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Relational Model

This lecture outlines the fundamental concepts of relational algebra, encompassing the various operations and queries used in database management. Key operations such as SELECT (σ), PROJECT (π), JOIN, INTERSECT, and RENAME are introduced, along with sample queries based on relational schemas like Student, Campus, and Employee. The lecture emphasizes how to retrieve data effectively from databases, showcasing practical examples like fetching student records based on GPA and department, and extracting employee information linked to projects. Understanding these operations is crucial for effective database querying.

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Relational Model

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  1. Relational Model Lecture Notes COP 4720 Lecture 9

  2. Outline of Lecture • Relational Algebra • Queries • Sec. 3.2 COP 4720

  3. 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) COP 4720

  4. 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 COP 4720

  5. Sample Schema for Exercises Student(ID, name, address, GPA, SAT) Campus(location, enrollment, rank) Apply(ID, location, date, major, decision) COP 4720

  6. Sample Queries PROJECTdate,decision(Apply) Campus X Apply COP 4720

  7. Sample Queries name, address (GPA>3.7  decision=‘No’ major=‘CS’ (Student Student.ID=Apply.ID Apply)) name, address ((Students) Students.ID= Not_Apply.ID (Not_Apply(ID (Students) - ID (Apply)))) 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. COP 4720

  8. 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) COP 4720

  9. Examples • 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 Employee result  projectfname,lname,address(research_emps) COP 4720

  10. Examples(refer to schema on previous slide) 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) ) b) Retrieve the names of all employees in dept. 5 who work more than 10 hrs./week on project X. COP 4720

  11. Same  (Employee) Fname=Dependent_Name(Dependent) Result  Fname,Lname(Same) ??????? c) List names of employees who have a dependent with the same first name. COP 4720

  12. Same  (Employee) Fname=Dependent_Name  ssn=essn(Dependent) Result  Fname,Lname(Same) c) List names of employees who have a dependent with the same first name. COP 4720

  13. d) Retrieve the names of employees who work on every project. Emp_proj  Project essn,pno (Works_on) T1  Projectessn (Emp_proj) T2  Project pno (Works_on) T3  T2 x T1 T4  T3 – Emp_proj T5  Project essn T4 Ans  T1 – T5 COP 4720

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