1. Introduction to SQL
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2. 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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3. A Bank Database name Acc# balance street depositor Customer Account
city
borrower
has
makes
name
Loan
Branch
Loan#
city
amount
assets
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4. 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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5. The Customer Table Customer-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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6. The Account Table Branch-name Account# Balance Oakland 101 2000
SJ-Main
205
7500
207
4500
Santa Clara
311
3100
SJ-West
251
850
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7. The Loan Table Branch-name loan# Amount Oakland 2301 5000 SJ-Main 5155
700
5709
9000
Santa Clara
1541
1800
SJ-West
4321
250
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8. The Depositor Table Customer-name Account# Bob 207 Carol 311 Ted 205
Alice
101
251
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9. 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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10. 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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11. 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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12. 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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13. 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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14. Queries on the Loan Table
Branch-name
Oakland
SJ-Main
Santa Clara
SJ-West
Loan = (branch-name, loan#, amount)
Find the names of all the branches in the Loan relation
select branch-name
from Loan
Note that the "where" clause is optional.
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15. More Queries on the Loan Table
An asterisk in the select clause denotes "all attributes"
select *
from Loan
where amount > 3000
Result contains only those rows that meet the criteria
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16. The where condition
Find the loan numbers for all loans made at the Oakland branch with loan amounts greater than 1200. 
select loan#
from Loan
where branch-name="Oakland" and amount>1200
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17. The from clause
Form the cross product of the tables listed in the FROM clause
Select A1, A2 from R1, R2
Result 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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18. arity(R) = k1 arity(R  S) = k1 + k2
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  S
A B C D D E F A B C D D' E F
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19. Cartesian Product Table A Table B Bill 4 John one Mary 5 Sue four Bill
two
A X B
Bill 4
John
one
Bill 4
Sue
four
Bill 4
Bill
two
Mary 5
John
one
Mary 5
Sue
four
Mary 5
Bill
two
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20. 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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21. Using the Cartesian Product
Find the name of customers with an account at the Oakland branch.
select customer-name
from Depositor, Account
where
Depositor.account# = Account.account#
and branch-name = "Oakland"
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22. Depositor join Account
Customer-name
Account#
Bob
207
Carol
311
Ted
205
Alice
101
251
Branch-name
Account#
Balance
Oakland
101
2000
SJ-Main
205
7500
207
4500
Santa Clara
311
3100
SJ-West
251
850
Customer-name
Account#
Branch-name
Balance
Alice
101
Oakland
2000
Ted
205
SJ-Main
7500
Bob
207
4500
Carol
311
Santa Clara
3100
251
SJ-West
850