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Oracle 1z1-071 certification exam covers a wide range of topics related to SQL database administration with the Oracle database. 1z1-071 exam tests your understanding of SQL syntax, data manipulation, data definition, and database objects creation. You will also be tested on your ability to retrieve data using SELECT statements, join tables, and aggregate functions. Additionally, the exam covers topics related to managing schema objects, managing data, and ensuring data integrity.

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Oracle 1z1-071 Exam is a computer-based test that consists of 73 multiple-choice questions. Candidates have 105 minutes to complete the exam. 1z1-071 exam is available in several languages, including English, Japanese, and Simplified Chinese. Candidates can take the exam at any authorized testing center or online.

Oracle Database SQL Sample Questions (Q19-Q24):

NEW QUESTION # 19
Evaluate the following SQL statement:
SQL> select cust_id, cust_last_name "Last name"
FROM customers
WHERE country_id = 10
UNION
SELECT cust_id CUST_NO, cust_last_name
FROM customers
WHERE country_id = 30
Identify three ORDERBYclauses either one of which can complete the query.

A. ORDER BY 2, 1
B. ORDER BY 2, cust_id
C. ORDER BY "CUST_NO"
D. ORDER BY "Last name"
E. ORDER BY CUST_NO

Answer: A,B,D

Explanation:
Explanation/Reference:
Explanation:
Using the ORDER BY Clause in Set Operations
-The ORDER BY clause can appear only once at the end of the compound query.
-Component queries cannot have individual ORDER BY clauses.
-The ORDER BY clause recognizes only the columns of the first SELECT query.
-By default, the first column of the first SELECT query is used to sort the output in an ascending order.

NEW QUESTION # 20
Examine the description of the EMP_DETAILS table given below:

Which two statements are true regarding SQL statements that can be executed on the EMP_DETAIL TABLE?

A. You cannot add a new column to the table with LONG as the data type.
B. You can alter the table to include the NOT NULL constraint on the EMP_IMAGE column.
C. An EMP_IMAGE column cannot be included in the ORDER BY clause.
D. An EMP_IMAGE column can be included in the GROUP BY clause.

Answer: A,C

Explanation:
Explanation

NEW QUESTION # 21
Examine the description of the MEMBERS table;
SELECT city,last_name LNAME FROM members ...
You want to display all cities that contain the string AN. The cities must be returned in ascending order, with the last names further sorted in descending order.
Which two clauses must you add to the query?

A. ORDER BY 1,2.
B. WHERE city IN ('%AN%')
C. ORDER BY last_name DESC,city ASC
D. WHERE city='%AN%;
E. CORADER BY 1, LNAME DESC
F. WHERE city LIKE '%AN%;

Answer: E,F

Explanation:
To achieve the desired output for the query from the MEMBERS table:
* C. ORDER BY 1, LNAME DESC: This clause correctly sorts the results first by the first column (city) in ascending order by default, and then by the alias LNAME (last_name) in descending order.
* E. WHERE city LIKE '%AN%': This clause correctly filters the rows to include only those cities containing the string 'AN' anywhere in the city name, using the LIKE operator which is suitable for pattern matching.
Incorrect options:
* A: This would order both columns in ascending order, which does not meet the requirement for last_name to be in descending order.
* B: This misplaces the order priorities and uses explicit column names that contradict the sorting requirement specified.
* D: Incorrect syntax for a LIKE clause; equality operator cannot be used with wildcards.
* F: The IN operator is incorrectly used here with a pattern, which is not valid syntax.

NEW QUESTION # 22
Examine the structure of the EMPLOYEEStable.

There is a parent/child relationship between EMPLOYEE_IDand MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_IDis 123.
Which query provides the correct output?

A. SELECT m.last_name, e.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.manager_id = m.manager_id)
WHERE e.employee_id = 123;
B. SELECT e.last_name, e.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.employee_id = m.employee_id)
WHERE e.employee_id = 123;
C. SELECT e.last_name, m.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123;
D. SELECT e.last_name, m.manager_id
FROM employees e RIGHT OUTER JOIN employees m
on (e.manager_id = m.employee_id)
AND e.employee_id = 123;

Answer: A

NEW QUESTION # 23
Examine the structure of the CUSTOMERS table: (Choose two.)

CUSTNO is the PRIMARY KEY.
You must determine if any customers' details have been entered more than once using a different CUSTNO, by listing all duplicate names.
Which two methods can you use to get the required result?

A. Left outer-join with self-join
B. Subquery
C. Right outer-join with self-join
D. Self-join
E. Full outer-join with self-join

Answer: B,D

NEW QUESTION # 24
......

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