CIS 671Query Languages1 CIS 671 Introduction to Database Systems II Introduction S. Parthasarathy.

CIS 671 Query Languages 1

CIS 671Introduction to Database Systems II

Introduction

S. Parthasarathy


CIS 671 Course Outline• Review of basic database material (CIS

670)• Object-Oriented and Object Relational

Database Systems• Data Warehousing and Data Mining• Active Databases• Introduction to Storage Structures• Emerging Technologies


RelationsRelation schema R or R(A1, A2, …, An)

•fixed set of attributes A1, A2, …, An

•each attribute Aj corresponds to exactly one of the underlying domains Dj (j = 1, 2, …, n), not necessarily distinct.

Example: EMP( EMPNO, NAME, DNO, JOB, MGR, SAL, COMMISSION)note: EMPNO and MGR take values from the same domain, the set of valid employee numbers

N, the number of attributes in the relation scheme is called the degree of the relation.Degree 1 unary degree 3 ternarydegree 2 binary degree n n-ary


Keys• A relation is a set of tuples.

• All elements of a set are distinct.

• Hence all tuples must be distinct.

• There may also be a subset of the attributes with the property that values must be distinct. Such a set is called a superkey.

• SK a set of attributes

• t1 and t2 tuples

• t1 [SK] t2 [SK]

Guaranteed by the “real world”.


Candidate Key -a minimal superkey

• A set of attributes, CK is a superkey,

• but no proper subset is a superkey.Example

EMPNO

NAME ???

Primary Key - one arbitrarily chosen candidate key.

ExampleEMPNO


Theoretical Properties of Relations

• Tuples are unordered.

• There are no tuple duplicates.

• Attributes are unordered

• All attribute values are atomic.

• Implementations vary in terms of how closely they follow the above properties.


Relational Database Schema S

• Set of Relation Schemas S = {R1, R2, …, Rp}

• Set of integrity constraints, IC– Integrity constraints will be discussed next.


Relational Database (Instance) DB of Schema S

• Set of relation instances, DB

• DB = {r1, r2, …, rp}

• Each rk is an instance of relation RK

• The rk’s satisfy the integrity constraints, IC


Database Example - SchemaEMP( EMPNO, NAME, DNO, JOB, MGR, SAL, COMMISSION)

EMPNO from domain EmployeeNumbers...Integrity constraints to be added

DEPT( DNO, DNAME, LOC)DNO from domain DepartmentNumberIntegrity constraints to be added


Database Example- Database (instance)

EMP( EMPNO, NAME, DNO, JOB, MGR, SAL, COMMISSION)1234 Smith D2 J1 4567 30K 102222 Jones D2 J2 4567 25K4567 Blue D1 J4 9999 40K9999 Green D3 J5 100K

DEPT( DNO, DNAME, LOC)

D1 xxx yyyD2 aaa yyyD3 bbb zzz


Relational Integrity Constraints

• Not included in Codd’s original definition of relations.

• Not supported in initial commercial relational database systems.

• Now supported in all major relational database systems.


Integrity Rules:Applicable to All Databases

• Key constraint

• Entity integrity

• Referential integrity


The Three Integrity Rules

• Key constraintCandidate keys are specified for each relation

schema.

Candidate key values must be unique for every tuple in any relation instance of that schema.


The Three Integrity Rules, Continued

• Entity integrity:– No attribute participating in the primary key is

allowed to accept null values.

Justification: we must be able to identify each tuple uniquely.


Foreign Keys EMP( EMPNO, NAME, DNO, JOB, MGR, SAL, COMMISSION)

DEPT( DNO, DNAME, LOC)

DNO in EMP - should be allowed only if that DNO appears as primary key in relation DEPT.

MGR in EMP - should be allowed only if that MGR appears as primary key in relation EMP.

The attributes DNO and MGR called foreign keys in relation EMP.


The Three Integrity Rules, Continued

• Referential integrity:– If relation schema R1 includes a foreign key, FK,

matching the primary key of relation schema R2 , then a value of FK in a tuple t1 of R1 must either

• Be equal to the value of PK in some tuple t2 of R2 ,

i.e. T1 [FK] = t2 [PK], or• Be wholly null (i.e. Each attribute value participating in

that FK value must be null).

Note: R1 and R2 need not be distinct.

Justification: if some tuple t1 references some tuple t2 , then tuple t2 must exist.


Implications of the Three Integrity Rules

• System could reject the operation.

• System could accept the operation, but perform some additional operations so as to reach a new legal state.

What should happen if an operation on the database is about to cause the violation of one of the integrity rules? i.e. about to put the database into an “illegal” state.


Query Languages Review

Srinivasan Parthasarathy


SQL - Parts of the Language

• Data Definition Language (DDL)– create table– create index

• Data Manipulation Language (DML)– select (retrieve)– update– insert– delete


Select - Basic Form(select from where)

Cartesian product followed by select and project.select project-listfrom Cartesian-product-listwhere select-condition(s)

Abstract example: Given tables R(A,B) and S(B,C)select R.A, R.B, S.Cfrom R, Swhere R.A > 10

BUT - BUT - DuplicatesDuplicates NOT eliminated. Bag vs. Set. NOT eliminated. Bag vs. Set.


EMP( EMPNO, … , DNO, JOB, . . .)100 D3 electrician200 D3 plumber300 D3 electrician400 D1 electrician500 D1 plumber600 D1 carpenter700 D2 electrician800 D2 carpenter900 D2 electrician

CASE STUDY EXAMPLE


Select as a JOINCartesian product followed by

select (“join” & “select” conditions) and project.select project-listfrom Cartesian-product-listwhere join-conditionand select-condition

Abstract example: Given tables R(A,B) and S(B,C)select R.A, R.B, S.Cfrom R, Swhere R.B = S.B /* join condition */ and R.A > 10 /* “select” condition */

How is this related to these relational operators? Select (), Project (), Join: Natural (*)


Using EMP and DEPTFrom Relational Algebra to SQL

• List the names, employee numbers, department numbers

and locations for all clerks.select NAME, EMPNO, E.DNO, LOC from EMP E, DEPT Dwhere E.DNO = D.DNO /* join condition */and JOB = ‘Clerk’ /* “select” condition */

.

EMP( EMPNO, NAME, DNO, JOB, MGR, SAL, COMMISSION)DEPT( DNO, DNAME, LOC)


• Duplicates in project - must use explicit distinctList the different department numbers in the EMP

table (eliminate duplicates).select distinct DNOfrom EMP

• Specify sort orderList employee number, name, and salary of employees

in department 50.select EMPNO, NAME, SALfrom EMPwhere DNO = 50order by EMPNO


• UnionList the numbers of those departments which

have an employee named ‘Smith’ or are located in ‘Columbus’.select DNOfrom EMPwhere ENAME = ‘Smith’ unionselect DNOfrom DEPTwhere LOC = ‘Columbus’

DuplicatesDuplicates ARE eliminated ARE eliminated by default.by default.

union union allall - leaves duplicates - leaves duplicates


Functions and Groups

• List the departments (DNO) and the average salary of each.

select DNO, avg(SAL)from EMP E, DEPT Dwhere E.DNO = D.DNOgroup by DNO

• List the departments (DNO, DNAME) in which the average employee salary < $25,000.

select DNO, DNAMEfrom EMP E, DEPT Dwhere E.DNO = D.DNOgroup by DNO, DNAMEhaving avg(SAL) < 25000


Nested Select: No analog in Relational Algebra • List names of employees in departments 25,

47 and 53.select NAMEfrom EMPwhere DNO in (25, 47, 53)

• List names of employees who work in departments in Ann Arbor.

select NAMEfrom EMPwhere DNO in ( select DNO

from DEPTwhere LOC =

‘Ann Arbor’ )


Null Values

All of the following conditions are always false.null > 25 null < 25 null = 25 null <> 25null >= 25 null <= 25 null = null null <> null

However we can use the following:select NAMEfrom EMPwhere SAL < 35000 or SAL is null


select DNOfrom DEPTwhere exists

( select *from EMP ED3where ED3.DNO = ‘D3’ and exists

( select *from EMP EYwhere EY.JOB = ED3.JOB

and EY.DNO = DEPT.DNO))

exists,

Q. Find the numbers of those departments that have employees who can do some job that is done by an employee in department D3. Answer: D1, D2, D3

The order of the two “selects” does not matter.

Ok, lets see why this works

Since there is an external reference to DEPT within the secondNested select we will execute the nested select for each DEPT Tuple. Assume the DEPT table contains only three tuples corresponding to D1, D2 and D3 in that order.

The first tuple we will evaluate is the DEPT.DNO = D1

The first nested select will simply highlight the tuples indepartment D3. The second nested select will point to tuples related to the tuplecurrently pointed to within the department table.

ED3.EMPNO, ED3.DNO, ED3.JOB,..)

100 D3 electrician

200 D3 plumber

300 D3 electrician

400 D1 electrician

500 D1 plumber

600 D1 carpenter

700 D2 electrician

800 D2 carpenter

900 D2 electrician

EY.EMPNO, EY.DNO, EY.JOB,..)

100 D3 electrician

200 D3 plumber

300 D3 electrician

400 D1 electrician

500 D1 plumber

600 D1 carpenter

700 D2 electrician

800 D2 carpenter

900 D2 electrician

select DNOfrom DEPTwhere exists





for all,

EMP( EMPNO, … , DNO, JOB, . . .)100 D3 electrician200 D3 plumber300 D3 electrician400 D1 electrician500 D1 plumber600 D1 carpenter700 D2 electrician800 D2 carpenter900 D2 electrician

Q. Find the numbers of those departments that have employees who can do all the jobs that are done by an employee in department D3. Answer: D1, but not D2


select DNOfrom DEPTwhere for all




Q. Find the numbers of those departments that have employees who can do all the jobs that are done by an employee in department D3. Answer: D1,D3, but not D2

However no for all exists in SQL.


select DNOfrom DEPTwhere not exists

( select *from EMP ED3where ED3.DNO = ‘D3’ and not exists



Q. Find the numbers of those departments that have employees who can do all the jobs that are done by an employee in department D3. Answer: D1,D3, but not D2

However no for all exists in SQL.Use two not exists. (see page 207 fora good list of mathematical logicoperations/tricks)

Ok, lets see why this works

Since there is an external reference to DEPT within the secondNested select we will execute the nested select for each DEPT Tuple. Assume the DEPT table contains only three tuples corresponding to D1, D2 and D3 in that order.

The first tuple we will evaluate is the DEPT.DNO = D1

The first nested select will simply highlight the tuples indepartment D3. The second nested select will point to tuples related to the tuplecurrently pointed to within the department table.

ED3.EMPNO, ED3.DNO, ED3.JOB,..)

100 D3 electrician

200 D3 plumber

300 D3 electrician

400 D1 electrician

500 D1 plumber

600 D1 carpenter

700 D2 electrician

800 D2 carpenter

900 D2 electrician

EY.EMPNO, EY.DNO, EY.JOB,..)

100 D3 electrician

200 D3 plumber

300 D3 electrician

400 D1 electrician

500 D1 plumber

600 D1 carpenter

700 D2 electrician

800 D2 carpenter

900 D2 electrician









and EY.DNO = DEPT.DNO) ) and DNO <> ‘D3’

Q12. Find the numbers of those departments that have employees who can do all the jobs that are done by an employee in department D3. Answer: D1, but not D2

Eliminate department D3.


SQL:1999 (SQL 3) Recursive Closure

Q 16. List all the superiors of EMPNO 500.

600, 950, 980

Q 17. List all those supervised by EMPNO 900.

700, 800, 100, 200, 300,

400

How to express these queries?

980

900 950

700 800 600 850

100 200 300 400 500


with recursiveSUPERIORS(EMPNO, MGR) as

(select EMPNO, MGR from EMP where EMPNO = 500

union all select SUPERIORS.EMPNO, EMP.MGR from SUPERIORS, EMP where EMP.EMPNO = SUPERIOR.MGR)

select MGRfrom SUPERIORS

Q 16. Given EMP (EMPNO, MGR, ...),list all the superiors of EMPNO 500.

Generate SUPERIORS (EMPNO, MGR)

Just the superiors 600of 500. 950

980

Initial table

The recursion


Q 16. List all the superiors of EMPNO 500. SUPERIORS (EMPNO, MGR)

Initial table

Second addition

First addition

EMP( EMPNO, MGR,...)100 700 200 700300 800400 800500 600600 950700 900800 900850 950900 980950 980980 980

500 600

500 950

500 980


Entity-Relationship (ER) Model(Peter P.-S. Chen)

Review

CIS 671S. Parthasarathy


Helpful for conceptualizing the Real World• Entity: a thing that exists

– e.g. person, automobile, department, employee

• Entity Set: a group of similar entities– e.g. all persons, all automobiles, all employees

• Relationship: association between entities– e.g. a person is assigned to a department

• Relationship Set : set of similar relationships

• Attribute: property of an entity or relationship– e.g. person - name, address

• Domain: set of values allowed for an attribute


ExampleEmployees E#, ENAME, ADDRESSDepartments D#, DNAMEProjects P#, PNAME

Constraints (cardinality)1. Employees may be assigned to only 1 department at a time.2. Employees may be assigned to several projects at once,

each with an associated %time.

Constraints (participation)3. Employees must be assigned to a department.4. Employees need not be assigned to any projects.


Complete Picture

Project

AssignedTo

Employee

Department

IsIn

%TIME

N

1

M

N

D#

DNAME

P#

E#

ENAME ADDRESS

PNAME


Example: Relationships and Attributes

ProjectEmployee AssignedTo

Employee DepartmentIsIn

%TIME%TIME

N 1

MN

2.

1.

2. Employees may be assigned to several projects at once, each with an associated %time.

1. Employees may be assigned to only 1 department at a time.


Example: Relationships and Attributes

ProjectEmployee AssignedTo

Employee DepartmentIsIn

%TIME%TIME

N 1

MN

4.Partial

3. Total

3. Employees must be assigned to a department.

• Total: Each entity must be included at least once in the relationship.

4. Employees need not be assigned to any projects.

• Partial: Each entity instance need not be included at least once in the relationship.


Entity-Relationship Enhancements: Attributes

1. Simple (atomic) vs. Composite Attributes– Simple

– Composite

Employee Employee

Employee

LName

FName

Name

E#

LName

FName


2. Single-valued vs. Multi-valued Attributes– Multi-valued

– Multi-valued as Entity


Student

Major_Program

Major

Major

StudentHas

MajorN M



3. Derived Attributes - Include in Department the average salary of the employees in the department.

Employee Department

Salary AvgSal

MemberOf


Name

Entity-Relationship Enhancements: Entities

• Weak Entity Type, Identifying Relationship Type, Partial Key

• E.g. Represent all the dependents of each employee given his or her name. No ID number. Problem: Names are not unique across employees.

Employee DependentsOf

Dependent

NameSSN

1 N


Complete Picture

Project

AssignedTo

Employee

Department

IsIn

%TIME

N

1

M

N

D#

DNAME

P#

E#

ENAME ADDRESS

PNAME


As Relations

• Entities– Department(D#, DNAME)– Employee(E#, ENAME, ADDRESS)– Project(P#, PNAME)

• Relationships– Is_In(E#, D#) 1:N– Assigned_To(E#, P#, %TIME) N:M


As Relations: Replacing Employee and Is_In with Employee’

• Entities– Department(D#, DNAME)– Employee’(E#, ENAME, ADDRESS, D#)– Project(P#, PNAME)

• Relationships– Assigned_To(E#, P#, %TIME) N:M

– [Is_In(E#, D#)]


Enhanced-ER (EER) Model

• Subclasses & Superclasses– Specialization & Generalization– Type Inheritance

• Categories


• Superclasses & Subclasses• Generalization & Specialization• Inheritance

Example: University Database

Person

Employee

Student

Grad Student

Undergrad Student

SSN

Salary

Name

Class

EmpID

DegreeProgram

MajorDept

d

o

U

UU

U

d disjoint

o overlap

total

partial

Usuperclass subclass

Superclass instance must always exist.


Example: Meeting Locations

MeetingLocationPlace

Park

Organization

Date

Day*

MeetsAtCapacity

RoomNumber*

Name*Street

Address

Time*

OrganizationName*

NorthSouthCoordinates

ParkName*

EaseWestCoordinates

Building

IsIn

Room

U U

U union• Categories

CIS 671Query Languages1 CIS 671 Introduction to Database Systems II Introduction S. Parthasarathy.

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