ORACLE 1Z1-071 PDF VCE | 1Z1-071 VALID EXAM GUIDE

Oracle 1z1-071 PDF VCE | 1z1-071 Valid Exam Guide

Oracle 1z1-071 PDF VCE | 1z1-071 Valid Exam Guide

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Oracle 1z1-071 (Oracle Database SQL) certification exam is designed to test the skills and knowledge of individuals who work with Oracle Database SQL. 1z1-071 exam is intended for database administrators, developers, and anyone else who works with Oracle Database SQL. Those who pass the exam will receive the Oracle Database SQL certification, which demonstrates their proficiency in working with Oracle Database SQL.

Oracle 1z1-071 (Oracle Database SQL) Certification Exam is recognized globally as a standard for SQL database management. Candidates who pass the exam are recognized as Oracle Certified Professionals and are highly sought after by employers in the IT industry. Oracle Database SQL certification provides an edge to professionals who want to demonstrate their expertise in SQL database management and opens up new job opportunities and career advancement prospects.

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1z1-071 Valid Exam Guide - 1z1-071 Valid Test Forum

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Oracle Database SQL Sample Questions (Q55-Q60):

NEW QUESTION # 55
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. 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,
  • B. (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);
  • C. 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,
  • D. (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

Answer: A


NEW QUESTION # 56
Which three statements are true regarding the SQL WHERE and HAVING clauses? (Choose three.)

  • A. The HAVING clause is used to exclude one or more aggregated results after grouping data.
  • B. The WHERE clause is used to exclude rows before grouping data.
  • C. The WHERE and HAVING clauses cannot be used together in a SQL statement.
  • D. The HAVING clause conditions can have aggregating functions.
  • E. The HAVING clause conditions can use aliases for the columns.

Answer: A,B,D


NEW QUESTION # 57
Which statement is true about aggregate functions?

  • A. The AVG function implicitly converts NULLS to zero.
  • B. Aggregate functions can be nested to any number of levels.
  • C. The MAX and MIN functions can be used on columns with character data types.
  • D. Aggregate functions can be used in any clause of a SELECT statement.

Answer: B


NEW QUESTION # 58
Which two statements are true regarding the EXISTS operator used in the correlated subqueries? (Choose two.)

  • A. The outer query continues evaluating the result set of the inner query until all the values in the result set are processed.
  • B. It is used to test whether the values retrieved by the outer query exist in the result set of the inner query.
  • C. The outer query stops evaluating the result set of the inner query when the first value is found.
  • D. It is used to test whether the values retrieved by the inner query exist in the result of the outer query.

Answer: B,C

Explanation:
References:
http://www.techonthenet.com/oracle/exists.php


NEW QUESTION # 59
View the Exhibit and examine the structure of CUSTOMERS table.
Using the CUSTOMERS table, you need to generate a report that shows an increase in the credit limit by 15% for all customers. Customers whose credit limit has not been entered should have the message "Not Available" displayed.
Which SQL statement would produce the required result?

  • A. SELECT NVL (TO CHAR(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
  • B. SELECT NVL(cust_credit_limit), 'Not Available') "NEW CREDIT"FROM customers;
  • C. SELECT TO_CHAR (NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;
  • D. SELECT NVL(cust_credit_limit * .15), 'Not Available') "NEW CREDIT"FROM customers;

Answer: A


NEW QUESTION # 60
......

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