CIS611 Lab Assignment 1 Part 4: Solution SS Chung

4. Update the following new changes into the database:

1) Joyce English with Ssn = 453453453 got married with John.

2) Jenifer Wallace with Ssn = 987654321 just had a new daughter named Erica.

3) Jenifer Wallace with Ssn = 987654321 is just assigned to a new project number ‘10’ to work

on.

Add these new entries into Dependent, Works_On tables in your database then Select * from

Dependent and Select * from Works_On to show the updated table contents.

5. Write SQL Select statements to retrieve data in the followings:

Q1:

Make a list of all project numbers that Research department employees are working on,

either as a worker or as a manager of the department that controls the projects.

(SELECT W.Pno FROM DEPARTMENT D, WORKS_ON W

WHERE D.Dname = 'research' and w.essn = D.mgrssn)

UNION /*as employee */

(SELECT W.Pno

FROM WORKS_ON W, department D, EMPLOYEE E

WHERE E.Dno = D.Dnumber AND

W.Essn = E.Ssn AND D.Dname = 'research');

/*Left Q: for projects as a manager of Research dept -- It returns 2,3,10,20*/ SELECT Distinct W.Pno

FROM DEPARTMENT D, WORKS_ON W

WHERE D.Dname = 'research' and w.essn = D.mgrssn;

/*Right Q: for projects as employees in Research dept -- It returns 1,2,3,10,20*/

SELECT Distinct W.Pno FROM WORKS_ON W, department D,

EMPLOYEE E

WHERE E.Dno = D.Dnumber AND

W.Essn = E.Ssn AND D.Dname = 'research';

Q1 in Relational Algebra:

MGRSSN_RESEARCH_DEPT ← MGRSSN ( DNAME=' Research' (DEPARTMENT))

RESEARCH_MGR_PROJ ← PNO (WORKS_ON ESSN=MGRSSN MGRSSN_RESEARCH_DEPT)

DNUMBER_RESEARCH_DEPT ← DNUMBER ( DNAME=' Research' (DEPARTMENT))

SSN_EMP_RESEARCH_DEPT ← SSN (EMPLOYEE DNO=DNUMBER DNUMBER_RESEARCH_DEPT)

RESEARCH_EMP_PROJ ← PNO (WORKS_ON ESSN=SSN SSN_EMP_RESEARCH_DEPT)

RESULT ← (RESEARCH_MGR_PROJ ∪ RESEARCH_EMP_PROJ)

/*another version because of LeftQ for Projects as a manager of Research dept. It

returns 1,2,3,10,20 */

(SELECT P.Pnumber

FROM PROJECT P, DEPARTMENT D

WHERE P.Dnum = D.Dnumber AND

D.Dname = 'research')

UNION /*as employee */ (SELECT W.Pno

FROM WORKS_ON W, department D,

EMPLOYEE E WHERE E.Dno = D.Dnumber AND

W.Essn = E.Ssn AND

D.Dname = 'research');

/*LeftQ --for projects as a manager -- It returns 1,2,3 for research dept projects

*/

SELECT Distinct P.Pnumber

FROM PROJECT P, DEPARTMENT D

WHERE P.Dnum = D.Dnumber AND

D.Dname = 'research';

/*Right Q: projects as Research department employees: 1,2,3,10,20*/

SELECT Distinct W.Pno FROM WORKS_ON W, department D,

EMPLOYEE E

WHERE E.Dno = D.Dnumber AND

W.Essn = E.Ssn AND

D.Dname = 'research';

DNUMBER_RESEARCH_DEPT ← DNUMBER ( DNAME=' Research' (DEPARTMENT))

RESEARCH_MGR_PROJ ← PNO (PROJECT DNUM=DNUMBER DNUMBER_RESEARCH_DEPT)

SSN_EMP_RESEARCH_DEPT ← SSN (EMPLOYEE DNO=DNUMBER DNUMBER_RESEARCH_DEPT)

RESEARCH_EMP_PROJ ← PNO (WORKS_ON ESSN=SSN SSN_EMP_RESEARCH_DEPT)

RESULT ← (RESEARCH_MGR_PROJ ∪ RESEARCH_EMP_PROJ)

Q2:

Get SSN and the last name of married female employees who work on three or more

projects --Q1

SELECT E.SSN, E.LNAME FROM EMPLOYEE E

WHERE E.Sex = ‘F’ and

E.SSN IN (SELECT Dp.ESSN FROM Dependent Dp

WHERE Dp.relationship = 'Spouse')

AND E.SSN IN (SELECT W.ESSN

FROM WORKS_ON W

GROUP BY W.ESSN

HAVING COUNT(W.Pno) >= 3);

-- OR Q2

SELECT E.SSN, E.LNAME

FROM EMPLOYEE E, dependent Dp

WHERE E.Sex = ‘F’ and E.ssn = Dp.essn and Dp.relationship = 'Spouse' AND E.SSN IN (SELECT W.ESSN

FROM WORKS_ON W

GROUP BY W.ESSN

HAVING COUNT(W.Pno) >= 3);

-- Q2

SELECT E.SSN, E.LNAME FROM EMPLOYEE E, Dependent Dp

WHERE E.Sex = ‘F’ and E.ssn = Dp.essn and Dp.relationship = ‘spouse’ AND

(SELECT COUNT(*) FROM WORKS_ON W

WHERE E.SSN = W.ESSN ) >= 3;

-- OR

-- Q3

SELECT E.SSN, E.LNAME FROM EMPLOYEE E, Dependent Dp,

(SELECT ESSN

FROM WORKS_ON

GROUP BY ESSN

HAVING COUNT(PNO) >= 3) AS THREEMOREPROJECT(ESSN) WHERE E.Sex = ‘F’ and E.ssn = Dp.essn and Dp.relationship = 'Spouse' AND

E.SSN = THREEMOREPROJECT.ESSN;

-- OR

-- Q4 select e.ssn, e.Lname

from EMPLOYEE e, Dependent Dp

where E.Sex = ‘F’ and E.ssn = Dp.essn and Dp.relationship = 'Spouse' AND e.Ssn IN (select wo.Essn

from WORKS_ON wo

group by wo.Essn having count(wo.Pno) >= 3);

Q2 in Relational Algebra:

SSN_Female_Emp � SSN ( ( SEX = ‘F’ (EMPLOYEE) ))

ESSN_Married_Emp � ESSN ( ( Relationship = ‘Spouse’ (DEPENDENT) ))

SSN_Married_Female_Emp � SSN (Female_Emp SSN=ESSN Married_Emp)

ESSN_3MoreProjects �( Count(pno) >= 3 (ESSN F COUNT(pno)WORKS_ON))

SSN_MarriedFemaleEmp_3MoreProject �

SSN(SSN_Married_Female_Emp SSN=ESSN ESSN_3MoreProjects)

Name_MarriedFemaleEmp_3MoreProject �

SSN,LNAME (EMPLOYEE * SSN_MarriedFemaleEmp_3MoreProject)

Q3. Write SQL using View instead of Subquery to express the query Q2.

Create View ThreeMOREPROJECT As

SELECT ESSN FROM WORKS_ON

GROUP BY ESSN

HAVING COUNT(PNO) >= 3;

Q3 Using View: SELECT E.SSN, E.LNAME

FROM EMPLOYEE E, ThreeMOREPROJECT

WHERE E.Sex = 'F' and E.SSN = ThreeMOREPROJECT.ESSN;

Or

Q3 Using View:

select e.ssn, e.Lname

from EMPLOYEE e, Dependent Dp where E.Sex = ‘F’ and E.ssn = Dp.essn and Dp.relationship = 'Spouse' AND

e.Ssn IN (select Essn

from ThreeMOREPROJECT);

Q4:

Get the Last and the First name of married employees who only have daughters.

(Hint: Indentify the following three sets in SQL first then write a main SQL using them.

Married = Select ESSN From Dependent Where relationship = ‘spouse’;

Girls = Select ESSN From Dependent Where relationship = ‘daughter’;

Boys = Select ESSN From Dependent Where relationship = ‘son’;

SELECT E.fname, E.lname FROM Employee E, Dependent D

WHERE D.essn = E.ssn AND D. relationship = 'spouse' AND

D.essn IN (SELECT ESSN FROM Dependent D2

WHERE D. essn =D2.essn and D2.relationship = 'daughter')

AND

D.essn NOT IN (SELECT D3.ESSN FROM Dependent D3

WHERE D. essn =D3.essn and D3.relationship = 'son');

OR

Q4:

Select E.fname, E.lname

From EMPLOYEE E, DEPENDENT D Where E.ssn = D.essn and D.relationship = 'spouse' AND

EXISTS (Select D2.ESSN

From Dependent D2 Where D. essn =D2.essn and relationship = 'daughter')

AND

NOT EXISTS (Select D3.ESSN

From Dependent D3 Where D. essn =D3.essn and D3.relationship = 'son');

Q4 in Relational Algebra:

MARRIED_EMP (ESSN) ← ESSN ((EMPLOYEE) SSN=ESSN Relationship=’Married’ (DEPENDENT))

MARRIED_W_DAUGHTER_EMP (ESSN) ←

ESSN (MARRIED_EMP ESSN=ESSN ( Relationship=’Daughter’(DEPENDENT)) )

MARRIED_W_SON_EMP (ESSN) ←

ESSN (MARRIED_EMP ESSN=ESSN ( Relationship=’Son’(DEPENDENT)) )

DAUGHTER_ONLY_EMP_SSNS ← (MARRIED_W_DAUGHTER_EMP - MARRIED_W_SON_EMP )

RESULT ← FNAME, LNAME (EMPLOYEE * DAUGHTER_ONLY_EMP_SSNS)) )

Q4 in Relational Tuple Calculus:

{e.LNAME, e.FNAME | EMPLOYEE (e) AND

(∃ (d)) (DEPENDENT(d) AND e.ssn = d.essn AND d.relationship = ‘spouse’ AND

(∃ (d1)) (DEPENDENT(d1) AND e..ssn = d1.essn AND d1.relationship = ‘daughter’ )AND

(NOT ∃ (d2)) (DEPENDENT(d2) AND e.ssn = d2.essn AND d2.relationship = ‘son’))}