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Oracle Database SQL Sample Questions (Q169-Q174):

NEW QUESTION # 169
The CUSTOMERS table has a CUST_CREDT_LIMIT column of data type number.
Which two queries execute successfully?

* A. SLECT NVL(TO_CHAR(cust_credit_limit * .15),'Not available') from customers;
* B. SELECT NVL2(cust_credit_limit,TO_CHAR(cust_credit_limit * .15),'NOT Available') FROM customers;
* C. SELECT NVL(cust_credit_limit * .15,'Not Available') FROM customers;
* D. SELECT TO_CHAR(NVL(cust_credit_limit * .15,'Not Available')) FROM customers;
* E. SELECT NVL2(cust_credit_limit * .15,'Not Available') FROM customers;
Answer: A,B

NEW QUESTION # 170
View the exhibit and examine the structure of the PROMOTIONS table.
You have to generate a report that displays the promo name and start date for all promos that started after the last promo in the 'INTERNET' category.
Which query would give you the required output?

* A. SELECT promo_name, promo_begin_date FROM promotionsWHERE
promo_begin_date> ALL (SELECT MAX (promo_begin_date)FROM promotions)
ANDpromo_category= 'INTERNET';
* B. SELECT promo_name, promo_begin_date FROM promotionsWHERE
promo_begin_date IN (SELECT promo_begin_dateFROM promotionsWHERE
promo_category= 'INTERNET');
* C. SELECT promo_name, promo_begin_date FROM promotionsWHERE
promo_begin_date > ALL (SELECT promo_begin_dateFROM promotionsWHERE
promo_category = 'INTERNET');
* D. SELECT promo_name, promo_begin_date FROM promotionsWHERE
promo_begin_date> ANY (SELECT promo_begin_dateFROM promotionsWHERE
promo_category= 'INTERNET');
Answer: C

NEW QUESTION # 171
Examine the structure of the EMPLOYEEStable.

You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.
Which query would retrieve the required result?
SELECT last_name, hire_date, salary

* A. FROM employees
WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id
'100')
UNION
SELECT last_name, hire_date, salary
FROM employees
WHERE hire_date > SYSDATE-180
INTERSECT
SELECT last_name, hire_date, salary
FROM employees
WHERE salary > 10000;
(SELECT last_name, hire_date, salary
* B. FROM employees
WHERE salary > 10000
UNION ALL
SELECT last_name, hire_date, salary
FROM employees
WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100) INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE-180; SELECT last_name, hire_date, salary
* C. FROM employees
WHERE salary > 10000
UNION ALL
SELECT last_name, hire_date, salary
FROM employees
WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id
100))
UNION
SELECT last_name, hire_date, salary
FROM employees
WHERE hire_date > SYSDATE-180;
* D. FROM employees
WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100) UNION ALL (SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE-180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000); SELECT last_name, hire_date, salary
Answer: A

NEW QUESTION # 172
Examine the structure of the employees table.

There is a parent/child relationship betweenEMPLOYEE_IDandMANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager asthe employee whoseEMPLOYEE_ID123.
Which query provides the correct output? A)



C)

D)


* A. Option B
* B. Option D
* C. Option A
* D. Option C
Answer: A

NEW QUESTION # 173
Examine the structure of the EMPLOYEEStable.
Name Null? Type
- ---------------- ----- ------ ------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
EMAIL NOT NULL VARCHAR2(25)
PHONE NUMBER VARCHAR2(20)
HIRE_DATE NOT NULL DATE
JOB_ID NOT NULL VARCHAR2(10)
SALARY NUMBER(8,2)
COMMISSION_PCT NUMBER(2,2)
MANAGER_ID NUMBER(6)
DEPARTMENT_ID NUMBER(4)
There is a parent/child relationship between EMPLOYEE_IDand MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_IDis 123.
Which query provides the correct output?

* A. SELECT e.last_name, m.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123;
* B. SELECT e.last_name, m.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123;
* C. SELECT e.last_name, e.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.employee_id)
WHERE e.employee_id = 123;
* D. SELECT m.last_name, e.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123;
Answer: B

NEW QUESTION # 174
......

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