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Oracle Database SQL Sample Questions (Q353-Q358):

NEW QUESTION # 353
View the exhibit and examine the structures of the EMPLOYEES and DEPARTMENTS tables.
EMPLOYEES
NameNull?Type
---------------------- -------------
EMPLOYEE_IDNOT NULLNUMBER(6)
FIRST_NAMEVARCHAR2(20)
LAST_NAMENOT NULLVARCHAR2(25)
HIRE_DATENOT NULLDATE
JOB_IDNOT NULLVARCHAR2(10)
SALARYNUMBER(10,2)
COMMISSIONNUMBER(6,2)
MANAGER_IDNUMBER(6)
DEPARTMENT_IDNUMBER(4)
DEPARTMENTS
NameNull?Type
---------------------- -------------
DEPARTMENT_IDNOT NULLNUMBER(4)
DEPARTMENT_NAMENOT NULLVARCHAR2(30)
MANAGER_IDNUMBER(6)
LOCATION_IDNUMBER(4)
You want to update EMPLOYEES table as follows:
You issue the following command:
SQL> UPDATE employees
SET department_id
(SELECT department_id
FROM departments
WHERE location_id = 2100),
(salary, commission)
(SELECT 1.1*AVG(salary), 1.5*AVG(commission)
FROM employees, departments
WHERE departments.location_id IN(2900, 2700, 2100))
WHERE department_id IN
(SELECT department_id
FROM departments
WHERE location_id = 2900
OR location_id = 2700;
What is outcome?

A. It generates an error because a subquery cannot have a join condition in a UPDATE statement.
B. It executes successfully and gives the desired update
C. It generates an error because multiple columns (SALARY, COMMISSION) cannot be specified together in an UPDATE statement.
D. It executes successfully but does not give the desired update

Answer: D

NEW QUESTION # 354
View the Exhibit and examine the structure of the PRODUCT table.

Which two tasks would require subqueries? (Choose two.)

A. display the number of products whose list prices are more than the average list price
B. display the total number of products supplied by supplier 102 and have product status as 'OBSOLETE'
C. display the minimum list price for each product status
D. display all suppliers whose list price is more than 1000
E. display all products whose minimum list price is more than the average list price of products having the status 'orderable'

Answer: A,E

NEW QUESTION # 355
Evaluate these commands which execute successfully:

Which two statements are true about the ORD_ITEMStable and the ORD_SEQsequence? (Choose two.)

A. If sequence ORD_SEQis dropped then the default value for column ORD_NOwill be NULL for rows inserted into ORD_ITEMS.
B. Any user inserting rows into table ORD_ITEMSmust have been granted access to sequence ORD_SEQ.
C. Sequence ORD_SEQis guaranteed not to generate duplicate numbers.
D. Sequence ORD_SEQcycles back to 1 after every 5000 numbers and can cycle 20 times.
E. Column ORD_NOgets the next number from sequence ORD_SEQwhenever a row is inserted into ORD_ITEMSand no explicit value is given for ORD_NO.

Answer: B,E

NEW QUESTION # 356
Examine the data in the EMPLOYEES table:

Which statement will compute the total annual compensation for each employee?

A. SELECT last _ name, (monthly _ salary * 12) + (monthly_ commission _ pct * 12) AS FROM employees:
B. select last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 *monthly_ commission_ pct) AS annual_ camp FROM employees
C. SELECT last _ NAME (monthly_ salary + monthly _commission _ pct) * 12 AS annual_ comp FROM employees;
D. SELECT last _ name, (monthly_ salary * 12) + (monthly_ salary * 12 * NVL (monthly_ commission _pct, 0)) AS annual _comp

Answer: D

Explanation:
The correct way to compute the total annual compensation, which includes the monthly salary and the monthly commission (if any), is:
Option C: SELECT last_name, (monthly_salary * 12) + (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) AS annual_comp FROM employees; This statement takes the monthly salary and multiplies it by 12 to get the annual salary, and then adds the annual commission which is the monthly salary multiplied by the commission percentage (if any, else 0) and then by 12.
Options A, B, and D are incorrect because:
Option A: Does not handle the case where the commission percentage is NULL which would result in NULL for the entire expression when added to the monthly salary.
Option B: Does not consider that the commission percentage might be NULL which could lead to incorrect calculations (or NULL values if commission is NULL).
Option D: Incorrectly adds the monthly commission percentage directly to the annual salary without considering that the percentage needs to be applied to the salary.

NEW QUESTION # 357
View the Exhibit and examine the structure of the EMP table which is not partitioned and not an index-organized table.

Evaluate this SQL statement:
ALTER TABLE emp
DROP COLUMN first_name;
Which two statements are true? (Choose two.)

A. The FIRST_NAME column can be dropped even if it is part of a composite PRIMARY KEY provided the CASCADE option is added to the SQL statement.
B. The FIRST_NAME column would be dropped provided at least one column remains in the table.
C. The FIRST_NAME column would be dropped provided it does not contain any data.
D. The drop of the FIRST_NAME column can be rolled back provided the SET UNUSED option is added to the SQL statement.

Answer: B,D

NEW QUESTION # 358
......

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