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Oracle 1z0-071 Exam Overview:

Certification Vendor:	Oracle
Exam Name:	Oracle Database SQL
Exam Number:	1Z0-071
Real Exam Qty:	63 - 78
Exam Format:	Multiple Choice, Multiple Select
Certificate Validity Period:	18 months
Related Certifications:	Oracle Database Developer Certified Associate Oracle Database Administrator Certified Associate
Exam Duration:	120 minutes
Exam Price:	US$245
Available Languages:	Japanese, Korean, Simplified Chinese, English, Traditional Chinese, Spanish
Passing Score:	63%
Recommended Training:	Oracle Database 19c: SQL Workshop Oracle Database SQL Fundamentals
Exam Registration:	Oracle University Registration Pearson VUE Registration
Sample Questions:	Oracle 1z0-071 Sample Questions
Exam Way:	Online proctored or Onsite at Pearson VUE test centers
Pre Condition:	No mandatory prerequisites; basic knowledge of SQL and relational databases recommended
Official Syllabus URL:	https://education.oracle.com/pexam_1Z0-071

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Oracle 1z0-071 exam, also known as the Oracle Database SQL exam, is designed to test the knowledge and skills of individuals in the area of SQL for Oracle databases. 1z0-071 Exam is ideal for individuals who are looking to validate their understanding of SQL and wish to obtain a certification in this area.

Oracle Database SQL Sample Questions (Q305-Q310):

NEW QUESTION # 305
View the exhibit and examine the structure of the EMPLOYEES table.

You want to select all employees having 100 as their MANAGER_ID manages and their manager.
You want the output in two columns: the first column should have the employee's manager's LAST_NAME and the second column should have the employee's LAST_NAME.
Which SQL statement would you execute?

A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id=100;
B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id=100;
C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id=100;
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id AND e.manager_id=100

Answer: C

NEW QUESTION # 306
Which two statements are true?

A. Neither CASE nor DECODE is a function.
B. AIl conditions evaluated using DECODE can also be evaluated using CASE.
C. CASE is a function and DECODE is not.
D. DECODE is a function and CASE is not.
E. Both CASE and DECODE are functions.
F. All conditions evaluated using CASE can also be evaluated using DECODE.

Answer: B,D

Explanation:
A). True. The DECODE function can evaluate conditions, but it is limited to equality checks. On the other hand, CASE can evaluate these and a broader range of conditions using different comparison operators.
D). True. DECODE is a function in Oracle SQL that allows for simple conditional query transformation.
CASE is not a function but a statement that provides more flexibility and readability when handling conditional logic in SQL queries.
DECODE is more limited compared to CASE since it can't perform logical operations other than equality.

NEW QUESTION # 307
View the exhibit and examine the structure of the EMPLOYEES table.

You want to select all employees having 100 as their MANAGER_ID manages and their manager.
You want the output in two columns: the first column should have the employee's manager's LAST_NAME and the second column should have the employee's LAST_NAME.
Which SQL statement would you execute?

A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id=100;
B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id=100;
C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id=100;
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id AND e.manager_id=100

Answer: C

NEW QUESTION # 308
View the Exhibit and examine the structure in the EMPLOYEES tables.

Evaluate the following SQL statement:
SELECT employee_id, department_id
FROM employees
WHERE department_id= 50 ORDER BY department_id
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=90
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=10;
What would be the outcome of the above SQL statement?

A. The statement would execute successfully but it will ignore the ORDER BY clause and display the rows in random order.
B. The statement would execute successfully and display all the rows in the ascending order of DEPARTMENT_ID.
C. The statement would not execute because the positional notation instead of the column name should be used with the ORDER BY clause.
D. The statement would not execute because the ORDER BY clause should appear only at the end of the SQL statement, that is, in the last SELECT statement.

Answer: D

NEW QUESTION # 309
Which two statements are true about Data Manipulation Language (DML) statements?

A. An UPDATE....SET.... statement can modify multiple rows based on only a single condition on a table.
B. A DELETE FROM..... statement can remove rows based on only a single condition on a table.
C. An INSERT INTO...VALUES.. statement can add multiple rows per execution to a table.
D. A DELETE FROM..... statement can remove multiple rows based on multiple conditions on a table.
E. An UPDATE... SET... statement can modify multiple rows based on multiple conditions on a table.
F. An INSERT INTO... VALUES..... statement can add a single row based on multiple conditions on a table.

Answer: D,E

Explanation:
Explanation
References:
http://www.techonthenet.com/sql/and_or.php

NEW QUESTION # 310
......

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