Figure 3 One possible database state for the COMPANY relational database schema

Question

Figure 3 One possible database state for the COMPANY relational database schema EMPLOYEE Bdate B Smith 123456789 1965-01-09 231 Fondren, Houston, TX M 30000 John Frankin T Wong 333445555 1955-12-08 Alicia rS Wallace 987654321 1941 06-20 291 Benry BelaeX43000 890665555 455555 638 Voss. Houston, TX M 40000 888665555 JZelaya 999887777 1968-01-19 3321 Castle TXF25000 RKNorayan 000884444 1002091e HmT M 38000 20445598 loyceA Engish 453453453 1972-0731 5631 Rice Hoason. TXF 25000 AhmadV Jabbar982987987 1969-03-29 980 Dutas Houstion TXM 25000 9876543 amo E IBorg 888665555 1932-11-10, 450st nMouton. TX M 55000 NULL 445555E DE PARTMENT DEPT LOCATIONS Drame Dumber 338455 9826543

Explanation / Answer

1.

Select Fname,Minit,Lname,Dependent_name from Employee inner join Dependent on Employee.Ssn = Dependent.Essn;

2.

Select Dname,Pname from Department inner join Project on Department.Dnumber = Project.Dnum;

3.

Select Dname,concat(Fname,Lname) as Manager ,Pname from Department inner join Project on Department.Dnumber = Project.Dnum inner join Employee on Department.Mgr_ssn = Employee.ssn;

4.

Select e1.Fname,e1.Lname,e2.Fname,e2.Lname from Employee e1 inner join Employee e2 on e1.Super_ssn = e2.ssn;

5.

Select Dnumber,count(Pnumber) as Projects from Department,Project where Department.Dnumber = Project.Dnum group by Dnumber order by Projects desc;

6.

Select Super_ssn,count(ssn) as EmployeesManaged from Employee group by super_ssn order by EmployeeManager desc Limit 1;

Do ask if any doubt. Please upvote

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.