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Oracle Database SQL Sample Questions (Q47-Q52):

NEW QUESTION # 47
See the Exhibit and examine the structure of the PROMOTIONS table:

Using the PROMOTIONS table,
you need to find out the average cost for all promos in the range $0-2000 and $2000-5000 in category A.
You issue the following SQL statements:

What would be the outcome?

A. It executes successfully and gives the required result.
B. It generates an error because multiple conditions cannot be specified for the WHEN clause.
C. It generates an error because NULL cannot be specified as a return value.
D. It generates an error because CASE cannot be used with group functions.

Answer: A

Explanation:
CASE Expression
Facilitates conditional inquiries by doing the work of an IF-THEN-ELSE statement:
CASE expr WHEN comparison_expr1 THEN return_expr1
[WHEN comparison_expr2 THEN return_expr2
WHEN comparison_exprn THEN return_exprn
ELSE else_expr]
END

NEW QUESTION # 48
View the exhibit and examine the structure of ORDERSand CUSTOMERStables.
ORDERS
Name
Null? Type
ORDER_ID NOT NULL NUMBER(4)
ORDER_DATE NOT NULL DATE
ORDER_MODE VARCHAR2(8)
CUSTOMER_ID NOT NULL NUMBER(6)
ORDER_TOTAL NUMBER(8, 2)
CUSTOMERS
Name Null? Type
CUSTOMER_ID NOT NULL NUMBER(6)
CUST_FIRST_NAME NOT NULL VARCHAR2(20)
CUST_LAST_NAME NOT NULL VARCHAR2(20)
CREDIT_LIMIT NUMBER(9,2)
CUST_ADDRESS VARCHAR2(40)
Which INSERT statement should be used to add a row into the ORDERStable for the customer whose CUST_LAST_NAMEis Robertsand CREDIT_LIMITis 600? Assume there exists only one row with CUST_LAST_NAME as Roberts and CREDIT_LIMIT as 600.

A. INSERT INTO orders (order_id, order_date, order_mode,
( SELECT customer id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), order_total);
VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);
B. INSERT INTO (SELECT o.order_id, o.order_date, o.order_mode, c.customer_id, o.order_total
FROM orders o, customers c
WHERE o.customer_id = c.customer_id AND c.cust_last_name='Roberts' AND
c.credit_limit=600)
VALUES (1,'10-mar-2007', 'direct', (SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
C. INSERT INTO orders
VALUES (1,'10-mar-2007', 'direct',
( SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), 1000);
D. INSERT INTO orders (order_id, order_date, order_mode,
( SELECT customer_id
FROM customers
WHERE cust_last_name='Roberts' AND credit_limit=600), order_total);
VALUES (1,'10-mar-2007', 'direct', &customer_id, 1000);

Answer: C

NEW QUESTION # 49
Examine the data in the CUSTOMERStable:

You want to list all cities that have more than one customer along with the customer details.
Evaluate the following query:

Which two JOIN options can be used in the blank in the above query to give the correct output? (Choose two.) LEFT OUTER JOIN

A. NATURAL JOIN
B. JOIN
C. FULL OUTER JOIN
D. RIGHT OUTER JOIN

Answer: A,C

NEW QUESTION # 50
Choose the best answer.
Examine the description of the EMPLOYEES table:

Which query is valid?

A. SELECT depe_id,join_date,SUM(salary) FROM employees GROUP BY dept_id:
B. SELECT dept_id,AVG(MAX(salary)) FROM employees GROUP BY dapt_id;
C. SELECT dept_id, join_date,SUM(salary) FROM employees GROUP BY dept_id, join_date;
D. SELECT dept_id,MAX(AVG(salary)) FROM employees GROUP BY dept_id;

Answer: C

NEW QUESTION # 51
View the exhibit and examine the description of the DEPARTMENTS and EMPLOYEES tables.

You wrote this SQL statement to retrieve EMPLOYEE_ID, FIRST_NAME, and DEPARTMENT NAME, for all employees:
SELECT employee_id, first_name, department_name
FROM employees
NATURAL JOIN departments;
The desired output is not obtained after executing the above SQL statement. What could be the reason for this?

A. The EMPLOYEES and DEPARTMENTS tables have more than one column with the same column name and data type.
B. The DEPARTMENTS table is not used before the EMPLOYEES table in the FROM clause.
C. The NATURAL JOIN clause is missing the USING clause.
D. The table prefix is missing for the column names in the SELECT clause.

Answer: A

Explanation:
Explanation
Natural join needs only one column to be the same in each table. The EMPLOYEES and DEPARTMENTS tables have two columns that are the same (Department_ID and Manager_ID)

NEW QUESTION # 52
......

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