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Oracle 1z1-071 certification exam covers a broad range of topics related to SQL programming language, such as data definition language (DDL), data manipulation language (DML), and data control language (DCL). In addition, the exam also covers various topics related to SQL functions, such as single-row functions, group functions, and conversion functions. 1z0-071 Exam also assesses the candidate's knowledge of schema objects, such as tables, views, indexes, and sequences.

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Oracle 1z0-071, also known as the Oracle Database SQL Certification Exam, is an industry-recognized certification that validates the skills and knowledge of professionals in SQL development and database management. Oracle Database SQL certification exam is designed for individuals who want to demonstrate their ability to write SQL statements, manage database objects, and perform other database-related tasks.

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Oracle Database SQL Sample Questions (Q377-Q382):

NEW QUESTION # 377
Examine the description of the ORDER_ITEMS table:

Examine this incomplete query:
SELECT DISTINCT quantity * unit_price total_paid FROM order_items ORDER BY <clause>; Which two can replace <clause> so the query completes successfully?

A. quantity, unit_price
B. total_paid
C. quantity
D. quantity * unit_price
E. product_id

Answer: B,D

NEW QUESTION # 378
View the exhibit and examine the structure of ORDERSand CUSTOMERStables.

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.
INSERT INTO (SELECT o.order_id, o.order_date, o.order_mode, c.customer_id,

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

Answer: B

NEW QUESTION # 379
Evaluate the following SQL commands:

The command to create a table fails. Identify the reason for the SQL statement failure.

A. You cannot use ORD_NO and ITEM_NO columns as a composite primary key because ORD_NO is also the FOREIGN KEY.
B. You cannot use SYSDATE in the condition of a CHECK constraint.
C. You cannot use the NEXTVAL sequence value as a DEFAULT value for a column.
D. You cannot use the BETWEEN clause in the condition of a CHECK constraint.

Answer: B

NEW QUESTION # 380
Examine the structure of the MEMBERS table:

Examine the SQL statement:
SQL > SELECT city, last_name LNAME FROM MEMBERS ORDER BY 1, LNAME DESC; What would be the result execution? (Choose the best answer.)

A. It displays all cities in ascending order, within which the last names are further sorted in descending order.
B. It fails because a column number and a column alias cannot be used together in the ORDER BY clause.
C. It fails because a column alias cannot be used in the ORDER BY clause.
D. It displays all cities in descending order, within which the last names are further sorted in descending order.

Answer: A

NEW QUESTION # 381
Examine the description of the CUSTOMERS table:

Which two SELECT statements will return these results:
CUSTOMER_ NAME
--------------------
Mandy
Mary

A. SELECT customer_ name FROM customers WHERE customer name LIKE 'Ma%' ;
B. SELECT customer_ name FROM customers WHERE customer_ name='*Ma*';
C. SELECT customer_ name FROM customers WHERE UPPER (customer name) LIKE 'MA&';
D. SELECT customer_ name FROM customers WHERE customer name LIKE 'Ma*';
E. SELECT customer_ name FROM customers WHERE UPPER (customer_ name ) LIKE 'MA*. ;
F. SELECT customer_ name FROM customers WHERE customer_ name LIKE ' % a % ' ;
G. SELECT customer_ name FROM customers WHERE customer_ name KIKE .*Ma*';

Answer: A,F

NEW QUESTION # 382
......

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