/* company.sql Script to define and populate the example database COMPANY from C
ID: 3757322 • Letter: #
Question
/*
company.sql
Script to define and populate the example database COMPANY from Chapter 3
of Elmasri and Navathe (pp.70-75 in the 6th edition), with some modifications
*/
-- drop tables
DROP TABLE EMPLOYEE CASCADE CONSTRAINTS;
DROP TABLE DEPARTMENT CASCADE CONSTRAINTS;
DROP TABLE DEPT_LOCATIONS CASCADE CONSTRAINTS;
DROP TABLE PROJECT CASCADE CONSTRAINTS;
DROP TABLE WORKS_ON CASCADE CONSTRAINTS;
DROP TABLE DEPENDENT CASCADE CONSTRAINTS;
-- create and populate tables
CREATE TABLE EMPLOYEE
(
Fname VARCHAR(20),
Minit CHAR(1),
Lname VARCHAR(20),
Ssn CHAR(9),
Bdate DATE,
Address VARCHAR(30),
Sex CHAR(1),
Salary NUMBER(5),
Super_Ssn CHAR(9),
Dno NUMBER(1),
PRIMARY KEY (Ssn),
FOREIGN KEY (Super_ssn)
REFERENCES EMPLOYEE (Ssn)
);
INSERT INTO EMPLOYEE VALUES ('James', 'E', 'Borg', '888665555', DATE '1937-11-10', '450 Stone, Houston, TX', 'M', 55000, NULL, 1);
INSERT INTO EMPLOYEE VALUES ('Jennifer', 'S', 'Wallace', '987654321', DATE '1941-06-20', '291 Berry, Bellaire, Tx', 'F', 37000, '888665555', 4);
INSERT INTO EMPLOYEE VALUES ('Franklin', 'T', 'Wong', '333445555', DATE '1955-12-08', '638 Voss, Houston, TX', 'M', 40000, '888665555', 5);
INSERT INTO EMPLOYEE VALUES ('John', 'B', 'Smith', '123456789', DATE '1965-01-09', '731 Fondren, Houston, TX', 'M', 30000, '333445555', 5);
INSERT INTO EMPLOYEE VALUES ('Alicia', 'J', 'Zelaya', '999887777', DATE '1968-01-19', '3321 castle, Spring, TX', 'F', 25000, '987654321', 4);
INSERT INTO EMPLOYEE VALUES ('Ramesh', 'K', 'Narayan', '666884444', DATE '1920-09-15', '975 Fire Oak, Humble, TX', 'M', 38000, '333445555', 5);
INSERT INTO EMPLOYEE VALUES ('Joyce', 'A', 'English', '453453453', DATE '1972-07-31', '5631 Rice, Houston, TX', 'F', 25000, '333445555', 5);
INSERT INTO EMPLOYEE VALUES ('Ahmad', 'V', 'Jabbar', '987987987', DATE '1969-03-29', '980 Dallas, Houston, TX', 'M', 22000, '987654321', 4);
INSERT INTO EMPLOYEE VALUES ('Melissa', 'M', 'Jones', '808080808', DATE '1970-07-10', '1001 Western, Houston, TX', 'F', 27500, '333445555', 5);
CREATE TABLE DEPARTMENT
(
Dname VARCHAR(20),
Dnumber NUMBER(1),
Mgr_ssn CHAR(9),
Mgr_start_date DATE,
PRIMARY KEY (Dnumber),
FOREIGN KEY (Mgr_ssn)
REFERENCES EMPLOYEE (Ssn)
);
INSERT INTO DEPARTMENT VALUES ('Research', 5, '333445555', DATE '1988-05-22');
INSERT INTO DEPARTMENT VALUES ('Administration', 4, '987654321', DATE '1995-01-01');
INSERT INTO DEPARTMENT VALUES ('Headquarters', 1, '888665555', DATE '1981-06-19');
-- this alter is needed to allow PROJECT and DEPARTMENT to reference each other
ALTER TABLE EMPLOYEE ADD FOREIGN KEY (Dno) REFERENCES DEPARTMENT (Dnumber);
CREATE TABLE DEPT_LOCATIONS
(
Dnumber NUMBER(1),
Dlocation VARCHAR(20),
PRIMARY KEY (Dnumber, Dlocation),
FOREIGN KEY (Dnumber)
REFERENCES DEPARTMENT (Dnumber)
);
INSERT INTO DEPT_LOCATIONS VALUES (1, 'Houston');
INSERT INTO DEPT_LOCATIONS VALUES (4, 'Stafford');
INSERT INTO DEPT_LOCATIONS VALUES (5, 'Bellaire');
INSERT INTO DEPT_LOCATIONS VALUES (5, 'Sugarland');
INSERT INTO DEPT_LOCATIONS VALUES (5, 'Austin');
CREATE TABLE PROJECT
(
Pname VARCHAR(20),
Pnumber NUMBER(2),
Plocation VARCHAR(20),
Dnum NUMBER(1),
PRIMARY KEY (Pnumber),
FOREIGN KEY (Dnum)
REFERENCES DEPARTMENT (Dnumber)
);
INSERT INTO PROJECT VALUES ('ProductX', 1, 'Bellaire', 5);
INSERT INTO PROJECT VALUES ('ProductY', 2, 'Sugarland', 5);
INSERT INTO PROJECT VALUES ('ProductZ', 3, 'Houston', 5);
INSERT INTO PROJECT VALUES ('Computerization', 10, 'Stafford', 4);
INSERT INTO PROJECT VALUES ('Reorganization', 20, 'Houston', 1);
INSERT INTO PROJECT VALUES ('Newbenefits', 30, 'Stafford', 4);
CREATE TABLE WORKS_ON
(
Essn CHAR(9),
Pno NUMBER(2),
Hours NUMBER(3,1),
PRIMARY KEY (Essn, Pno),
FOREIGN KEY (Essn)
REFERENCES EMPLOYEE (Ssn),
FOREIGN KEY (Pno)
REFERENCES PROJECT(Pnumber)
);
INSERT INTO WORKS_ON VALUES ('123456789', 1, 32.0);
INSERT INTO WORKS_ON VALUES ('123456789', 2, 8.0);
INSERT INTO WORKS_ON VALUES ('453453453', 1, 20.0);
INSERT INTO WORKS_ON VALUES ('453453453', 2, 20.0);
INSERT INTO WORKS_ON VALUES ('333445555', 1, 10.0);
INSERT INTO WORKS_ON VALUES ('333445555', 2, 10.0);
INSERT INTO WORKS_ON VALUES ('333445555', 3, 5.0);
INSERT INTO WORKS_ON VALUES ('333445555', 10, 10.0);
INSERT INTO WORKS_ON VALUES ('333445555', 20, 10.0);
INSERT INTO WORKS_ON VALUES ('333445555', 30, 5.0);
INSERT INTO WORKS_ON VALUES ('999887777', 30, 30.0);
INSERT INTO WORKS_ON VALUES ('999887777', 10, 10.0);
INSERT INTO WORKS_ON VALUES ('987987987', 10, 35.0);
INSERT INTO WORKS_ON VALUES ('987987987', 30, 5.0);
INSERT INTO WORKS_ON VALUES ('987654321', 30, 20.0);
INSERT INTO WORKS_ON VALUES ('987654321', 20, 15.0);
INSERT INTO WORKS_ON VALUES ('888665555', 20, 10.0);
CREATE TABLE DEPENDENT
(
Essn CHAR(9),
Dependent_name VARCHAR(20),
Sex CHAR(1),
Bdate DATE,
Relationship VARCHAR(10),
PRIMARY KEY (Essn, Dependent_name),
FOREIGN KEY (Essn)
REFERENCES EMPLOYEE (Ssn)
);
INSERT INTO DEPENDENT VALUES ('333445555', 'Alice', 'F', DATE '1986-04-05', 'Daughter');
INSERT INTO DEPENDENT VALUES ('333445555', 'Theodore', 'M', DATE '1983-10-25', 'Son');
INSERT INTO DEPENDENT VALUES ('333445555', 'Joy', 'F', DATE '1958-05-03', 'Spouse');
INSERT INTO DEPENDENT VALUES ('987654321', 'Abner', 'M', DATE '1988-01-04', 'Son');
INSERT INTO DEPENDENT VALUES ('987654321', 'Jennifer', 'F', DATE '1988-01-04', 'Daughter');
INSERT INTO DEPENDENT VALUES ('123456789', 'John', 'M', DATE '1988-02-28', 'Son');
INSERT INTO DEPENDENT VALUES ('123456789', 'Alice', 'F', DATE '1988-12-30', 'Daughter');
INSERT INTO DEPENDENT VALUES ('123456789', 'Elizabeth', 'F', DATE '1967-05-05', 'Spouse');
INSERT INTO DEPENDENT VALUES ('453453453', 'Joyce', 'F', DATE '1990-04-05', 'Daughter');
-- display contents of tables
SELECT * FROM EMPLOYEE;
SELECT * FROM DEPARTMENT;
SELECT * FROM DEPT_LOCATIONS;
SELECT * FROM PROJECT;
SELECT * FROM WORKS_ON;
SELECT * FROM DEPENDENT;
CSC 355 Database Systems
Fall Quarter 2018
Assignment 3 (9/19)
Due Sunday, September 30th @ 11:59pm.
Reading: Sections 4.3, 4.4, and 5.1 of Elmasri & Navathe 6th Edition. Chapters 2, 4, and 6 of Price's SQL book may be helpful as a reference, though they contain more material than we have covered. To find the Price book, you can search for “Oracle 12c” at library.depaul.edu (then you can choose “view ebook”)
For both Parts 1 and 2, please submit your SQL code. Submitting your output is optional.
First, download the script file company.sql from the course website and run it in SQLDeveloper to construct the example database illustrated in Chapter 3 of Elmasri & Navathe (pp.70-75). Inspect the schemas and tables in SQLDeveloper to make sure you understand the structure of the database. (Note that some changes have been made to the database state -- it differs somewhat from the state that is illustrated in Figure 3.6 on p.72.)
In a separate .sql file (do not modify company.sql), write a script that contains the following SQL queries. Add a comment before each query in your script file to label the queries 1 through 16 (e.g., the comment '-- 1.' on a line before the first query, the comment '-- 2.' on a line before the second query, etc).
(7 pts) List the last names of all department managers who have no dependents.
(8 pts) Find the average salary for employees who have exactly 3 dependents.
(8 pts) For each department whose average salary is greater than $42,000, retrieve the department name and the number of employees in that department.
(8 pts) Retrieve the names of all employees who work in the department that has the employee with the lowest salary among all employees.
Explanation / Answer
If you have any doubts, please give me comment...
-- List the last names of all department managers who have no dependents.
SELECT LNAME, FNAME
FROM EMPLOYEE
WHERE EXISTS (SELECT *
FROM DEPARTMENT
WHERE SSN=MGRSSN) AND NOT EXISTS ( SELECT *
FROM DEPENDENT
WHERE SSN=ESSN)
-- Find the average salary for employees who have exactly 3 dependents.
SELECT AVG(SALARY)
FROM (SELECT ESSN, SALARY
FROM EMPLOYEE, DEPENDENTS
WHERE SSN = ESSN
GROUP BY SSN, SALARY
HAVING COUNT(*)>3) AS T;
-- For each department whose average salary is greater than $42,000, retrieve the department name and the number of employees in that department.
SELECT Dname, no_emp
FROM Department D, (
SELECT Dno, COUNT(*) no_emp
FROM EMPLOYEE
GROUP BY Dno
HAVING AVG(Salary) >42000
) T
WHERE D.Dnum = Dno;
-- Retrieve the names of all employees who work in the department that has the employee with the lowest salary among all employees.
SELECT Fname, Minit, Lname
FROM EMPLOYEE
WHERE Salary <=(SELECT MIN(Salary) FROM EMPLOYEES);
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.