Lecture 6 Structured Query Language SQL Lecture 6 Structured Query Language SQL Instructor: Haya Sammaneh.

Lecture 6Structured Query Language

SQL

Lecture 6Structured Query Language

SQL

Instructor: Haya Sammaneh

Tables

• create table account (account_number varchar(15) not null, branch_name varchar(15) not null, balance number not null, primary key(account_number));

• create table branch (branch_name varchar(15) not null , branch_city varchar(15) not null, assets number not null, primary key(branch_name));

• account(account_number, branch_name, balance);• branch(branch_name, branch_city, assets);

Tables

• create table customer (customer_name varchar(15) not null , customer_street varchar(12) not null, customer_city varchar(15) not null, primary key(customer_name));

• create table loan (loan_number varchar(15) not null , branch_name varchar(15) not null, amount number not null, primary key(loan_number));

• customer(customer_name, customer_street, customer_city );• Loan(loan_number, branch_name, amount);

Tables• create table depositor (customer_name varchar(15) not null,

account_number varchar(15) not null, primary key(customer_name, account_number), foreign key(account_number) references account(account_number),

foreign key(customer_name) references customer(customer_name));

• create table borrower (customer_name varchar(15) not null, loan_number varchar(15) not null,

primary key(customer_name, loan_number), foreign key(customer_name) references customer(customer_name),

foreign key(loan_number) references loan(loan_number));

• depositor(customer_name, account_number);• borrower(customer_name, loan_number);

Basic StructureBasic Structure

• A typical SQL query has the form:select A1, A2, …, An

from R1, R2, …, Rm

where condition- Ai represent attributes- Ri represent relations

This query is equivalent to the relational algebra expression:

A1, A2, …, An(P(R1 R2 … Rm))

• The result of an SQL query is a relation.

ProjectionProjection

• The select corresponds to the projection operation of the relational algebra. It is used to list the attributes desired in the result of a query.

• Find the names of all branches in the loan relationselect branch-namefrom loan

Equivalent to: branch-name(loan)

• An asterisk in the select clause denotes “all attributes”select *from loan

Duplicate RemovalDuplicate Removal

• SQL allows duplicates in relations as well as in query results.

• Use select distinct to force the elimination of duplicates.Find the names of all branches in the loan relation, and remove duplicates

select distinct branch-namefrom loan

• The keyword all specifies that duplicates not be removed.

select all branch-namefrom loan

force the DBMS to remove duplicates

force the DBMS not to remove duplicates

Arithmetic Operations on Retrieved Results

Arithmetic Operations on Retrieved Results

• The select clause can contain arithmetic expressions involving the operators,,, and , and operating on constants or attributes of tuples.

• The query:select branch-name, loan-number, amount * 100 from loan

would return a relation which is the same as the loan relations, except that the attribute amount is multiplied by 100

The where ClauseThe where Clause

• The where clause specifies conditions that tuples in the relations in the from clause must satisfy.

• Find all loan numbers for loans made at the Nablus branch with loan amounts greater than $1200.

select loan-numberfrom loanwhere branch-name=“Nablus” and amount >1200

• SQL allows logical connectives and, or, and not. Arithmetic expressions can be used in the comparison operators.

The where Clause (Cont.)The where Clause (Cont.)

• SQL includes the between operator • Find the loan number of those loans with loan

amounts between $90,000 and $100,000 (that is, $90,000 and $100,000)

select loan-numberfrom loanwhere amount between 90000and

100000

The from ClauseThe from Clause• The from clause corresponds to the Cartesian product

operation of the relational algebra. • Find the Cartesian product borrower loan

select *from borrower, loan

It is rarely used without a where clause.

• Find the name and loan number of all customers having a loan at the Nablus branch.select distinct customer-name, borrower.loan-number from borrower, loanwhere borrower.loan-number = loan.loan-number and

branch-name = “Nablus”

Table

loanbranch-nameloan-number amount

Nablus

Nablus

Perryridge

L-170

L-260

L-230

3000

1700

4000

borrowercust-name loan-number

Jones

Hayes

Smith

L-170

L-230

L-155

branch-nameloan-number amount

Nablus

Nablus

L-170

L-230

3000

4000

cust-name

Jones

Smith

Loan Borrower

The Rename OperationThe Rename Operation

• Renaming relations and attributes using the as clause:old-name as new-name

• Find the name and loan number of all customers having a loan at the Nablus branch; replace the column name loan-number with the name loan-id.

select distinct customer-name, borrower.loan-number as loan-idfrom borrower, loanwhere borrower.loan-number = loan.loan-number and

branch-name = “Nablus”

Tuple Variables/Alias Tuple Variables/Alias

• Tuple variables are defined in the from clause via the use of the “as” clause.

• Find the customer names and their loan numbers for all customers having a loan at some branch.

select distinct customer-name, T.loan-numberfrom borrower as T, loan as Swhere T.loan-number = S.loan-number

String OperationsString Operations

• Character attributes can be compared to a pattern: % matches any substring.

• Find the name of all customers whose street includes the substring ‘Main’. (Eg Mainroad, Smallmain Road, AMainroad,…)

select customer-namefrom customerwhere customer-street like “%Main%”

• How to match the name “Main%”: (Eg abcMain%, MainMain%,…)

Ordering the Display of TuplesOrdering the Display of Tuples

• List in alphabetic order the names of all customers having a loan at Nablus branch

select distinct customer-namefrom borrower, loanwhere borrower.loan-number = loan.loan-number and

branch-name = “Nablus”order by customer-name

• order by customer-name desc, amount ascdesc for descending order; asc for ascending order (default)

Set operationsSet 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)

SQLAggregate Functions

Aggregate FunctionsAggregate Functions

• Operates on a column of a relation, and return a value

avg: average valuemin: minimum valuemax: maximum valuesum: sum of valuescount: number of values

Aggregate Functions(cont.)Aggregate Functions(cont.)

• Find the average account balance at the Nablus branch.

select avg(balance)from accountwhere branch-name=“Nablus”

account select balancefrom accountwhere branch-name =“Nablus”

Avg()120,000

balance

Aggregate Functions(cont.)Aggregate Functions(cont.)

• Find the numbers of tuples in the customer relation.select count(*)from customer

– remember * stands for all attributes– compare to:

select count(customer-city)from customer

• Find the number of depositors in the bankselect count (distinct customer-name)from depositor

Aggregate functions - Group by Aggregate functions - Group by

• Find the number of accounts for each branch. select branch-name, count( distinct account-number)from accountgroup by branch-name

• For each group of tuples with the same branch-name, applyaggregate function count and distinct to account-number

branch-name count-account-no

Perryridge Brighton Redwood

2 2 1

branch-name account-number balance

PerryridgeBrightonPerryridgeBrightonRedwood

a-102a- 217a-201a-215a-222

400750900750700

branch-name account-number balance

Perryridge Perryridge Brighton Brighton Redwood

a-102 a-201 a-217 a-215 a-222

400 900 750 750 700

account table

Null valuesNull 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.• More precisely,

– Any comparison with null returns unknown (caution: Oracle treats it as false!)

– (true or unknown) = true, (false or unknown) = unknown (unknown or unknown) = unknown,(true and unknown) = unknown, (false and unknown) = false (unknown and unknown) = unknown

– Result of where clause predicate is treated as false if it evaluates to unknown

Null Values (cont.)Null Values (cont.)• Find all loan numbers which appear in the loan relation with null

values for amount.

select loan-numberfrom loanwhere amount is null

• Total of all loan amountsselect sum(amount)from loan

Above statement ignores null amounts;

Check for each borrowerif he is also a depositor

Example Nested QueryExample Nested Query

• Find all customers who have both an account and a loan in the bank.

select distinct customer-namefrom borrowerwhere customer-name in (select customer-name

from depositor)

Example QueryExample Query

• Find all customers who have a loan at the bank but do not have an account at the bank.

select distinct customer-namefrom borrowerwhere customer-name not in (select customer-name

from depositor)

ViewsViews

• Provide a mechanism to hide certain data from the view of certain users. To create a view we use the command:

create view view-name as <query expression>where:– <query expression> is any legal SQL query– the name of the view is represented by view-

name

Example QueriesExample Queries

• A view consisting of branches and their customerscreate view all-customer as

(select branch-name, customer-namefrom depositor, accountwhere depositor.account-number = account.name-number)

union(select branch-name, customer-namefrom borrower, loanwhere borrower.loan-number = loan.loan-number)

• Find all customers of the Nablus branchselect customer-namefrom all-customerwhere branch-name = “Nablus”