L10_PPT_IVSem

8/18/2019 L10_PPT_IVSem

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Database Management System

Lecture 10Lecture 10DDL andDDL and TupleTuple calculuscalculus

8/18/2019 L10_PPT_IVSem


Data Definition LanguageData Definition Language

• The schema for each relation, including

Allows the specification of:

attribute types.• Integrity constraints

• u or za on n orma on or eac re a on.

• Non-standard SQL extensions also allows ecification of

– The set of indices to be maintained foreach relations.

– The physical storage structure of eachrelation on disk.

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Create Table ConstructCreate Table Construct

•table command:

create table r (A 1 D 1, A 2 D 2, ..., A n D n ,- 1 ,

...,(integrity-constraintk))

– r s e name o e re a on

– each A i is an attribute name in the schemaof relation r

– D i is the data type of attribute A i Example:

crea e a e ranc (branch_name char(15),branch_city char(30),

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asse s n eger

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Domain Types in SQL Domain Types in SQL

• -. ,specified length n.

• varchar(n). Variable length character strings, with- .

• int. Integer (a finite subset of the integers that ismachine-dependent).

. -of the integer domain type).

• numeric(p,d). Fixed point number, with user-specified,

point.

• real, double precision. Floating point and double-

, -dependent precision.

• float(n). Floating point number, with user-specified

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.

• More are covered in Chapter 4.

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Integrity Constraints on TablesIntegrity Constraints on Tables

• not nu

• primary key (A 1, ..., A n )

xamp e: ec are ranc _name as e pr mary ey or ranc

.create table branch

branch name char 15 , _

branch_city char(30) not null,

assets integer,

primary key (branch_name))

primary key declaration on an attribute automatically ensures

not null in SQL-92 onwards, needs to be explicitly stated in

-

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Basic Insertion and Deletion of TuplesBasic Insertion and Deletion of Tuples

• Newly created table is empty

• Add a new tuple to account

insert into account values ('A-9732', 'Perryridge', 1200)

– Insertion fails if any integrity constraint is

violated

• Delete all tuples from account

delete from account

Note: Will see later how to delete selected tuples

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Drop and Alter Table ConstructsDrop and Alter Table Constructs

• he drop table command deletes all informationabout the dropped relation from the database.

• The alter table command is used to add

alter table r add A D where A is the name of the attribute to be addedto relation r and D is the domain of A.

– All tuples in the relation are assigned null asthe value for the new attribute.

• The alter table command can also be used to

drop attributes of a relation:

a ter ta e r rop Awhere A is the name of an attribute of relation r

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–databases

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Basic Query StructureBasic Query Structure

• A t ical S L uer has the form:

select A 1, A 2, ..., A n

, , ..., m

where P

– A i represents an attr ute

– R i

represents a relation

– .

• This query is equivalent to the relational algebraexpression. ))((

21,,, 21 mP A A A r r r

n×××∏ K

K

σ

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.

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The select ClauseThe select Clause

• he select clause list the attributes desired in the resultof a query

– corresponds to the projection operation of the

• Example: find the names of all branches in the loan relation:

se ect ranc _name

from loan

• In the relational algebra, the query would be:

branch_name (loan )

• NOTE: SQL names are case insensitive (i.e., you may use- - .

– E.g. Branch_Name ≡ BRANCH_NAME ≡ branch_name

– Some people use upper case wherever we use bold

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ont.

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The select Clause (Cont.)The select Clause (Cont.)

results.

• To force the elimination of duplicates, insert theeywor st nct a ter se ect.

• Find the names of all branches in the loan relations,and remove duplicates

select distinct branch_name

from loan

• The keyword all specifies that duplicates not beremoved.

select all branch_name

from loan

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The select Clause (Cont.)The select Clause (Cont.)

• An asterisk in the select clause denotes “allattributes”

select *from loan

expressions involving the operation, +, –, ∗, and

/, and operating on constants or attributes oftuples.

• E.g.:

_ , _ ,amount ∗ 100from loan

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The where ClauseThe where Clause

result must satisfy

– Corresponds to the selection predicate of the

relational algebra.• To find all loan number for loans made at the

erryr ge ranc w oan amoun s grea erthan $1200.

select loan_number from loan

where branch_name = ' Perryridge' and

• Comparison results can be combined using thelogical connectives and, or, and not.

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The from ClauseThe from Clause

query

– Corresponds to the Cartesian product operation of

the relational algebra.• Find the Cartesian product borrower X loan

select ∗

from borrower, loan

,having a loan at the Perryridge branch.

select customer_name, borrower.loan_number, amount

,

where borrower.loan_number = loan.loan_number and

branch_name = 'Perryridge'

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The Rename OperationThe Rename Operation

• SQL allows renaming relations and attributesusing the as clause:

old-name as new-name • E.g. Find the name, loan number and loan

amoun o a cus omers; rename e co umn

name loan_number as loan_id.

select customer_name, borrower.loan_number as

loan_id, amount

,

where borrower.loan_number = loan.loan_number

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Tuple VariablesTuple Variables

• uple variables are defined in the from clause via theuse of the as clause.

• Find the customer names and their loan numbers and.

select customer_name, T.loan_number, S.amount from borrower as T, loan as S

Find the names of all branches that have greater assets than

some branch located in Brooklyn.

w ere . oan_num er = . oan_num er

select distinct T.branch_name

from branch as T, branch as S

where T.assets > S.assets and S.branch_city = ' Brooklyn'

eywor as s op ona an may e om eborrower as T ≡ borrower T

Some database such as Oracle require as to be omitted

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Example InstancesExample Instances s i d b i d d a y

5 8 1 0 3 1 1 / 1 2 / 9 6

s id s n a m e ra tin g a g e

2 2 d u s tin 7 4 5 .0S1

• We will use these

instances of theSailors and Reserves

u e r .

5 8 ru s ty 1 0 3 5 .0

re at ons n ourexamples.

• If the key for the

s id s n a m e r a tin g a g e

2 8 y u p p y 9 3 5 .0

eserves re at oncontained only theattributes sid and

3 1 l u b b e r 8 5 5 .54 4 g u p p y 5 3 5 .0

5 8 r u s t 1 0 3 5 .0

,semantics differ?

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L10_PPT_IVSem

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