ORACLE RELIABLE 1Z0-071 EXAM SIMULATIONS EXAM 100% PASS | 1Z0-071 NEW DUMPS PDF

Oracle Reliable 1z0-071 Exam Simulations Exam 100% Pass | 1z0-071 New Dumps Pdf

Oracle Reliable 1z0-071 Exam Simulations Exam 100% Pass | 1z0-071 New Dumps Pdf

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The Oracle 1z0-071 exam covers topics such as SQL fundamentals, data manipulation, data definition, and control and management of transactions. It is a comprehensive exam that tests the candidate's ability to work with Oracle databases and write SQL queries.

The Oracle 1z0-071 is a certification exam that proves the candidate’s knowledge of SQL technology and SQL concepts required to participate in any database project. The target audience for this exam is the IT professionals with some experience in working with these technologies. The fundamental knowledge of general computing concepts and command line interfaces will be an advantage as well.

The Oracle 1z0-071 Exam covers a range of topics, including SQL fundamentals, data retrieval using SELECT statements, data manipulation, and database objects. It also includes advanced SQL topics, such as subqueries, set operators, and data aggregation. 1z0-071 exam is intended for professionals who have a good understanding of SQL and have experience working with databases.

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Oracle Database SQL Sample Questions (Q196-Q201):

NEW QUESTION # 196
You execute the following commands:

For which substitution variables are you prompted for the input?

  • A. Only 'hiredate'
  • B. None, because no input required
  • C. Both the substitution variables 'hiredate' and 'mgr_id
  • D. Only 'mgr_id'

Answer: C


NEW QUESTION # 197
Examine the description of EMPLOYEES table:
Which three queries return all rows for which SALARY+COMMISSION is greate than 20000?

  • A. SELECT * FROM employees WHERE salary+NVL(commission,0)>=20000;
  • B. SELECT * FROM employees WHERE salary+NULLF(commission,0)>=20000;
  • C. SELECT * FROM employees WHERE NVL(salary+commission,0)>==20000;
  • D. SELECT * FROM employees WHERE NVL(salary+commission,0)>=20000;
  • E. SELECT * FROM employees WHERE NVL2(salary)+commission,salary+commission,
  • F. SELECT * FROM employees WHERE salary+NVL2(commission,commission,0)>=20000;

Answer: A,E,F


NEW QUESTION # 198
View the exhibit and examine the description of the DEPARTMENTS and EMPLOYEES tables.

The retrieve data for all the employees for their EMPLOYEE_ID, FIRST_NAME, and DEPARTMENT NAME, the following SQL statement was written:
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 DEPARTMENTS table is not used before the EMPLOYEES table in the FROM clause.
  • B. The table prefix is missing for the column names in the SELECT clause.
  • C. The NATURAL JOIN clause is missing the USING clause.
  • D. The EMPLOYEES and DEPARTMENTS tables have more than one column with the same column name and data type.

Answer: D

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 # 199
View the exhibit and examine the structure of the EMPLOYEES table.

You want to display all employees and their managers having 100 as the MANAGER_ID. You want the output in two columns: the first column would have the LAST_NAME of the managers and the second column would have LAST_NAME of the employees.
Which SQL statement would you execute?

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

Answer: C


NEW QUESTION # 200
The following are the steps for a correlated subquery, listed in random order:
* The WHERE clause of the outer query is evaluated.
* The candidate row is fetched from the table specified in the outer query.
* This is repeated for the subsequent rows of the table, till all the rows are processed.
* Rows are returned by the inner query, after being evaluated with the value from the candidate row in the outer query.
Which is the correct sequence in which the Oracle server evaluates a correlated subquery?

  • A. 4, 1, 2, 3
  • B. 4, 2, 1, 3
  • C. 2, 4, 1, 3
  • D. 2, 1, 4, 3

Answer: C

Explanation:
Explanation
References:
http://rajanimohanty.blogspot.co.uk/2014/01/correlated-subquery.html


NEW QUESTION # 201
......

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