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Oracle 1z0-071 Exam Syllabus Topics:

Section	Objectives
Relational Database Concepts	- Tables, rows, columns, and relational model basics - Data types and database objects overview
Data Manipulation and Schema Objects	- DDL operations 1. CREATE, ALTER, DROP tables and constraints - DML operations 1. INSERT, UPDATE, DELETE, MERGE
Subqueries and Advanced SQL	- Correlated subqueries and nested queries - Single-row and multi-row subqueries
SQL Functions	- Single-row functions 1. Character, numeric, date functions - Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions
SQL Retrieval	- SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting - Joins and set operations 1. UNION, INTERSECT, MINUS 2. INNER, OUTER, CROSS joins

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Oracle Database SQL Sample Questions (Q149-Q154):

NEW QUESTION # 149
The PRODUCTS table has the following structure.

Evaluate the following two SQL statements:
SQL>SELECT prod_id, NVL2 (prod_expiry_date, prod_expiry_date + 15, ' ') FROM products; SQL>SELECT prod_id, NVL (prod_expiry_date, prod_expiry_date + 15) FROM products; Which statement is true regarding the outcome?

A. Both the statements execute and give the same result
B. Both the statements execute and give different results
C. Only the first SQL statement executes successfully
D. Only the second SQL statement executes successfully

Answer: B

Explanation:
Explanation
Using the NVL2 Function
The NVL2 function examines the first expression. If the first expression is not null, the NVL2 function returns the second expression. If the first expression is null, the third expression is returned.
Syntax
NVL2(expr1, expr2, expr3)
In the syntax:
Expr1 is the source value or expression that may contain a null
Expr2 is the value that is returned if expr1 is not null
Expr3 is the value that is returned if expr1 is null

NEW QUESTION # 150
Examine this Statement which returns the name of each employee and their manager, SELECT e.last name AS emp,,m.last_name AS mgr FROM employees e JOIN managers m ON e.manager_ id = m. employee_ id ORDER BY emp; You want to extend the query to include employees with no manager. What must you add before JOIN to do this?

A. LEFT OUTER
B. RIGHT OUTER
C. FULL OUTER
D. CROSS

Answer: A

Explanation:
To include employees with no manager in the query results, a LEFT OUTER JOIN should be used. This type of join returns all records from the left table (employees), and the matched records from the right table (managers). The result is NULL from the right side if there is no match.
Here's the modified query:
SELECT e.last_name AS emp, m.last_name AS mgr FROM employees e LEFT OUTER JOIN managers m ON e.manager_id = m.employee_id ORDER BY emp; This ensures that even if an employee does not have a manager (i.e., e.manager_id is NULL or there is no corresponding m.employee_id), that employee will still be included in the results.

NEW QUESTION # 151
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. WHERE city IN ('%AN%')
B. WHERE city='%AN%;
C. WHERE city LIKE '%AN%;
D. ORDER BY last_name DESC,city ASC
E. ORDER BY 1,2.
F. CORADER BY 1, LNAME DESC

Answer: C,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 # 152
Examine the description of the EMPLOYEES table:

Which query is valid?

A. SELECT dept_id, MAX(AVG(salary)) FROM employees GROUP BY dept_id;
B. SELECT dept_id, join_date, SUM(salary) FROM employees GROUP BY dept_id, join_date;
C. SELECT dept_id, join_date, SUM(salary) FROM employees GROUP BY dept_id;
D. SELECT dept_id, AVG(MAX(salary)) FROM employees GROUP BY dept_id;

Answer: B

NEW QUESTION # 153
Examine the description of the EMPLOYEES table:

Which two queries return rows for employees whose manager works in a different department?

A. SELECT emp.*
FROM employees emp
LEFT JOIN employees mgr
ON emp.manager_ id = mgr.employee_ id
AND emp. department id < > mgr. department_ id;
B. SELECT emp.*
FROM employees emp
WHERE NOT EXISTS (
SELECT NULL
FROM employees mgr
WHERE emp.manager id = mgr.employee_ id
AND emp.department_id<>mgr.department_id
);
C. SELECT emp. *
FROM employees emp
JOIN employees mgr
ON emp. manager_ id = mgr. employee_ id
AND emp. department_ id<> mgr.department_ id;
D. SELECT emp. *
FROM employees emp
WHERE manager_ id NOT IN (
SELECT mgr.employee_ id
FROM employees mgr
WHERE emp. department_ id < > mgr.department_ id
);
E. SELECT emp. *
FROM employees emp
RIGHT JOIN employees mgr
ON emp.manager_ id = mgr. employee id
AND emp. department id <> mgr.department_ id
WHERE emp. employee_ id IS NOT NULL;

Answer: C,E

NEW QUESTION # 154
......

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