onnect
Advanced Database Concepts
Final Project
Name:-Fatimah Abdullah Al-Jomaie
ID:-200800688
Section:-201
Due Date Dec 31, 2012
Individual work
Part 1 (Normalization)
Answer ALL of the following Questions. (4 marks)
1- A car registration database that keeps information only on currently registered cars
uses the following table. There is a globally unique serial number for each car but the
license plate number is only unique within each country. A car with a license plate in
a certain country must be registered in an office located in that country. Each
registration office serves one country.
Cars (make, model, year, serialNo, color, licensePlateNo, country, registrationOffice,
ownerId)
a) Write down a set of functional dependencies that covers all the dependencies
in this scenario.
Answer
FD1: serialNo
make, model, year, color, licensePlateNo, country,
registrationOffice, ownerId.
FD2: licensePlateNo, country
serialNo.
FD3: registrationOffice
country.
b) Determine all the candidate keys for the relation Cars (Minimum two keys
Answer
1- SerialNo
2- (licensePlateNo, country)
c) Select each of the candidate keys identified in turn to be the primary key. In
each case, and based on each primary key, what is the highest normal form to
which the relation Cars conforms? Explain your answer in
Answer
serialNo as the primary key, the relation Cars is in the 2NF(Second Normal Form)
because its primary key is simple, and all relations with simple non-composite
primary keys are in 2NF.
But the relation Cars is not in 3NF (Third normal form) because a nonprimary
attribute is transitively dependent on the primary key. That means
serialNo
registrationOffice and registrationOffice
country, and
nonprimary attribute (country) is transitively dependent on the primary key
(serialNo).
(licensePlateNo, country) as the primary key, the relation Cars is in 3NF because
licensePlateNo or country is a determinant of any other attribute. That means every
non-primary attribute is fully functionally dependent on the primary key. This
condition makes the relation in the 2NF.
And (licensePlateNo, country)
registrationOffice and registrationOffice
— Country, it means no nonprimary key attribute is transitively dependent on the
primary key. It means the relation Cars is in 3NF.
d) Normalize the relation Cars to the next higher normal form. Indicate to which
normal form(s) the resulting relations now conform? And why?
Answer
The first relation cars is in the 2NF to do it in 3NF it should be no nonprimary key
attribute is transitively dependent on the primary key.
serialNo
registrationOffice and registrationOffice
country, and this has a
transitive dependent it means that a nonprimary attribute (country) is transitively
dependent on the primary key (serialNo), then, the relation Cars is not in 3NF.
To do it in 3NF it should create another list to remove any transitive dependence
Like this
Cars (make, model, year, serialNo, color, licensePlateNo, country, ownerId)
RegistrationOffices (registrationOffice, country)
The second relation cars are already in the Third normal form also it can do it like
this
Cars (make, model, year, serialNo, color, licensePlateNo, country, ownerId)
RegistrationOffices (registrationOffice, country)
Part 2 (SQL Statements)
Answer ALL of the following Questions. (6 marks)
The Hospital relational headings
Team (TeamCode, TelephoneNo, StaffNo)
ConsistsOf (StaffNo, TeamCode)
Doctor (StaffNo, DoctorName, Position)
Specialist (StaffNo, Specialism)
Patient (PatientId, PatientName, Gender, Height, Weight, StaffNo, WardNo)
Ward (WardNo, WardName, NumberOfBeds)
Nurse (StaffNo, NurseName, WardNo)
Supervises (StaffNo, Supervisor)
Treatment (StaffNo, PatientID, StartDate, Reason)
Prescription (PrescriptionNo, Quantity, DailyDosage, StaffNo, PatientId, StartDate,
DrugCode)
Drug (DrugCode, DrugName, Type, Price)
Use the above ERD to answer the following questions:
a- Give a list of all drug types without repetition in alphabetical order
Answer
Select distinct type
from drug
order by type ASC
b- For each Nurse, give a list of his/her StaffNo and the name of his/her
supervisor
Answer
Select s.staff_no,n.nurse_name
from nurse u, supervises s, nurse n
where u.staff_no = s.staff_no and s.supervisor = n.staff_no
c- List the names of doctors who are not specialist
Answer
select doctor_name
from doctor
where not exists( select staff_no from Specialist
where Specialist.staff_no= doctor.staff_no)
Part 3 (PLSQL)
Answer ALL of the following Questions. (10 marks)
A University has two types of Employees, full time and part-time. Full-time
employees receive a monthly salary whereas part-time employees have an hourly rate.
Initially, the two types were mutually exclusive, meaning that no employee can be
both full-time and part-time at the same time. The company used the following tables:
Full-Time-Employee (EmpId, Name, Address, Salary)
Part-Time-Employee (EmpId, Name, Address, HourlyRate)
Later it was decided to allow full-time employees to work on a part time basis in their
spare time and so can be both full-time and part-time employees. To reduce
redundancy, the company decided to change the database schema to the following:
Employee (EmpId, Name, Address)
FT (EmpId, Salary)
PT (EmpId, Hourly-Rate)
Using cursors create a SQL procedure to convert the old schema into the new one. In
order to maintain old application programs, your procedure should also create two
views, called Full-Time-Employee and Part-Time-Employee with identical structure
of the old tables with the same names.
Answer
create table full_time_employee
(
emp_id
int not null,
emp_name
varchar(25),
address
varchar(25),
salary
int,
primary key (emp_id)
)
insert into full_time_employee(emp_id, emp_name, address, salary)
values (123,'Ahamed','Qatif', 500)
insert into full_time_employee(emp_id, emp_name, address, salary)
values (456,'Ali', 'Dammam', 800)
insert into full_time_employee(emp_id, emp_name, address, salary)
values (789, 'Mohammed','Jaddah', 600)
create table part_time_employee
(
emp_id
int Not null,
emp_name
varchar(25),
address
varchar(25),
hourlyRate
int,
primary key (emp_id)
)
insert into part_time_employee (emp_id, emp_name, address, hourlyRate)
values (234, 'Malik', 'Riyadh', 12)
insert into part_time_employee (emp_id, emp_name, address, hourlyRate)
values (567, 'Abdullah', 'Dammam', 15)
insert into part_time_employee (emp_id, emp_name, address, hourlyRate)
values (891, 'Jone', 'Qatif', 18)
create table Employee
(
emp_id
int not null,
emp_name varchar(25),
address
varchar(25),
primary key (emp_id)
)
create table FT
(
emp_id
int not null,
salary
int,
primary key (emp_id)
)
create table PT
(
emp_id
int not null,
hourlyRate int,
primary key (emp_id)
)
CREATE OR REPLACE PROCEDURE emp_fp AS
CURSOR f_t_emp_cur IS
SELECT ep_no, ep_name, ep_address, ep_sale from full_time_employee;
ep_no
f_t_emp.ep_no%type;
ep_name
f_t_emp.ep_name%type;
ep_address f_t_emp.ep_address%type;
ep_sale
f_t_emp.ep_sale%type;
CURSOR p_t_emp_cur IS
SELECT ep_no, ep_name, ep_address, ep_hR from part_time_employee;
ep_no
p_t_emp.ep_no%type;
ep_name
p_t_emp.ep_name%type;
ep_address p_t_emp.ep_address%type;
ep_hR
p_t_emp.ep_hR%type;
BEGIN
OPEN f_t_emp_cur;
OPEN p_t_emp_cur;
FETCH f_t_emp_cur INTO ep_no, ep_name, ep_address, ep_sale;
FETCH p_t_emp_cur INTO ep_no, ep_name, ep_address, ep_hR;
WHILE f_t_emp_cur %found LOOP
insert into Employee (emp_id, emp_name, address) values(ep_no, ep_name
,ep_address);
insert into FT (emp_id, salary) values (ep_no, ep_name, ep_address, ep_sale);
WHILE p_t_emp_cur %found LOOP
insert into Employee (emp_id, emp_name, address) values(ep_no, ep_name
,ep_address);
insert into PT (emp_id, hourlyRate) values (ep_no, ep_name, ep_address, ep_hR);
FETCH f_t_emp_cur INTO ep_no, ep_name, ep_address, ep_sale;
FETCH p_t_emp_cur INTO ep_no, ep_name, ep_address, ep_hR;
END LOOP;
CLOSE f_t_emp_cur ;
CLOSE p_t_emp_cur ;
END;