Relational Model Concepts

The relational model represents the database as a collection of relations. Each relation resembles a table of values. A table is called a relation. A row is called a tuple. A column header is called an attribute. The data type describing the type that can appear in each column is called a domain. A domain (D) is a set of atomic values.

Basic Structure Employee - Relation name Empno, ename, salary, dno – Attribute All records are Tuples

Query Language Categories –Procedural Example Relational Algebra –Non Procedural Example Relational Calculus

Relation Algebra It is procedural language It consists of set of operations that take one or more relations as input and produces new relation as output. The operation can be divided into three parts Basic operations Additional operations Extended operations

Basic operations Select (  ) Project (  ) Union (U) Rename (  ) Set difference (-) Cartesian product (x)

Additional operations Set intersection (  ) Natural join Division (  ) Assignment (  )

Extended operations Aggregate operations Outer join

Basic operations The Select operation (  ) –It is used to select a subset of the tuples (records) from a relation that satisfy a selection condition. Syntax  (R) Note : R is table name Example 1)  dno=10 (emp) 2)  ename=‘raja’ (emp) Note : Display eno,ename,salar dno (all heading)

Project operation The required attributes alone can be listed from a relation (selects certain columns from the table) Syntax  (R) Note : R is table name Example 1)  eno,salary (emp) display eno, salary (two column) 2)  ename  dno=10 (emp) Note : Display ename column only

The union operation (U) RUS is all tuples that are either in R or in S. Duplicate tuples are eliminated. Syntax RUS Example  ename (emp1) U  ename (emp2)

The Set difference operation ( - ) It is used to find tuples that are in one relation but are not in another. Syntax R - S Example  ename (emp) U  ename (empit)

Cartesian Product operation (X) It allows us to combine information from any 2 relations. Syntax R X S = { tq/t  R and q  s} Example  dno=10 (emp X dept)

Rename operation (  ) Allow us to name the relational algebra expressions Allow us to refer to a relation by more than one name. Example  temp (  eno,salary (emp))

Join operation R=(A,B,C,D) s=(E,B,D) –(A,B,C,D,E Column heading) R IXI S=(A,B,C,D,E) Example R IXI S =  r.A, r.B, r.C, r.D, s.E (  r.B =s.B and r.D=s.D (RXS)) equal  r.B =s.B and r.D=s.D (R IXI S))

Division operation ( r  s) R=(A1,A2,A3, …… Am, B1,B2,B3,…….Bn) S=(B1,B2,…..Bn) The result of r  s is r  s =(A1,A2,A3, …… Am)

Assignment Operation (  ) It can be assigned to a temporary relation variable (S). Example Temp (  )  eno,salary (emp) Temp1 (  )  eno,salary  salary>10000 (emp)

Aggregate functions Aggregate function takes a collection of values and returns a single values as a result. AVG – average value MIN – minimum value MAX – Maximum value SUM – Sum of values COUNT – number of record

Syntax G Aggregate_function (attribute_name) (relation name) Example G Sum(salary) (emp) dno G Sum(salary) (emp) Find sum of salary for employees to each department.

Difference between join (IXI) and Cartesian product (X) Join (IXI)Cartesian product (X) Only combinations of tuples satisfying the joint condition appear in the result. All combinations of tuples are included in the result.

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) Consider the relation database, where the primary keys are underlined. Give an expression in relational algebra, SQL query of the following a)Find the names of all employees who work for First Bank Corporation b)Find the names and cities of residence of all employees who work for First Bank Corporation c)Find the names, street address, cities of residence of all employees who work for bank and earn more than Rs. 10,000 per annum. d)Find the names of all employees in this database who live in the same city as the company for which they work. e)Find the names of all employees who live in the same city and on the same street as do their managers. f)Find the names of all employees in this database who do not work for First Bank Corporation

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) a) Find the names of all employees who work for First Bank Corporation Relation Algebra  person_name (  company_name="First Bank Corporation” (works)) SQL Query SQL> Select person_name from works where company_name=“First Bank Corporation”;

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) b) Find the names and cities of residence of all employees who work for First Bank Corporation Relation Algebra  person_name, city (  company_name="First Bank Corporation” ^ employee.person_name=works.person_name (Employee X works)) SQL Query SQL> Select person_name, city from employee, works where company_name=“First Bank Corporation” and employee.person_name=works.person_name ;

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) c) Find the names, street address, cities of residence of all employees who work for bank and earn more than Rs. 10,000 per annum. Relation Algebra  person_name,street,city (  company_name="First Bank Corporation” ^ Salary>10000 ^ employee.person_name=works.person_name (Employee X works)) SQL Query SQL> Select person_name, street, city from employee, works where company_name=“First Bank Corporation” and salary>10000 and employee.person_name=works.person_name ;

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) d) Find the names of all employees in this database who live in the same city as the company for which they work. Relation Algebra  person_name (  employee.city=company.city ^ employee.person_name =works.person_name^company.company_name=works.company_name (Employee X works X company)) SQL Query SQL> Select person_namefrom employee, works, company where employee.city=company.city and works.comapany_name = company.company_name;

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) e)Find the names of all employees who live in the same city and on the same street as do their managers. Relation Algebra  person_name (  employee.person_name=manges.person_name ^ company.city=employee.city (Employee X manages X company)) SQL Query SQL> Select person_name from employee,manages, company where employee.person_name=manages.person_name And company.city=employee.city;

employee (person_name, street, city) works (person_name, company_name, salary) Company (company_name,city) manages (person_name, manager_name) f) Find the names of all employees in this database who do not work for First Bank Corporation Relation Algebra  person_name (  employee.person_name<> First Bank Corporation” (Works)) SQL Query SQL> Select person_name from works where company_name<> “First Bank Corporation”;