10/3/2017

Chapter 6 The Relational Algebra and Calculus 10/3/2017 Chapter 6 The Relational Algebra and Calculus Copyright © 2004 Ramez Elmasri and Shamkant Navathe

Recap of Relational Algebra Operations

Database State for COMPANY All examples discussed below refer to the COMPANY database shown here.

Q4: For every project located in ‘stafford’, list the project no, the controlling depart no, and the department manager’s last name, address and birth date. STAFFORD_PROJS   Plocation=’stafford’ (project) CONTR_DEPT  (STAFFORD_PROJS DNUM= DNUMBER DEPARTMENT) PROJ_DEPT_MGR (CONTR_DEPT MGR-SSN= SSN EMPLOYEE) RESULT   PNUMBER,DNUM, LNAME, ADDRESS,BDATE (PROJ_DEPT_MGR)

Examples of in Relational Algebra Q5: Find the names of employees who work on all the projects controlled by department no 5. ADEPT5_PROJS(PNO)   PNUMBER( DNUM=’5’ PROJECT) EMP_PROJ(SSN,PNO)   ESSN,PNO(WORKS_ON) RESULT_EMP_SSNS  (EMP_PROJ ÷ DEPT5_PROJS) RESULT   LNAME, FNAME (RESULT_EMP_SSNS * EMPLOYEE)

Examples of in Relational Algebra Q7. List the names of all employee with two or more dependants Here we use the aggregate function count to count the number of employees to be more than 2 T1(ssn,no_of_dependant) ESSN ʒ Count dependant_name(DEPENDANT) T2σ NO_of_dependant >= 2 (T1) RESULTLNAME,FNAME (T2*EMPLOYEE)

Examples of in Relational Algebra Q8. List the names of managers who have at least one dependant MGRS(ssn) MGR_SSN(DEPARTMENT) EMPS_WITH_DEPS(SSN) ESSN(DEPENDANT) MGRS_WITH_DEPS (MGRS∩EMPS_WITH_DEPS) RESULTLNAME,FNAME (MGRS_WITH_DEPS*EMPLOYEE)

Examples of Queries in Relational Algebra STUDENT Course STUDENT Consider the relations student ,course, and studentCourse as follows: Course

Examples of Queries in Relational Algebra Q1. Get student names for the students who have completed course c1. Here we can make natural join because both tables have common attribute we need in join condition RESULT   S_name(Student *C_course=“c1” StudentCourse) Q2. Get student names for the students who have studied by at least book named “introduction to compilers” RESULT  s_name( s_code( c_code((σ c_book=‘introduction to compilers’(Course))* StudentCourse)) * Student)

Examples of Queries in Relational Algebra Q3. Get student names for the students who have completed all courses RESULT  s_name(( s_code,c_code(StudentCourse)÷  C_code(course))*(student) Q4. Get student codes for the students who have completed at least all those courses completed by student s2 RESULT  s_code( s_code,c_code(studentCourse)÷  c_code(σ s_code=‘s2’(studentCourse)))

Examples of Queries in Relational Algebra Q5. Get student names for the students who have not completed course c2 Here we firstly get students who have completed the course c2 and make the difference between the result and all student in student realtion RESULT  s_name(( s_code (Student) –  s_code (σ c_code=‘c2’(StudentCourse)))* Student) Q6. get all pairs of student names such that the two students are in the same year. Here we take acopy from table student and rename it as stud1 And then we make join between both relaions RESULT stud1.s_name,stud1.s_code( σ((stud1.s_Code=student.s_code)&(stud1.s_year=student.s_year)) (ρ stud1(student)) * (student))