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Oracle Database SQL Sample Questions (Q195-Q200):

NEW QUESTION # 195
View the Exhibit and examine the data in the PRODUCT_INFORMATION table.

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

A. displaying the number of products whose list prices are more than the average list price
B. displaying the minimum list price for each product status
C. displaying all supplier IDs whose average list price is more than 500
D. displaying all the products whose minimum list prices are more than average list price of products having the status orderable
E. displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE

Answer: A,D

NEW QUESTION # 196
View the exhibit and examine the structure in ORDERS and ORDER_ITEMS tables.

You need to create a view that displays the ORDER_ID, ORDER_DATE, and the total number of items in each order.
Which CREATE VIEW statement would create the views successfully?

A. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id) ||"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_dateWHITH CHECK OPTION;
B. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
C. CREATE OR REPLACE VIEW ord_vu (order_id, order_date)AS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)"NO OF ITEMS"FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;
D. CREATE OR REPLACE VIEW ord_vuAS SELECT o.order_id, o.order_date, COUNT (i.line_item_id)FROM orders o JOIN order_items iON (o.order_id = i.order_id)GROUP BY o.order_id, o.order_date;

Answer: B

NEW QUESTION # 197
Examine the data in the CUST NAME column of the CUSTOMERS table:
CUST_NAME
------------------------------
Renske Ladwig
Jason Mallin
Samuel McCain
Allan MCEwen
Irene Mikkilineni
Julia Nayer
You want to display the CUST_NAME values where the last name starts with Mc or MC. Which two WHERE clauses give the required result?

A. WHERE INITCAP (SUBSTR(cust_name, INSTR(cust_name,'') +1)) IN ('MC%','Mc%)
B. WHERE SUBSTR(cust_name,INSTR(cust_name,'') +1) LIKE'Mc%' OR'MC%'
C. WHERE INITCAP(SUBSTR(cust_name, INSTR(cust_name,'') +1)) LIKE'Mc%'
D. WHERE SUBSTR(cust_name, INSTR(cust_name,'') +1) LIKE'Mc%'
E. WHERE UPPER (SUBSTR(cust_name, INSTR(cust_name, '') +1)) LIKE UPPER('MC%')

Answer: C,E

NEW QUESTION # 198
View the exhibit and examine the structures of the EMPLOYEESand DEPARTMENTStables.
EMPLOYEES
Name Null? Type
- ---------------- ----- -------------
EMPLOYEE_ID NOT NULL NUMBER(6)
FIRST_NAME VARCHAR2(20)
LAST_NAME NOT NULL VARCHAR2(25)
HIRE_DATE NOT NULL DATE
JOB_ID NOT NULL VARCHAR2(10)
SALARY NUMBER(10,2)
COMMISSION NUMBER(6,2)
MANAGER_ID NUMBER(6)
DEPARTMENT_ID NUMBER(4)
DEPARTMENTS
Name Null? Type
----------------- ----- -------------
DEPARTMENT_ID NOT NULL NUMBER(4)
DEPARTMENT_NAME NOT NULL VARCHAR2(30)
MANAGER_ID NUMBER(6)
LOCATION_ID NUMBER(4)
You want to update EMPLOYEEStable as follows:
Update only those employees who work in Boston or Seattle (locations 2900 and 2700).

Set department_idfor these employees to the department_idcorresponding to London

(location_id 2100).
Set the employees' salary in location_id2100 to 1.1 times the average salary of their department.

Set the employees' commission in location_id2100 to 1.5 times the average commission of their

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

Answer: D

NEW QUESTION # 199
Examine this statement:

Which two things must be changed for it to execute successfully?

A. One of the LONG columns must be changed to a VARCHAR2 or CLOB.
B. The foreign key constraint on DEPT_ID must be defined at the table level instead of the column level.
C. The word CONSTRAINT in the foreign key constraint on DEPT_ID must be changed to FOREIGN KEY.
D. The primary key constraint on BMP_ID must have a name.
E. The NOT NULL constraint on ENAME must be defined at the column level instead of the table level.

Answer: A,E

Explanation:
The statement is trying to create a table with columns of different data types and constraints. Here's what needs to be corrected:
C: In Oracle, the LONG data type is used for character data of variable length up to 2 Gigabytes, but it is deprecated, and you should use CLOB or VARCHAR2 instead. Furthermore, a table cannot have more than one LONG column.
D: The NOT NULL constraint should be specified at the column level, not at the table level. The correct syntax for creating a table with a NOT NULL constraint is to define it inline with the column definition, like this:
ename VARCHAR2(15) CONSTRAINT ename_nn NOT NULL,
The other options are incorrect:
A: The foreign key constraint syntax is correct; the word CONSTRAINT is followed by the constraint name and then the REFERENCES clause.
B: The foreign key constraint can be defined at the column level.
E: While it's a good practice to name constraints, it is not mandatory for the primary key constraint to have a name; Oracle will generate one if it's not provided.
Reference:
Oracle Documentation on CREATE TABLE: SQL Language Reference - CREATE TABLE Oracle Documentation on Data Types: SQL Language Reference - Data Types

NEW QUESTION # 200
......

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