BI RONIN: SQL interview Questions and Answers

Click Here!OBIEE Interview Questions and answers
Click Here!OBIEE Interview Questions And Answers for 4 years Experienced
OBIEE Interview Questions And Answers for Experienced 2016 Click Here !

Practice 1 Solutions : Questions on SELECT statement

1. The following SELECT statement executes successfully:

Ans: True

SELECT last_name, job_id, salary AS Sal

FROM employees;

2. The following SELECT statement executes successfully:

Ans: True

SELECT *

FROM job_grades;

3. There are four coding errors in this statement. Can you identify them?

SELECT employee_id, last_name

sal x 12 ANNUAL SALARY

FROM employees;

– The EMPLOYEES table does not contain a column called sal. The column is called

SALARY.

– The multiplication operator is *, not x, as shown in line 2.

– The ANNUAL SALARY alias cannot include spaces. The alias should read

ANNUAL_SALARY or be enclosed in double quotation marks.

– A comma is missing after the column, LAST_NAME.

4. Show the structure of the DEPARTMENTS table. Select all data from the DEPARTMENTS table.

DESCRIBE departments

SELECT *

FROM departments;

5. Show the structure of the EMPLOYEES table. Create a query to display the last name, job code,

hire date, and employee number for each employee, with employee number appearing first.

Provide an alias STARTDATE for the HIRE_DATE column. Save your SQL statement to a file

named lab1_7.sql.

DESCRIBE employees

SELECT employee_id, last_name, job_id, hire_date StartDate

FROM employees;

6. Create a query to display unique job codes from the EMPLOYEES table.

SELECT DISTINCT job_id

FROM employees;

If you have time, complete the following exercises:

7. Copy the statement from lab1_7.sql into the iSQL*Plus Edit window. Name the column

headings Emp #, Employee, Job, and Hire Date, respectively. Run your query again.

SELECT employee_id "Emp #", last_name "Employee",

job_id "Job", hire_date "Hire Date"

FROM employees;

8. Display the last name concatenated with the job ID, separated by a comma and space, and name

the column Employee and Title.

SELECT last_name||’, ’||job_id "Employee and Title"

FROM employees;

If you want an extra challenge, complete the following exercise:

9. Create a query to display all the data from the EMPLOYEES table. Separate each column by a

comma. Name the column THE_OUTPUT.

SELECT employee_id || ’,’ || first_name || ’,’ || last_name

|| ’,’ || email || ’,’ || phone_number || ’,’|| job_id

|| ’,’ || manager_id || ’,’ || hire_date || ’,’ ||

salary || ’,’ || commission_pct || ’,’ || department_id THE_OUTPUT FROM employees;

* * * * * * * * * * * * * *

Practice 2 Solutions : Restricting and Sorting Data

1. Create a query to display the last name and salary of employees earning more than $12,000.

Place your SQL statement in a text file named lab2_1.sql. Run your query.

SELECT last_name, salary

FROM employees

WHERE salary > 12000;

2. Create a query to display the employee last name and department number for employee number

176.

SELECT last_name, department_id

FROM employees

WHERE employee_id = 176;

3. Modify lab2_1.sql to display the last name and salary for all employees whose salary is not

in the range of $5,000 and $12,000. Place your SQL statement in a text file named

lab2_3.sql.

SELECT last_name, salary

FROM employees

WHERE salary NOT BETWEEN 5000 AND 12000;

4. Display the employee last name, job ID, and start date of employees hired between February 20,

1998, and May 1, 1998. Order the query in ascending order by start date.

SELECT last_name, job_id, hire_date

FROM employees

WHERE hire_date BETWEEN ’20-Feb-1998’ AND ’01-May-1998’

ORDER BY hire_date;

Introduction to Oracle9i: SQL A-5

Practice 2 Solutions (continued)

5. Display the last name and department number of all employees in departments 20 and 50 in

alphabetical order by name.

SELECT last_name, department_id

FROM employees

WHERE department_id IN (20, 50)

ORDER BY last_name;

6. Modify lab2_3.sql to list the last name and salary of employees who earn between $5,000

and $12,000, and are in department 20 or 50. Label the columns Employee and Monthly

Salary, respectively. Resave lab2_3.sql as lab2_6.sql. Run the statement in

lab2_6.sql.

SELECT last_name "Employee", salary "Monthly Salary"

FROM employees

WHERE salary BETWEEN 5000 AND 12000

AND department_id IN (20, 50);

7. Display the last name and hire date of every employee who was hired in 1994.

SELECT last_name, hire_date

FROM employees

WHERE hire_date LIKE ’%94’;

8. Display the last name and job title of all employees who do not have a manager.

SELECT last_name, job_id

FROM employees

WHERE manager_id IS NULL;

9. Display the last name, salary, and commission for all employees who earn commissions. Sort

data in descending order of salary and commissions.

SELECT last_name, salary, commission_pct

FROM employees

WHERE commission_pct IS NOT NULL

ORDER BY salary DESC, commission_pct DESC;

Introduction to Oracle9i: SQL A-6

Practice 2 Solutions (continued)

If you have time, complete the following exercises.

10. Display the last names of all employees where the third letter of the name is an a.

SELECT last_name

FROM employees

WHERE last_name LIKE ’__a%’;

11. Display the last name of all employees who have an a and an e in their last name.

SELECT last_name

FROM employees

WHERE last_name LIKE ’%a%’

AND last_name LIKE ’%e%’;

If you want an extra challenge, complete the following exercises:

12. Display the last name, job, and salary for all employees whose job is sales representative or

stock clerk and whose salary is not equal to $2,500, $3,500, or $7,000.

SELECT last_name, job_id, salary

FROM employees

WHERE job_id IN (’SA_REP’, ’ST_CLERK’)

AND salary NOT IN (2500, 3500, 7000);

13. Modify lab2_6.sql to display the last name, salary, and commission for all employees

whose commission amount is 20%. Resave lab2_6.sql as lab2_13.sql. Rerun the

statement in lab2_13.sql.

SELECT last_name "Employee", salary "Monthly Salary",

commission_pct

FROM employees

WHERE commission_pct = .20;

* * * * * * * * * * * * * *

Practice 3 Solutions : Single-Row Functions

1. Write a query to display the current date. Label the column Date.

SELECT sysdate "Date"

FROM dual;

2. For each employee, display the employee number, last_name, salary, and salary increased by 15%

and expressed as a whole number. Label the column New Salary. Place your SQL statement in

a text file named lab3_2.sql.

SELECT employee_id, last_name, salary,

ROUND(salary * 1.15, 0) "New Salary"

FROM employees;

3. Run your query in the file lab3_2.sql.

SELECT employee_id, last_name, salary,

ROUND(salary * 1.15, 0) "New Salary"

FROM employees;

4. Modify your query lab3_2.sql to add a column that subtracts the old salary from

the new salary. Label the column Increase. Save the contents of the file as lab3_4.sql.

Run the revised query.

SELECT employee_id, last_name, salary,

ROUND(salary * 1.15, 0) "New Salary",

ROUND(salary * 1.15, 0) - salary "Increase"

FROM employees;

5. Write a query that displays the employee’s last names with the first letter capitalized and all other

letters lowercase and the length of the name for all employees whose name starts with J, A, or M.

Give each column an appropriate label. Sort the results by the employees’ last names.

SELECT INITCAP(last_name) "Name",

LENGTH(last_name) "Length"

FROM employees

WHERE last_name LIKE ’J%’

OR last_name LIKE ’M%’

OR last_name LIKE ’A%’

ORDER BY last_name;

6. For each employee, display the employee’s last name, and calculate the number of months

between today and the date the employee was hired. Label the column MONTHS_WORKED. Order

your results by the number of months employed. Round the number of months up to the closest

whole number.

Note: Your results will differ.

SELECT last_name, ROUND(MONTHS_BETWEEN

(SYSDATE, hire_date)) MONTHS_WORKED

FROM employees

ORDER BY MONTHS_BETWEEN(SYSDATE, hire_date);

7. Write a query that produces the following for each employee:

<employee last name> earns <salary> monthly but wants <3 times

salary>. Label the column Dream Salaries.

SELECT last_name || ’ earns ’

|| TO_CHAR(salary, ’fm$99,999.00’)

|| ’ monthly but wants ’

|| TO_CHAR(salary * 3, ’fm$99,999.00’)

|| ’.’ "Dream Salaries"

FROM employees;

If you have time, complete the following exercises:

8. Create a query to display the last name and salary for all employees. Format the salary to be 15

characters long, left-padded with $. Label the column SALARY.

SELECT last_name,

LPAD(salary, 15, ’$’) SALARY

FROM employees;

9. Display each employee’s last name, hire date, and salary review date, which is the first Monday

after six months of service. Label the column REVIEW. Format the dates to appear in the format

similar to “Monday, the Thirty-First of July, 2000.”

SELECT last_name, hire_date,

TO_CHAR(NEXT_DAY(ADD_MONTHS(hire_date, 6),’MONDAY’),

’fmDay, "the" Ddspth "of" Month, YYYY’) REVIEW

FROM employees;

10. Display the last name, hire date, and day of the week on which the employee started. Label

the column DAY. Order the results by the day of the week starting with Monday.

SELECT last_name, hire_date,

TO_CHAR(hire_date, ’DAY’) DAY

FROM employees

ORDER BY TO_CHAR(hire_date - 1, ’d’);

If you want an extra challenge, complete the following exercises:

11. Create a query that displays the employees’ last names and commission amounts. If an

employee does not earn commission, put “No Commission.” Label the column COMM.

SELECT last_name,

NVL(TO_CHAR(commission_pct), ’No Commission’) COMM

FROM employees;

12. Create a query that displays the employees’ last names and indicates the amounts of their

annual salaries with asterisks. Each asterisk signifies a thousand dollars. Sort the data in

descending order of salary. Label the column EMPLOYEES_AND_THEIR_SALARIES.

SELECT rpad(last_name, 8)||’ ’|| rpad(’ ’, salary/1000+1, ’*’)

EMPLOYEES_AND_THEIR_SALARIES

FROM employees

ORDER BY salary DESC;

13. Using the DECODE function, write a query that displays the grade of all employees based on the

value of the column JOB_ID, as per the following data:

JOB GRADE

AD_PRES A

ST_MAN B

IT_PROG C

SA_REP D

ST_CLERK E

None of the above 0

SELECT job_id, decode (job_id,

’ST_CLERK’, ’E’,

’SA_REP’, ’D’,

’IT_PROG’, ’C’,

’ST_MAN’, ’B’,

’AD_PRES’, ’A’,

’0’)GRADE

FROM employees;

14. Rewrite the statement in the preceding question using the CASE syntax.

SELECT job_id, CASE job_id

WHEN ’ST_CLERK’ THEN ’E’

WHEN ’SA_REP’ THEN ’D’

WHEN ’IT_PROG’ THEN ’C’

WHEN ’ST_MAN’ THEN ’B’

WHEN ’AD_PRES’ THEN ’A’

ELSE ’0’ END GRADE

FROM employees;

* * * * * * * * * * * * * *

Practice 4 Solutions : Displaying Data from Multiple Tables

1. Write a query to display the last name, department number, and department name for all

employees.

SELECT e.last_name, e.department_id, d.department_name

FROM employees e, departments d

WHERE e.department_id = d.department_id;

2. Create a unique listing of all jobs that are in department 80. Include the location of the

department in the output.

SELECT DISTINCT job_id, location_id

FROM employees, departments

WHERE employees.department_id = departments.department_id

AND employees.department_id = 80;

3. Write a query to display the employee last name, department name, location ID, and city of all

employees who earn a commission.

SELECT e.last_name, d.department_name, d.location_id, l.city

FROM employees e, departments d, locations l

WHERE e.department_id = d.department_id

AND

d.location_id = l.location_id

AND e.commission_pct IS NOT NULL;

4. Display the employee last name and department name for all employees who have an a

(lowercase) in their last names. Place your SQL statement in a text file named lab4_4.sql.

SELECT last_name, department_name

FROM employees, departments

WHERE employees.department_id = departments.department_id

AND last_name LIKE ’%a%’;

5. Write a query to display the last name, job, department number, and department name for all

employees who work in Toronto.

SELECT e.last_name, e.job_id, e.department_id,

d.department_name

FROM employees e JOIN departments d

ON (e.department_id = d.department_id)

JOIN locations l

ON (d.location_id = l.location_id)

WHERE LOWER(l.city) = ’toronto’;

6. Display the employee last name and employee number along with their manager’s last name and

manager number. Label the columns Employee, Emp#, Manager, and Mgr#, respectively.

Place your SQL statement in a text file named lab4_6.sql.

SELECT w.last_name "Employee", w.employee_id "EMP#",

m.last_name "Manager", m.employee_id "Mgr#"

FROM employees w join employees m

ON (w.manager_id = m.employee_id);

7. Modify lab4_6.sql to display all employees including King, who has no manager.

Place your SQL statement in a text file named lab4_7.sql. Run the query in lab4_7.sql

SELECT w.last_name "Employee", w.employee_id "EMP#",

m.last_name "Manager", m.employee_id "Mgr#"

FROM employees w

LEFT OUTER JOIN employees m

ON (w.manager_id = m.employee_id);

If you have time, complete the following exercises.

8. Create a query that displays employee last names, department numbers, and all the

employees who work in the same department as a given employee. Give each column an

appropriate label.

SELECT e.department_id department, e.last_name employee,

c.last_name colleague

FROM employees e JOIN employees c

ON (e.department_id = c.department_id)

WHERE e.employee_id <> c.employee_id

ORDER BY e.department_id, e.last_name, c.last_name;

9. Show the structure of the JOB_GRADES table. Create a query that displays the name, job,

department name, salary, and grade for all employees.

DESC JOB_GRADES

SELECT e.last_name, e.job_id, d.department_name,

e.salary, j.grade_level

FROM employees e, departments d, job_grades j

WHERE e.department_id = d.department_id

AND e.salary BETWEEN j.lowest_sal AND j.highest_sal;

-- OR

SELECT e.last_name, e.job_id, d.department_name,

e.salary, j.grade_level

FROM employees e JOIN departments d

ON (e.department_id = d.department_id)

JOIN job_grades j

ON (e.salary BETWEEN j.lowest_sal AND j.highest_sal);

If you want an extra challenge, complete the following exercises:

10. Create a query to display the name and hire date of any employee hired after employee Davies.

SELECT e.last_name, e.hire_date

FROM employees e, employees davies

WHERE davies.last_name = ’Davies’

AND davies.hire_date < e.hire_date

-- OR

SELECT e.last_name, e.hire_date

FROM employees e JOIN employees davies

ON (davies.last_name = ’Davies’)

WHERE davies.hire_date < e.hire_date;

11. Display the names and hire dates for all employees who were hired before their managers, along

with their manager’s names and hire dates. Label the columns Employee, Emp

Hired, Manager, and Mgr Hired, respectively.

SELECT w.last_name, w.hire_date, m.last_name, m.hire_date

FROM employees w, employees m

WHERE w.manager_id = m.employee_id

AND w.hire_date < m.hire_date;

-- OR

SELECT w.last_name, w.hire_date, m.last_name, m.hire_date

FROM employees w JOIN employees m

ON (w.manager_id = m.employee_id)

WHERE w.hire_date < m.hire_date;

* * * * * * * * * * * * * *

Click Here to go OBIA 7.9.6.3/7.9.6.4 Implementation and Installation

Practice 5 Solutions : Aggregating Data Using Group Functions

Determine the validity of the following three statements. Circle either True or False.

1. Group functions work across many rows to produce one result.

True

2. Group functions include nulls in calculations.

False. Group functions ignore null values. If you want to include null values, use the NVL

function.

3. The WHERE clause restricts rows prior to inclusion in a group calculation.

True

4. Display the highest, lowest, sum, and average salary of all employees. Label the columns

Maximum, Minimum, Sum, and Average, respectively. Round your results to the nearest whole

number. Place your SQL statement in a text file named lab5_6.sql.

SELECT ROUND(MAX(salary),0) "Maximum",

ROUND(MIN(salary),0) "Minimum",

ROUND(SUM(salary),0) "Sum",

ROUND(AVG(salary),0) "Average"

FROM employees;

5. Modify the query in lab5_4.sql to display the minimum, maximum, sum, and average salary for

each job type. Resave lab5_6.sql to lab5_4.sql. Run the statement in lab5_5.sql.

SELECT job_id, ROUND(MAX(salary),0) "Maximum",

ROUND(MIN(salary),0) "Minimum",

ROUND(SUM(salary),0) "Sum",

ROUND(AVG(salary),0) "Average"

FROM employees

GROUP BY job_id;

6. Write a query to display the number of people with the same job.

SELECT job_id, COUNT(*)

FROM employees

GROUP BY job_id;

7. Determine the number of managers without listing them. Label the column Number of

Managers. Hint: Use the MANAGER_ID column to determine the number of managers.

SELECT COUNT(DISTINCT manager_id) "Number of Managers"

FROM employees;

8. Write a query that displays the difference between the highest and lowest salaries. Label the

column DIFFERENCE.

SELECT MAX(salary) - MIN(salary) DIFFERENCE

FROM employees;

If you have time, complete the following exercises.

9. Display the manager number and the salary of the lowest paid employee for that manager.

Exclude anyone whose manager is not known. Exclude any groups where the minimum

salary is $6,000 or less. Sort the output in descending order of salary.

SELECT manager_id, MIN(salary)

FROM employees

WHERE manager_id IS NOT NULL

GROUP BY manager_id

HAVING MIN(salary) > 6000

ORDER BY MIN(salary) DESC;

10. Write a query to display each department’s name, location, number of employees, and the

average salary for all employees in that department. Label the columns Name, Location,

Number of People, and Salary, respectively. Round the average salary to two decimal

places.

SELECT d.department_name "Name", d.location_id "Location",

COUNT(*) "Number of People",

ROUND(AVG(salary),2) "Salary"

FROM employees e, departments d

WHERE e.department_id = d.department_id

GROUP BY d.department_name, d.location_id;

If you want an extra challenge, complete the following exercises:

11. Create a query that will display the total number of employees and, of that total, the number of

employees hired in 1995, 1996, 1997, and 1998. Create appropriate column headings.

SELECT COUNT(*) total,

SUM(DECODE(TO_CHAR(hire_date, ’YYYY’),1995,1,0))"1995",

SUM(DECODE(TO_CHAR(hire_date, ’YYYY’),1996,1,0))"1996",

SUM(DECODE(TO_CHAR(hire_date, ’YYYY’),1997,1,0))"1997",

SUM(DECODE(TO_CHAR(hire_date, ’YYYY’),1998,1,0))"1998"

FROM employees;

12. Create a matrix query to display the job, the salary for that job based on department number, and

the total salary for that job, for departments 20, 50, 80, and 90, giving each column an appropriate

heading.

SELECT job_id "Job",

SUM(DECODE(department_id , 20, salary)) "Dept 20",

SUM(DECODE(department_id , 50, salary)) "Dept 50",

SUM(DECODE(department_id , 80, salary)) "Dept 80",

SUM(DECODE(department_id , 90, salary)) "Dept 90",

SUM(salary) "Total"

FROM employees

GROUP BY job_id;

Pages

SQL interview Questions and Answers