Oracle sql online test for Certification

Oracle SQL is the widely used in various Products. Having a knowledge in Oracle SQL is plus point.

This Oracle sql online test check the understanding on various topics like Oracle table, Oracle view ,constraints, type of index, sequences, subqueries , oracle joins ,materialized view, primary key constraints , single row function ,sql queries and many more

The below test is randomized sql test and select 15 questions from the set of many questions,so each time you try to attempt it, you will get different questions in different order and option would also be in different order, So it will be fun practicing with this tool. You can try as many times as you want till you feel comfortable about it

Instructions for online test

1.This quiz consists of 15 questions .It is recommended to complete the test in 30 min
2. It consists both single choice and more than one choice questions to test your knowledge
3. You can press the submit button at the end of your test and get the final score
4.You will also get the know the correct choice after you finished the test.

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1. The emp table contains these columns:
Last_name Varchar2 (25)
First_name Varchar2 (25)
Salary Number7, 2
You need to display the names of employees on more than an average salary of all
employees. Evaluate the SQL statement.
SELECT, LAST_NAME, FIRST_NAME from emp where salary< avg(salary);

Which change should you make to achieve the desired results?

Change the function in the Where clause.

Use a sub query in the where clause to compare the average salary value.

Move the function to the select clause and add a group clause.

Move the function to the select clause and add a group by clause and a having clause.

Question 1 of 15

2. Examine the structure of student_masters table:
Name Null Type
STU ID NOT NULL NUMBER(3)
NAME VARCHER2(25)
ADDRESS VARCHER2(50)
GRADUATION DATE
Currently the table is empty. You have decided that null values should not be allowed for the NAME column. Which statement restricts NULL values from being entered into column?

ALTER TABLE student_masters MODIFY(name varcher2(25) NOT NULL);

ALTER TABLE student_masters MODIFY CONSTRAINT name(NOT NULL);

ALTER TABLE student_masters ADD CONSTRAINT NOT NULL (name);

ALTER TABLE student_masters ADD CONSTRAINT name(NOT NULL);

Question 2 of 15

3. The emp table contains these columns:
FIRST-NAME VARCHER2(25)
COMISSION NUMBER(3,2)
Evaluate this SQL statement
SELECT first-name,commission FROM emp WHERE commission= (SELECT comission FROM emp WHERE UPPER(first-name)= ‘smith’)
Which statement will cause this statement to fail?

There is no employee with the first name smith.

The first name values in the data base are in the lower case.

smith has a null commission resolution.

smith has a zero commission resolution.

Question 3 of 15

4. You need to create a report to display the ship date and order totals of your order table. If the order has not been shipped your report must display not shipped. If the total is not available your report must say not available. In the order table the ship date column has a data type of date the total column has a data type of number. Which statement do you use to create this report?

Select order, shipdate “Not shipped”,total “Not available” FROM order;

Select order, TO_CHAR (shipdate, ‘Not shipped’) TO_CHAR (total, ‘Not available’) FROM order;

Select order, NVL (shipdate ‘Not shipped’),NVL (total, “Not available”) FROM order;

Select order, NVL (TO_CHAR (shipdate), ‘Not shipped’),NVL (TO_CHAR (total), ‘Not available’) FROM order;

Question 4 of 15

5. What should be done after each fetch statement in a PL/SQL block?

Test for rows using the cursor attribute.

Initialize the loop.

Open the cursor.

Close the cursor.

Question 5 of 15

6. Examine the following command:

CREATE TABLE pay_master
(employee-id Number(9)
CONSTRAINT pay-pk PRIMARY KEY,
2000-pay NUMBER(8,2)
manager-name VARCHAR2(25)
CONSTRAINT mgr-name-nn. NOT NULL,
pay-96 NUMBER(8,2));

Which two lines of this command will return an error?

Fifth

First

Second

Fourth

Sixth

Third

Question 6 of 15

7. You create the sal table with this command

CREATE TABLE sal (purchase-no NUMBER(9)
CONSTRAINT sale-purchase-no-pk PRIMARY KEY,
costumer-id NUMBER(9) CONSTRAINT sale-customer-id-nl NOT NULL);

Which index or indexes are created for this table?

An index is created for purchase_no column.

No indexes are created for this table.

An index is created for each column.

An index is created for the customer_no column.

Question 7 of 15

8. You have been instructed to create a report that shows different jobs in each division within your company. No duplicate roles can be displayed. Which of the following SELECT statements should be used?

SELECT no duplicate divisionno, job FROM worker;

SELECT distinct divisionno, job FROM worker;

CREATE report DISPLAY divisionno, job

SELECT divisionno, job FROM worker;

Question 8 of 15

9. You query a database with the following command:

SELECT dept_no, AVG (MONTHS_BETWEEN (SYSDATE, hire_date)) FROM worker_company
WHERE AVG (MONTHS_BETWEEN (SYSDATE, hire_date)) > 60
GROUP BY dept_no
ORDER BY AVG (MONTHS_BETWEEN (SYSDATE, hire_date) )

Where does the statement cause an error?

A group function cannot contain a single row function.

A WHERE clause cannot be used to restrict groups.

A SELECT clause cannot contain a group function.

An ORDER BY clause cannot contain a group function.

Question 9 of 15

10. How would you add a foreign key constraint on the dept_no column in the EMP table. Referring to the ID column in the DEPT table?

Use the ALTER TABLE command with the MODIFY clause on the EMP table.

Use the ALTER TABLE command with the ADD clause in the DEPT table.

Use the ALTER TABLE command with the ADD clause on the EMP table.

This task cannot be accomplished.

Use the ALTER TABLE command with the MODIFY clause on the DEPT table.

Question 10 of 15

11. Examine the WORKER table below:

NAME NULL TYPE
WORKER ID NOT NULL NUMBER(3)
NAME NOT NULL VARCHAR2(25)
PHONE NOT NULL VARCHAR2(9)
ADDRESS VARCHAR2(50)
POSITION UPDATE

There are hundreds of records in the EMPLOYEE TABLE. You need to modify the phone column to hold only number values. Which statement will modify the data type appropriately?

The data type of a column cannot be modified if there is data in the column.

ALTER WORKER table MODIFY COLUMN phone NUMBER(9)

ALTER TABLE worker MODIFY phone NUMBER(9)

A VARCHAR2 data type cannot be changed to a NUMBER data type for a column.

Question 11 of 15

12. You must permanently remove all data from the sales_data table, but will need the table structure in the future. What single command should be issued?

DROP TABLE sales_data

TRUNCATE TABLE sales_data KEEP STRUCTURE;

TRUNCATE TABLE sales_data

DELETE FROM sales_data

Question 12 of 15

13. You need to analyze how long your orders to be shipped from the date that the order is placed. To do this you must create a report that displays the customer number, date order, date shipped and the number of months in whole numbers from the time the order is placed to the time the order is shipped. Which statement produces the
required results?

SELECT custid, orderate, shipdate, ROUND(DAYS_BETWEEN(shipdate,orderate))/30 FROM order;

SELECT custid, orderate, shipdate,ROUND OFF (shipdate-orderate) “Time Taken” FROM order;

SELECT custid, orderate, shipdate, MONTHS_BETWEEN (shipdate,orderate) “Time Taken” FROM order;

SELECT custid, orderate, shipdate, ROUND(MONTHS_BETWEEN(shipdate,orderate)) “Time Taken”FROM order;

Question 13 of 15

14. You need to store currency data and you know that data will always have two digits to the right of the decimal points. However the number of digits to the left of the decimal place will vary greatly. Which data type would be most appropriate to store the data?

NUMBER

CHAR

NUMBER(T)

VARCHAR

Question 14 of 15

15. You want of display the details of all employees whose last names is john. But you are not sure in which case last names are stored. Which statement will list all the employees whose last name is john?

Select last name, first name. FROM emp WHERE LOWER (last name)= ‘john’;

Select last name, first nam FROM emp WHERE last name= ‘john’;

Select last name, first name FROM emp WHERE UPPER (last name)= ‘john’;

Select last name, first name FROM emp WHERE last name=UPPER (‘john’);

Question 15 of 15

I hope you like this Oracle sql online test and it helps in your examination preparation. Best of Luck for the Oracle certification Examination. Please do provide feedback on it. We will also be adding more quizes and questions in future.

Happy Learning!!!!

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