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COEN 178 Intro to Database Systems Lab 4 solution

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In this lab, you will create two new tables and load them with data from staff_2010 table that you have created in the earlier labs.
The following is the plan of what you will do today.
a) Create two new tables with foreign key constraints
b) Load the new tables with a subset of data from staff_2010 table (we will avoid the tedium of having to load them by hand with hardcoded values)
c) Test the enforcement of foreign key constraint.
d) Write a PLSQL procedure and a function to insert values for salary using a random number generator (this will come in handy if you want to generate “fake values” for demo purposes).

Part 1
Question 1 (10 pts)
Create a table, AlphaCoEmp as follows.
CREATE TABLE AlphaCoEmp(Name VARCHAR(25) Primary Key, Title VARCHAR(20) DEFAULT NULL, Salary Number(10,2) DEFAULT 0);
Inserting values:
You will load the table, AlphaCoEmp, with employee last names from Staff_2010.
Now run the query below to insert the last names from table, Staff_2010.

INSERT INTO AlphaCoEmp (name) SELECT last from Staff_2010;
Did you see any errors? The errors are because of the primary key constraint which is being violated by some duplicate last names in the Staff_2010 table. How do you fix this query so that it loads unique last names from Staff_2010?
(Hint: Think distinct, modify and rerun the query)

Yes, there are errors since Staff_2010 has multiple last names that are the same. To fix this, we have you use the “DISTINCT” key word in order to prevent this from happening. Thus, our query is now “INSERT INTO AlphaCoEmp (name) SELECT DISTINCT last from Staff_2010”.

Do a Select * from AlphaCoEmp and check the results displayed.

Question 2 (10 pts)
Let us define a table called Emp_Work as follows:
Create Table Emp_Work(name VARCHAR(25) Primary Key, Project VARCHAR(20) default NULL,
Constraint FK_AlphaCo
Foreign Key (name) REFERENCES AlphaCoEmp(name));
Note the foreign key constraint defined in this table with a name.
Let us insert employee names that start with A or G or S from the AlphaCoEmp table into Emp_Work table. We will use regular expressions to grab names that start with A or G or S. (You are free to use your choice for letters).
insert into Emp_Work(name) Select Name from AlphaCoEmp where
REGEXP_LIKE(name,'(^[ags])’,’i’);

Study the Regex that describes a pattern where the string starts with (^) a [ags] (a or g or s).
“^” denotes the start of a string ($ would signify the end of a string) and the “ags” shows that we’re looking for a word that begins with A, G, or S. The “i” means that it is not case-sensitive.
Now, run a query to display the names that start with a (or A) or g(or G) or s (or S) from Emp_Work table.
Select Name from AlphaCoEmp where REGEXP_LIKE(name,'(^[ags])’,’i’);
Pick a name from the displayed list and delete it from the table, AlphaCoEmp.
Example:
Delete from AlphaCoEmp
Where name=’Smith’;
a) What is the ‘i’ for?
The “i” denotes case-sensitivity. Or in other words, the characters that we are looking for are not case-sensitive.
b) Did your deletion work? Explain why it did not work.
The deletion did not succeed since it violated the child record integrity. In other words, the restraint prevented us from deleting Smith.

Question 3 (10 pts)
Let us change the Emp_Work table definition such that when a primary key that has a reference from a foreign key is deleted, it should delete the row with that foreign key as well.
We will use the alter Table command and define the constraint, on delete cascade.
Remember how you defined this table earlier.
Create Table Emp_Work(name VARCHAR(20) Primary Key, Project VARCHAR(20) default NULL,
Constraint FK_AlphaCo
Foreign Key (name) REFERENCES AlphaCoEmp(Name));

Now, we need to drop this constraint FK_AlphaCo and add the constraint with delete cascade. We do dropping and adding the new constraint with Alter Table statement.
Run this statement to drop the constraint.
Alter table Emp_Work
drop constraint FK_AlphaCo;

Add the new constraint as follows:
Alter table Emp_Work
add constraint FK_AlphaCo
FOREIGN KEY (name)
references AlphaCoEmp(name)
on delete cascade;

Assuming your table is successfully altered, try to delete the name you tried earlier, from AlphaCoEmp table.
Example:
DELETE from AlphaCoEmp
Where name=’Smith’;
a) Did you succeed this time?
Yes, the deletion was successful this time.
b) Check if the name “Smith” is in the Emp_Work table. It should have been deleted automatically if our constraint worked. Did it work?
It was successful. The query returned “no rows selected”.

Part 2
In this part, we will write a PLSQL procedures and functions, cursors and use a random number generator.
Question 4 (5 pts)
We will write a simple procedure to display a message passed as a parameter.
Create or Replace Procedure DisplayMessage(message in VARCHAR)
As
BEGIN
DBMS_OUTPUT.put_line(‘Hello ‘||message);

END;
/
Show Errors;

Create the procedure above. You can do it in one of two ways:
a) Copy and paste the code at SQL prompt.
b) Paste it into a text file (test.sql, for example) and give command start test.sql (must include the path name of the file) at the SQL prompt.
——————————————————————————–
Now, at the SQL prompt, call the procedure with command,
exec DisplayMessage(‘include a message’).

Question 5 (10 pts)
In this exercise, you will learn to write a procedure to insert salaries into AlphaCoEmp table. The salaries will be generated using a random number generator.

a) Let us try to generate a random integer in the range 10 to 100 in SQL.
Type the following query at SQL prompt.
Select ROUND(DBMS_RANDOM.value (10,100)) from DUAL;

b) You will write a procedure to insert salaries (random values in a range) for all employees in AlphaCoEmp table. Remember we have loaded a default value of 0 for salary, when we created the table.
Do a Select * from AlphaCoEmp table and check that the salaries are currently 0.
c) The procedure does the following:
• Since we want to set the salaries of all employees in the table, we need to write the query which fetches all the employees.
• Therefore, we need to use a cursor to hold and point to the rows fetched.
• For each row fetched (use a loop), we will generate a random number for a salary within the range (passed in as parameters) and update the table.

Create or Replace Procedure setSalaries(low in INTEGER, high in INTEGER)
As
Cursor Emp_cur IS
Select * from AlphaCoEmp;

— Local variables
l_emprec Emp_cur%rowtype;
l_salary AlphaCoEmp.salary%type;

BEGIN
for l_emprec IN Emp_cur
loop
l_salary := ROUND(dbms_random.value(low,high));

update AlphaCoEmp
set salary = l_salary
where name = l_emprec.name;

END LOOP;
commit;
END;
/
show errors;

Create this procedure (you can copy and paste it at the prompt in console window)
Once you get a clean compilation of the procedure, execute it with salary values of your choice.

Example:

exec setSalaries (50000,100000);

Now, do a select * on AlphaCoEmp table. You should be able to see the salaries (where previously there was a 0 for this column).

Question 6 (10 pts)
Now that you have salaries set in AlphaCoEmp table, write and run a SQL query to display the names of people (in AlphaCoEmp table) with salaries between a low and a high value of your choice (between 80000 and 100000, for example).

SELECT Name, salary FROM AlphaCoEmp WHERE salary BETWEEN 80000 and 100000;

Question 7 (20 pts)
In this exercise, you will complete a PLSQL function that returns the salary where name of the employee is passed in as a parameter. Remember, functions return something to the calling code whereas procedures are like void functions.
Complete the function below.
Create or Replace FUNCTION getEmpSalary(p_name in VARCHAR)

Return NUMBER IS

/* Define the local variables you need.
You need a variable to hold the salary returned */

l_salary ; — Give the data type

BEGIN
/* Select the salary into l_salary
Where employee name is p_name */

Select salary
from AlphaCoEmp

return l_salary;
END;
/
show errors;

Test your function using the query below. For the parameter, give an employee name of your choice that is in the AlphaCoEmp table.
Select getEmpSalary(‘’)
From Dual;
Did your function work correctly?
Yes, the function worked correctly by returning the salary for a person with the given name in the parameter. In my example, Gibbs had a salary of 97149.