select customer name, loan.loan number , amount

from loan, borrower

where loan.loan number = borrower .loan number

and branch name = ’Perryridge’

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Renaming

SQL allows renaming relations and attributes using the as

clause:

old-name as new-name

Example: find the name, loan number and loan amount ofall customers; rename the column loan number as loan id

select customer name, borrower .loan number as loan id ,amount

from borrower , loan

where borrower .loan number = loan.loan number

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Tuple Variables

Tuple variables are defined in the from clause via the useof the as clause.

Example: find the customer names and their loan numbersfor all customers having a loan

select customer name, B.loan number , L.amount

from borrower as B, loan as L

where B.loan number = L.loan number

Example: find the names of all branches that have greaterassets than some branch located in Brooklyn

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Tuple Variables







branch(branch name, branch city , assets)

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Tuple Variables







select distinct T .branch name

from branch as T , branch as S

where T .assets > S .assets and S .branch city = ’Brooklyn’

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String Operations

The operator like uses patterns that are described usingtwo special characters:

percent (%) - matches any substring.underscore ( ) - matches any character.

Example: find the names of all customers whose streetincludes the substring “Main”

select customer name

from customer

where customer street like ’%Main%’

To match the name “Main%”

like ’Main\%’ escape ’\’

SQL supports many other string operations

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Ordering the Display of Tuples

List in alphabetic order the names of all customers havinga loan in Perryridge branch

select distinct customer name

from borrower B, loan L


and branch name = ’Perryridge’order by customer name

We may specify desc for descending order or asc forascending order (the default)

order by customer name desc

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Set Operations

Find all customers who have a loan, an account, or both

(select customer name from depositor)union(select customer name from borrower)

Find all customers who have both a loan and an account

(select customer name from depositor)intersect(select customer name from borrower)

Find all customers who have an account but no loan

(select customer name from depositor)except(select customer name from borrower)

Each of the above operations automatically eliminates duplicates; to retain

all duplicates use the corresponding multiset versions: union all,

intersect all and except all

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Aggregate Functions

Operate on the multiset of values of a column of a relation, andreturn a value

avg, min, max, sum, count

Find the average account balance at the Perryridge branch

select avg(balance)from account

where branch name = ’Perryridge’

Find the number of tuples in the customer relation

select count(∗) from customer

Find the number of depositors in the bank

select count(distinct customer name)from depositor

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Aggregate Functions - group by

Find the number of depositors for each branch

select branch name, count(distinct customer name)from depositor , account

where depositor .account number = account.account number

group by branch name

Note: attributes in select clause outside of aggregate functionsmust appear in group by list

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Aggregate Functions - having

Find the names of all branches where the average accountbalance is more than $1 200

select branch name, avg(balance)from account


having avg(balance) > 1200

Note: predicates in the having clause are applied after the formation

of groups whereas predicates in the where clause are applied before

forming groups

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Null Values

It is possible for tuples to have the null value

The result of any arithmetic expression involving null isnull

Any comparison with null returns unknown

The predicates is null and is unknown can be used tocheck for null values and unknown results

select loan number

from loan

where amount is null

select loan number

from loan

where amount > 1000 is unknown

Note: result of where clause predicate is treated as false if itevaluates to unknown

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Null Values and Aggregates

Total all loan amounts

select sum(amount)from loan

Above statement ignores null amounts

Result is null if there is no non-null amount

All aggregate operations except count(∗) ignore tupleswith null values on the aggregated attributes.

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Nested Subqueries

A common use of nested subqueries is to perform tests forset membership, set comparisons, and set cardinality

Find all customers who have both an account and a loanat the bank


from borrower

where customer name in (select customer name

from depositor)

Find all customers who have a loan at the bank but do nothave an account at the bank


from borrower

where customer name not in (select customer name

from depositor)

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Basic SQLQueries


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Set Comparison

Find all branches that have greater assets than some branchlocated in Brooklyn

select branch name

from branch

where assets >some (select assets

from branch

where branch city = ’Brooklyn’)

>some, <some, =some, <>some

Find the names of all branches that have greater assets than allbranches located in Brooklyn

select branch name

from branch

where assets >all (select assets

from branch


>all, <all, =all, <>all

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Test for Empty Relations

The exists construct returns the value true if theargument subquery is nonempty

exists r ⇔ r 6= ∅

Example: find all customers who do not have an accountat the Perryridge branch

select D.customer name

from depositor D

where not exists (select T .customer name

from account A, depositor T

where D.customer name = T .customer name andT .account number = A.account number andA.branch name = ’Perryridge’)

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Division

Find all customers who have an account at all brancheslocated in Brooklyn

select distinct S .customer name

from depositor S

where not exists ((select branch name

from branch

where branch city = ’Brooklyn’)except(select R.branch name

from depositor T , account R

where T .account number = R.account number andS .customer name = T .customer name ))

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Basic SQLQueries


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Division

Find all customers who have an account at all brancheslocated in Brooklyn

select distinct S .customer name

from depositor S

where not exists ((select branch name

from branch

where branch city = ’Brooklyn’)except(select R.branch name

from depositor T , account R

where T .account number = R.account number andS .customer name = T .customer name ))

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Test for Absence of Duplicate Tuples

The unique construct tests whether a subquery has anyduplicate tuples in its result

Example: find all customers who have at most oneaccount at the Perryridge branch

select D.customer name

from depositor D

where unique (select T .customer name

from account A, depositor T

where D.customer name = T .customer name andT .account number = A.account number andA.branch name = ’Perryridge’)

What about all customers who have at least two accountsat the Perryridge branch?

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Derived Relations

SQL allows a subquery expression to be used in the fromclause

Find the average account balance of those branches wherethe average account balance is greater than $1200

select branch name, avg balance

from (select branch name, avg(balance)from account

group by branch name)as branch avg(branch name, avg balance)

where avg balance > 1200

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The with Clause

The with clause provides a way of defining a temporaryview whose definition is available only to the query inwhich the with clause occurs

Example: find all accounts with the maximum balance

with max balance(value) asselect max(balance)from account

select account number

from account, max balance

where account.balance = max balance.value

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Outline



3 Views


5 Joined Relations

6 Examples

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Views

In some cases, it is not desirable for all users to see theentire logical model

Consider a person who needs to know a customer’s loannumber but has no need to see the loan amount

select customer name, loan number

from borrower , loan

where borrower .loan number = loan.loan number

A view provides a mechanism to hide certain data fromthe view of certain users.

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View Definition

A view is defined using the create view statement whichhas 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 referto the virtual relation that the view generates.

View definition is not the same as creating a new relationby evaluating the query expressionRather, a view definition causes the saving of anexpression; the expression is substituted into queries usingthe view.

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View Examples

A view consisting of branches and their customers

create view all customer as(select branch name, customer name

from depositor D, account A

where D.account number = A.account number)union

(select branch name, customer name

from borrower B, loan L

where B.loan number = L.loan number)

Find all customers of the Perryridge branch


from all customer


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Outline



3 Views


5 Joined Relations

6 Examples

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Deleting Tuples

Delete all account tuples at the Perryridge branch

delete from account


Delete all accounts at every branch located in the city’Needham’

delete from account

where branch name in (select branch name

from branch

where branch city = ’Needham’)

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Insertion

Add a new tuple to account

insert into account

values (’A-9732’, ’Perryridge’, 1200)

or equivalently

insert into account(branch name, balance, account number)values (’Perryridge’, 1200, ’A-9732’)

Add a new tuple to account with balance set to null

insert into account

values (’A-777’, ’Perryridge’, null)

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Insertion (cont.)

Provide as a gift for all loan customers of the Perryridgebranch, a $200 savings account. Let the loan numberserve as the account number for the new savings account

insert into account

select loan number , branch name, 200from loan


insert into depositor

select customer name, loan number

from loan, borrower

where branch name = ’Perryridge’ andloan.account number = borrower .account number

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Updates

Increase all accounts with balances over $10,000 by 6%,all other accounts receive 5%

update account

set balance = balance ∗ 1.06where balance > 10000

update accountset balance = balance ∗ 1.05where balance <= 10000

update accountset balance =

casewhen balance <= 10000 then balance ∗ 1.05else balance ∗ 1.06

end

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Updates

Increase all accounts with balances over $10,000 by 6%,all other accounts receive 5%

update account

set balance = balance ∗ 1.06where balance > 10000

update accountset balance = balance ∗ 1.05where balance <= 10000

update accountset balance =

casewhen balance <= 10000 then balance ∗ 1.05else balance ∗ 1.06

end

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Update of a View

Create a view of all loan data in the loan relation, hidingthe amount attribute

create view branch loan asselect branch name, loan number

from loan

Add a new tuple to branch loan

insert into branch loan

values (’Perryridge’, ’L-307’)

This insertion must be represented by the insertion of thetuple

(’L-307’, ’Perryridge’, null)

into the loan relation

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Update of a View (cont.)

Some updates through views are impossible to translateinto updates on the database relations

create view avg bal asselect branch name, avg(balance)from account


insert into avg bal

values (’Downtown’, 300)

Most SQL implementations allow updates only on simpleviews (without aggregates) defined on a single relation

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Outline



3 Views


5 Joined Relations

6 Examples

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Join Operations

Join types

inner join

left outer join

right outer join

full outer join

Join conditions

natural

on <predicate>

using (A1, A2, . . . ,An)

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Join Operations (cont.)

loan

loan number branch name amount

L-170 Downtown 3000L-230 Redwood 4000L-260 Perryridge 1700

borrower

customer name loan number

Jones L-170Smith L-230Hayes L-155

select ∗ fromloan inner join borrower

on loan.loan number = borrower .loan number

loan number branch name amount customer name loan number

L-170 Downtown 3000 Jones L-170L-230 Redwood 4000 Smith L-170

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loan



borrower



select ∗ fromloan natural join borrower

loan number branch name amount customer name

L-170 Downtown 3000 JonesL-230 Redwood 4000 Smith

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loan



borrower



select ∗ fromloan left outer join borrower

on loan.loan number = borrower .loan number

loan number branch name amount customer name loan number

L-170 Downtown 3000 Jones L-170L-230 Redwood 4000 Smith L-230L-260 Perryridge 1700 null null

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loan



borrower



select ∗ fromloan natural right outer join borrower


L-170 Downtown 3000 JonesL-230 Redwood 4000 SmithL-155 null null Hayes

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loan



borrower



select ∗ fromloan full outer join borrower using(loan number)


L-170 Downtown 3000 JonesL-230 Redwood 4000 SmithL-260 Perryridge 1700 nullL-155 null null Hayes

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Outline



3 Views


5 Joined Relations

6 Examples

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Example Queries

Get data from all customers

account(account number , branch name, balance)branch(branch name, branch city , assets)

customer(customer name, customer street, customer city)loan(loan number , branch name, amount)

depositor(customer name, account number)borrower(customer name, loan number)

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JoinedRelations

Examples

Example Queries

Get data from all customers

select ∗ from customer

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JoinedRelations

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Example Queries

Get the name and city of all customers with a loan




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JoinedRelations

Examples

Example Queries

Get the name and city of all customers with a loan

select distinct c .customer name, customer city

from borrower b, customer c

where b.customer name = c .customer name

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JoinedRelations

Examples

Example Queries

Get the name and city of all customers with a loan at thePerryridge branch




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Examples

Example Queries

Get the name and city of all customers with a loan at thePerryridge branch

select distinct c .customer name, customer city

from borrower b, customer c, loan l

where b.customer name = c .customer name

and b.loan number = l .loan number

and branch name = ’Perryridge’

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the number of all accounts with a balance between $700and $900




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the number of all accounts with a balance between $700and $900

select account number

from account

where balance between 700 and 900

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all customers whose street name ends in ’Hill’




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all customers whose street name ends in ’Hill’


from customer

where customer street like ’%Hill’

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all customers with both an account and a loanat the Perryridge branch




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all customers with both an account and a loanat the Perryridge branch

select distinct b.customer name

from borrower b, loan l , depositor d , account a

where b.loan number = l .loan number

and b.customer name = d .customer name

and d .account number = a.account number

and a.branch name = ’Perryridge’and l .branch name = ’Perryridge’

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all customers with an account at any branchwhere customer Hayes has an account




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Example Queries

Get the name of all customers with an account at any branchwhere customer Hayes has an account

select distinct d .customer name

from depositor d , account a

where d .account number = a.account number

and branch name in(select branch name

from depositor d , account a


and d .customer name = ’Hayes’)

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all agencies with higher assets than allagencies in Brooklyn




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JoinedRelations

Examples

Example Queries

Get the name of all agencies with higher assets than allagencies in Brooklyn

select branch name

from branch

where assets > all

(select assets

from branch


DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all clients with a loan at the Perryridge branch




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of all clients with a loan at the Perryridge branch


from borrower b, loan l

where b.loan number = l .loan number

and l .branch name = ’Perryridge’order by b.customer name

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name and number of customers of all agencies




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name and number of customers of all agencies

select branch name, count(distinct customer name)from depositord , accounta



DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name and average balance of all agencies with anaverage balance above $700




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JoinedRelations

Examples

Example Queries

Get the name and average balance of all agencies with anaverage balance above $700

select branch name, avg(balance)from account


having avg(balance) > 700

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of the agencies with the highest average balance




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the name of the agencies with the highest average balance

select branch name

from account


having avg(balance) >= all(select avg(balance)from account

group by branch name)

DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get the total number of customers in the database




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JoinedRelations

Examples

Example Queries

Get the total number of customers in the database

select count(∗) from customer

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JoinedRelations

Examples

Example Queries

Get average balance of all customers who live in Harrison andhave, at least, two bank accounts




DatabaseSystem

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JoinedRelations

Examples

Example Queries

Get average balance of all customers who live in Harrison andhave, at least, two bank accounts

select avg(balance)from depositor d , account a, customer c

where d .customer name = c .customer name

and d .account number = a.account number

and customer city = ’Harrison’group by d .customer name

having count(distinct a.account number) >= 2

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Concepts



Views


JoinedRelations

Examples

End of Chapter 3

Date post:	27-Apr-2018
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Database System Concepts - ULisboa · Database System Concepts Chapter 3: SQL Departamento de...

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