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Introduction to SQL

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DB Tables and SQL

• The data is stored in the database in relations or tables

• Data types are constrained like in an excel spreadsheet

• Column headings are the entity attributes

• SQL is a language for extracting information from the tables

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A Bank Database

Customer Account

Loan Branch

street

name

Loan#

amount

name

city

city

assets

makes

borrower

depositor

has

Acc# balance

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Bank Database Schema

• Branch = (branch-name, branch-city, assets)• Customer = (customer-name, customer-street,

customer-city)• Account = (branch-name, account#, balance)• Depositor = (customer-name, account#)• Loan = (branch-name, loan#, amount)• Borrower = (customer-name, loan#)

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The Customer TableCustomer-name C-Street C-city

Bob 123 Third St San Jose

Carol 456 Main St Santa Clara

Ted 89 Blossom Ave Los Gatos

Alice 64 Longwalk Dr Oakland

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The Account Table

Branch-name Account# Balance

Oakland 101 2000

SJ-Main 205 7500

SJ-Main 207 4500

Santa Clara 311 3100

SJ-West 251 850

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The Loan Table

Branch-name loan# Amount

Oakland 2301 5000

SJ-Main 5155 700

SJ-Main 5709 9000

Santa Clara 1541 1800

SJ-West 4321 250

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The Depositor Table

Customer-name Account#

Bob 207

Carol 311

Ted 205

Alice 101

Bob 251

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Creating a Table• An SQL relation is defined with the create table

command:

create table <name> ( <column-desc-list>, <constraint-list> )

 

• Column-description-list is a comma separated list of column names and their domains.

• Constraint-list is a comma separated list of integrity constraints.

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Domains

char(n) Fixed length character string.

varchar(n) Variable length character string, max length n.

integer Integer (actual range is DB dependent)

date Dates with 4 digit year, 2 digit month an day.

time Day, hours, minutes, seconds

 • Null values are allowed in all domain types.

Declaring an attribute to be not null excludes the null value from the domain.

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The Customer Table

Example:

create table Customer(customer-name char(30), c-street char(30), c-city char(30),

primary key (customer-name) ) 

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Branch Table with Constraints

Example:

create table Branch (branch-name char(15), branch-city char(30), assets integer,

primary key (branch-name), check (assets >= 0) )

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The select statement

select A1, A2, … An

from r1, r2, … , rk where P

• Ai's are attribute names, r's are tables and P is a predicate (condition).

•  Cartesian product of the tables in the from clause is formed. Then, the criteria in the where clause is used to select rows from the product. Last, the columns and expressions in the select clause are projected to form the result.

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Queries on the Loan Table

• Loan = (branch-name, loan#, amount)• Find the names of all the branches in

the Loan relation

select branch-namefrom Loan

• Note that the "where" clause is optional.

Branch-name

Oakland

SJ-Main

SJ-Main

Santa Clara

SJ-West

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More Queries on the Loan Table

• An asterisk in the select clause denotes "all attributes"

select *from Loanwhere amount > 3000

 • Result contains only those rows that meet

the criteria

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The where condition

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

select loan#from Loanwhere branch-name="Oakland"

and amount>1200

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The from clause

• Form the cross product of the tables listed in the FROM clause

Select A1, A2 from R1, R2Result R = R1 R2

• pairs each tuple t1 of R1 with each tuple t2 of R2 and puts in R a tuple t1t2.

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Cartesian Product ()

arity(R) = k1 arity(R S) = k1 + k2

arity(S) = k2 card(R S) = card(R) card(S)

R S is the set all possible (k1 + k2)-tuples

whose first k1 attributes are a tuple in R

last k2 attributes are a tuple in S

R S R SA B C D D E F A B C D D' E F

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Cartesian ProductTable A

BillMary

45

two

Table B

A X B

JohnSue

onefour

Bill

Mary

4

5

Bill

Bill

Mary

4

5

Bill

Mary

4

5

John one

John one

Sue four

Sue fourtwoBill

twoBill

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Using the Cartesian Product

• Find the name of customers with an account at the Oakland branch.

• Account = (branch-name, account#, balance)• Depositor = (customer-name, account#)

• Cartesian product of Account and Depositor will match each row of Account with each row of Depositor – we need a way to retain only those rows with matching account#

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Using the Cartesian Product

• Find the name of customers with an account at the Oakland branch.

select customer-namefrom Depositor, Accountwhere Depositor.account# = Account.account# and branch-name = "Oakland"

Depositor join AccountCustomer-name Account#

Bob 207

Carol 311

Ted 205

Alice 101

Bob 251

Branch-name Account#

Balance

Oakland 101 2000

SJ-Main 205 7500

SJ-Main 207 4500

Santa Clara 311 3100

SJ-West 251 850

Customer-name Account# Branch-name Account#

Balance

Alice 101 Oakland 101 2000

Ted 205 SJ-Main 205 7500

Bob 207 SJ-Main 207 4500

Carol 311 Santa Clara 311 3100

Bob 251 SJ-West 251 850