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Oracle 1z0-071 exam, also known as the Oracle Database SQL exam, is a professional certification exam designed for individuals who want to demonstrate their skills in SQL development and database management. 1z1-071 exam is intended for database administrators, developers, and IT professionals who work with Oracle databases and want to validate their expertise in SQL.
The Oracle 1z1-071 Exam consists of 73 multiple-choice questions that are to be completed within 105 minutes. The questions are based on topics like SQL syntax, data retrieval, data manipulation, table creation, and indexing. 1z1-071 exam is available in several languages, including English, Japanese, and Spanish.
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Oracle 1z0-071 certification exam is a computer-based test that consists of 73 multiple-choice questions. 1z1-071 Exam has a duration of 105 minutes, and the passing score is 63%. 1z1-071 exam fee is $245, and candidates can register for the exam through the Oracle Certification website. To prepare for the exam, candidates can take advantage of various training resources, including online courses, practice exams, and study guides. Passing the Oracle 1z0-071 certification exam is an excellent way to validate your SQL programming skills and enhance your career opportunities in the database industry.
Oracle Database SQL Sample Questions (Q394-Q399):
NEW QUESTION # 394
Examine the description of the EMPLOYEES table:

Which statement increases each employee's SALARY by the minimum SALARY for their DEPARTM ENT_ID?
- A. UPDATE employees e1
SET salary = salary+(SELECT MIN (salary)
FROM employees e2) ; - B. UPDATE employees e1
SET salary = salary +
(SELECT MIN(e1. salary)
FROM employees e2
WHERE e1.department_id = e2 .department_id); - C. UPDATE employees e1
SET salary =(SELECT e2. salary + MIN(e2.salary)
FROM employees e2
WHERE e1.department_ id = e2. department_id GROUP BY e2. department_id) ; - D. UPDATE employees e1
SET salary=
(SELECT e1.salary + MIN(e2.salary)
FROM employees e2
WHERE e1. department_ id = e2.department_id);
Answer: D
Explanation:
The correct statement to increase each employee's salary by the minimum salary for their department is:
D . This option correctly increases each employee's salary by the minimum salary in their department. The subquery calculates the minimum salary for the department of the current row being updated (e1) in the outer UPDATE statement. The subquery does not use GROUP BY since it returns a single value - the minimum salary for the matching department.
A, B, and C are incorrect. Specifically:
A uses a subquery with an aggregate function combined with GROUP BY, which is not needed and will result in an error.
B incorrectly references MIN(e1.salary) within the subquery, which is incorrect since e1 is the alias for the outer query's table and should not be used within the subquery.
C sets the salary to itself plus the minimum salary across all employees, not filtered by the department.
Reference:
Oracle Database SQL Language Reference, 12c Release 1 (12.1): "UPDATE"
NEW QUESTION # 395
Examine the description or the CUSTOMERS table:

For Customers whose income level has a value, you want to display the first name and due amount as 5% of their credit limit. Customers whose due amount is null should not be displayed.
Which query should be used?
- A. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
- B. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level != NULL AND due_amount != NULL;
- C. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMOUNT FROM customers WHERE cust_income_level != NULL AND cust_credit_level != NULL;
- D. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level <> NULL AND due_amount <> NULL;
- E. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers WHERE cust_income_level IS NOT NULL AND due_amount IS NOT NULL;
Answer: A
Explanation:
In Oracle SQL, the correct way to check for non-null values is to use the IS NOT NULL condition. Using != NULL or <> NULL is incorrect because NULL represents the absence of any value, and you cannot use equality or inequality operators to check for NULL.
C . SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMOUNT FROM customers WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL; This query will correctly filter out any customers with a NULL income level or credit limit and then calculate the due amount as 5% of their credit limit for the remaining customers.
Reference:
Oracle Database SQL Language Reference 12c, especially sections on conditions and expressions that deal with NULL values.
NEW QUESTION # 396
Which three actions can you perform on an existing table containing data? (Choose three.)
- A. Add a new NOT NULL column with a DEFAULT value
- B. Change the default value of a column
- C. Define a default value that is automatically inserted into a column containing nulls
- D. Change a DATE column containing data to a NUMBER data type
- E. Add a new column as the table's first column
- F. Increase the width of a numeric column
Answer: B,E
NEW QUESTION # 397
The following are the steps for a correlated subquery, listed in random order:
1) The WHERE clause of the outer query is evaluated.
2) The candidate row is fetched from the table specified in the outer query.
3) The procedure is repeated for the subsequent rows of the table, till all the rows are processed.
4) Rows are returned by the inner query, after being evaluated with the value from the candidate row in the outer query. Identify the option that contains the steps in the correct sequence in which the Oracle server evaluates a correlated subquery.
- A. 4,2,1,3
- B. 4,1,2,3
- C. 2,4,1,3
- D. 2,1,4,3
Answer: C
NEW QUESTION # 398
View the Exhibit and examine the structure of the CUSTOMERS table.
Using the CUSTOMERS table, you must generate a report that displays a credit limit increase of 15% for all customers.
Customers with no credit limit should have "Not Available" displayed.
Which SQL statement would produce the required result?
- A. SELECT TO_CHAR(NVL(cust_credit_limit*.15), 'Not Available')) "NEW CREDIT" FROM customers
- B. SELECT NVL (cust_credit_limit*.15, 'Not Available') "NEW CREDIT" FROM customers
- C. SELECT NVL (cust_credit_limit, 'Not Available')*.15 "NEW CREDIT" FROM customers
- D. SELECT NVL (TO_CHAR(cust_credit_limit*.15), 'Not Available') "NEW CREDIT" FROM customers
Answer: B
NEW QUESTION # 399
......
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