Relational Model

COP 4720Lecture 9

Lecture Notes

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Outline of Lecture

• Relational Algebra• Queries

• Sec. 3.2

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Summary of Relational Algebra (1)

E ::= R

| SELECTC(R) or c(R)

| PROJECTA1,A2,...,An(R) or A1,A2,...,An

(R)

| E1 X E2

| E1 U E2

| E1 - E2

| RENAMER(A1,A2,...,Am)(E) or

R(A1,A2,...,Am)(E)

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Summary of Relational Algebra (2)

E ::= R

| E1 JOIN E2 or E1 E2

| E1 JOINC E2 or E1 C E2

| E1 INTERSECT E2 or E1 E2

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Sample Schema for Exercises

Student(ID, name, address, GPA, SAT)

Campus(location, enrollment, rank)

Apply(ID, location, date, major, decision)

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Sample Queries

PROJECTdate,decision(Apply)

Campus X Apply

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Sample Queries

Find Names and addresses of all students with GPA > 3.7 who applied to CS major and were rejected.

List name and address of all students who didn’t apply anywhere.name, address ((Students) Students.ID= Not_Apply.ID

(Not_Apply(ID (Students) - ID (Apply))))

name, address (GPA>3.7 decision=‘No’ major=‘CS’

(Student Student.ID=Apply.ID Apply))

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Schema for examples

Employee(fname,lname,ssn,bdate,address,sex,salary,superssn,dno)

Department(dname, dnumber,mgrssn,mgrstartdate)

Dept_Locations(dnumber, dlocation)

Project(pname,pnumber,plocation,dnum)

Works_on(essn,pno,hours)

Dependent(essn,dependent_name,sex,bdate,relationship)

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Examplesa) Retrieve the name and address of all employees who work for the

‘Research’ department.

research dname = ‘Research’(Department)

research_emps research dnumber = dno(Employee)

Result fname,lname,address(research_emps)

or research Selectdname = ‘Research’(Department)

research_emps research joindnumber=dno Employeeresult projectfname,lname,address(research_emps)

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Examples(refer to schema on previous slide)

b) Retrieve the names of all employees in dept. 5 who work more than 10 hrs./week on project X.

Option 1:Fname,Lname (Employee SSN=ESSN

(Pname=‘Project X’ hours>10 Dnum=5

(Project Pnumber=Pno Works_On) ) )

Option 2:Emp_Work_X Pname=‘Project X’ (Project Pnumber=Pno Works_On)

Emp-Work_10 Employee SSN=ESSN (hours>10 (Emp_Work_X) )

Result Fname,Lname (Dnum=5(Emp_Work_10) )

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c) List names of employees who have a dependent with the same first name.

Same (Employee) Fname=Dependent_Name(Dependent)

Result Fname,Lname (Same)

???????

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c) List names of employees who have a dependent with the same first name.

Same (Employee) Fname=Dependent_Name ssn=essn(Dependent)

Result Fname,Lname (Same)

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d) Retrieve the names of employees who work on every project.

Emp_proj Project essn,pno (Works_on)T1 Project essn (Emp_proj)T2 Project pno (Works_on)T3 T2 x T1T4 T3 – Emp_projT5 Project essn T4Ans T1 – T5