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Oracle 1z1-071 Exam Syllabus Topics:

Section	Objectives
Topic 1: Subqueries and Advanced SQL	- Single-row and multi-row subqueries - Correlated subqueries and nested queries
Topic 2: Data Manipulation and Schema Objects	- DML operations 1. INSERT, UPDATE, DELETE, MERGE - DDL operations 1. CREATE, ALTER, DROP tables and constraints
Topic 3: SQL Functions	- Single-row functions 1. Character, numeric, date functions - Conversion and conditional functions 1. TO_CHAR, TO_NUMBER, CASE expressions
Topic 4: SQL Retrieval	- Joins and set operations 1. UNION, INTERSECT, MINUS 2. INNER, OUTER, CROSS joins - SELECT statements and filtering data 1. WHERE clause and conditions 2. ORDER BY and row limiting
Topic 5: Relational Database Concepts	- Data types and database objects overview - Tables, rows, columns, and relational model basics

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Oracle Database SQL Sample Questions (Q424-Q429):

NEW QUESTION # 424
Examine the structure of the PROMOTIONS table: (Choose the best answer.)

Management requires a report of unique promotion costs in each promotion category.
Which query would satisfy this requirement?

A. SELECT promo_category, DISTINCT promo_cost FROM promotions
B. SELECT DISTINCT promo_cost, promo_category FROM promotions
C. SELECT DISTINCT promo_category, promo_cost FROM promotions ORDER BY 1
D. SELECT DISTINCT promo_cost, DISTINCT promo_category FROM promotions;

Answer: C

NEW QUESTION # 425
View the Exhibit and examine the data in the PRODUCT_INFORMATIONtable.

Which two tasks would require subqueries? (Choose two.)

A. displaying all the products whose minimum list prices are more than average list price of products having the status orderable
B. displaying the minimum list price for each product status
C. displaying the number of products whose list prices are more than the average list price
D. displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE
E. displaying all supplier IDs whose average list price is more than 500

Answer: A,C

NEW QUESTION # 426
Examine the data in the ENPLOYEES table:

Which statement will compute the total annual compensation tor each employee

A. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * NVL (monthly_commission_pct, 0)) AS annual_comp FROM employees
B. SELCECT last_namo, (monthly_salary * 12) + (monthly_commission_pct * 12) AS annual_comp FROM employees
C. SELCECT last_namo, (monthly_salary * 12) + (menthy_salary * 12 * monthly_commission_pct) AS annual_comp FROM employees
D. SECECT last_namo, (menthy_salary + monthly_commission_pct) * 12 AS annual_comp FROM employees;

Answer: A

Explanation:
The correct statement for computing the total annual compensation for each employee is option C. This is because the monthly commission is a percentage of the monthly salary (indicated by the column name MONTHLY_COMMISSION_PCT). To calculate the annual compensation, we need to calculate the annual salary (which is monthly_salary * 12) and add the total annual commission to it.
Here's the breakdown of the correct statement, option C:
* (monthly_salary * 12) computes the total salary for the year.
* NVL(monthly_commission_pct, 0) replaces NULL values in the monthly_commission_pct column with 0, ensuring that the commission is only added if it exists.
* (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) computes the annual commission by first determining the monthly commission (which is a percentage of the monthly salary), and then multiplying it by 12 to get the annual commission.
* Finally, (monthly_salary * 12) + (monthly_salary * 12 * NVL(monthly_commission_pct, 0)) adds the annual salary and the annual commission to get the total annual compensation.
The other options are incorrect:
* Option A is incorrect because it adds the monthly_commission_pct directly to the monthly_salary, which does not consider that monthly_commission_pct is a percentage.
* Option B is incorrect because it adds the commission percentage directly without first applying it to the monthly salary.
* Option D is incorrect because it does not handle the NULL values in the commission column, which would result in a NULL total annual compensation whenever the monthly_comission_pct is NULL.
References:
* Oracle Documentation on NVL function: NVL
* Oracle Documentation on Numeric Literals: Numeric Literals

NEW QUESTION # 427
Which two statements are true about the ORDER BY clause when used with a SQL statement containing a SET operator such as UNION?

A. Each SELECT statement in the compound query can have its own ORDER BY clause.
B. Only column names from the first SELECT statement in the compound query are recognized.
C. The first column in the first SELECT of the compound query with the UNION operator is used by default to sort output in the absence of an ORDER BY clause.
D. Each SELECT statement in the compound query must have its own ORDER BY clause.
E. Column positions must be used in the ORDER BY clause.

Answer: A,B

NEW QUESTION # 428
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 not execute because the positional notation instead of the column name should be used with the ORDER BY clause.
B. 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.
C. The statement would execute successfully but it will ignore the ORDER BY clause and display the rows in random order.
D. The statement would execute successfully and display all the rows in the ascending order of DEPARTMENT_ID.

Answer: B

NEW QUESTION # 429
......

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