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Oracle 1z0-071 Exam Syllabus Topics:

Section	Weight	Objectives
Relational Database Concepts	10%	- Theoretical and physical aspects of relational databases - Relationship between database and SQL - ERD and SQL statement mapping
Displaying Data from Multiple Tables	15%	- Cartesian products - Equijoins and non-equijoins - Self-joins and outer joins
Conversion Functions and Conditional Expressions	10%	- TO_CHAR, TO_NUMBER, TO_DATE functions - Implicit and explicit data type conversion - NVL, NULLIF, COALESCE and CASE expressions
Restricting and Sorting Data	15%	- WHERE clause and filtering conditions - ORDER BY clause and sorting rules - Substitution variables, DEFINE and VERIFY commands - Operator precedence
Managing Tables and Constraints		- Temporary and external tables - Data types, creating/altering/dropping tables - Constraints: primary key, foreign key, unique, check, not null
Controlling User Access		- Roles, granting and revoking privileges - System and object privileges
Using SET Operators		- Ordering results with set operations - UNION, UNION ALL, INTERSECT, MINUS operators
Retrieving Data Using SQL SELECT Statement	20%	- Column aliases, concatenation, literals and DISTINCT - SELECT statement syntax and usage - Arithmetic operations and NULL handling
Aggregating Data Using Group Functions	10%	- Handling NULLs in group functions - GROUP BY and HAVING clauses - SUM, AVG, COUNT, MAX, MIN functions
Managing Views		- Create, modify and drop views - Updatable and read-only views
Using Subqueries to Solve Queries		- Single-row and multi-row subqueries - Correlated subqueries for update/delete
Using Data Dictionary Views		- Retrieve metadata and object information
Using Single-Row Functions to Customize Output	20%	- Character, number, and date functions - Nested functions usage
Managing Indexes, Synonyms and Sequences		- Create and use synonyms - Create and use sequences - Create and maintain indexes
Working with Time Zones and Date/Time Data		- TIMESTAMP, INTERVAL data types - Time zone conversion functions
Manipulating Data and Managing Transactions		- INSERT, UPDATE, DELETE, MERGE statements - Multi-table inserts - Transaction control: COMMIT, ROLLBACK, SAVEPOINT

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Oracle Database SQL Sample Questions (Q415-Q420):

NEW QUESTION # 415
Which two statements are true regarding the execution of the correlated subqueries? (Choose two.)

A. The outer query executes only once for the result returned by the inner query.
B. The nested query executes after the outer query returns the row.
C. The nested query executes first and then the outer query executes.
D. Each row returned by the outer query is evaluated for the results returned by the inner query.

Answer: B,D

Explanation:
Explanation

NEW QUESTION # 416
Examine the data in the NEW_EMPLOYEES table:

Examine the data in the EMPLOYEES table:

You want to:
1. Update existing employee details in the EMPLOYEES table with data from the NEW EMPLOYEES table.
2. Add new employee detail from the NEW_ EMPLOYEES able to the EMPLOYEES table.
Which statement will do this:

A. MERGE INTO employees e
USING new employees ne
ON (e.employee_id = ne.employee_id)
WHEN FOUND THEN
UPDATE SET e.name =ne.name, e.job_id=ne.job_id, e.salary =ne.salary
WHEN NOT FOUND THEN
INSERT VALUES (ne.employee_id,ne.name,ne.job_id,ne.salary) ;
B. MERGE INTO employees e
USING new_employees n
WHERE e.employee_id = ne.employee_id
WHEN FOUND THEN
UPDATE SET e.name=ne.name,e.job_id =ne.job_id, e.salary=ne.salary
WHEN NOT FOUND THEN
INSERT VALUES (ne.employee_ id,ne.name,ne.job id,ne.salary) ;
C. MERGE INTO employees e
USING new employees ne
WHERE e.employee_id = ne.employee_ id
WHEN MATCHED THEN
UPDATE SET e.name = ne.name, e.job_id = ne.job_id,e.salary =ne. salary
WHEN NOT MATCHED THEN
INSERT VALUES (ne. employee_id,ne.name, ne.job_id,ne.salary) ;
D. MERGE INTO employees e
USING new_employees n
ON (e.employee_id = ne.employee_id)
WHEN MATCHED THEN
UPDATE SET e.name = ne.name, e.job id = ne.job_id,e.salary =ne. salary
WHEN NOT MATCHED THEN
INSERT VALUES (ne. employee_id,ne.name,ne.job_id,ne.salary);

Answer: D

Explanation:
The correct answer to perform the specified update and insert operations is to use the MERGE statement, which is designed to facilitate such "upsert" operations (update or insert).
A . The syntax is incorrect; the WHERE clause is not used with MERGE.
B . This is the correct syntax and use of the MERGE statement. It uses the ON clause to specify the join condition, and it provides actions for WHEN MATCHED and WHEN NOT MATCHED situations.
C . The keywords FOUND and NOT FOUND are not valid in the context of the MERGE statement in Oracle SQL.
D . Similar to option A, the syntax is incorrect because MERGE uses an ON clause, not a WHERE clause, and the keywords FOUND and NOT FOUND are incorrect.
The correct syntax of a MERGE statement includes defining a condition using the ON clause to determine how rows from the source and target tables are matched, then specifying the actions for when rows match (WHEN MATCHED) and when they do not match (WHEN NOT MATCHED).
Reference:
Oracle Documentation on MERGE: https://docs.oracle.com/database/121/SQLRF/statements_9016.htm#SQLRF01606

NEW QUESTION # 417
Examine the description or the BOOKS_TRANSACTIONS 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 AMOUNT
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 AMOUNT
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 IS NOT NULL
AND due_amount IS NOT NULL;
D. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level !=NULL
AND cust credit_level !=NULL;
E. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level !=NULL
AND due_amount !=NULL;

Answer: A

Explanation:
D: True. This query selects the first name and calculates the due amount as 5% of the credit limit from the customers table where the income level is not null and the credit limit is also not null. The IS NOT NULL operator is used to check for non-null values correctly.
The IS NOT NULL operator is the correct way to check for non-null values in SQL. The <> NULL and != NULL comparisons are incorrect because NULL is not a value that can be compared using these operators; it represents the absence of a value.
There is no due_amount column in the customers table according to the structure provided, so any WHERE clause referencing due_amount is irrelevant and incorrect. The due_amount is a derived column in the SELECT clause.
The correct syntax to check for non-null values is IS NOT NULL, which is used in the WHERE clause of the correct answer.
Reference:
The use of IS NOT NULL in the WHERE clause to filter out null values is documented in Oracle's SQL reference.
NULL comparisons must be done with IS NULL or IS NOT NULL for correct logical evaluation in SQL.

NEW QUESTION # 418
Evaluate the following SQL commands:

The command to create a table fails. Identify the reason for the SQL statement failure.

A. You cannot use ORD_NO and ITEM_NO columns as a composite primary key because ORD_NO is also the FOREIGN KEY.
B. You cannot use the NEXTVAL sequence value as a DEFAULT value for a column.
C. You cannot use the BETWEEN clause in the condition of a CHECK constraint.
D. You cannot use SYSDATE in the condition of a CHECK constraint.

Answer: D

NEW QUESTION # 419
View the Exhibit and examine the structure of the PRODUCT_INFORMATION table. (Choose the best answer.)

PRODUCT_ID column is the primary key.
You create an index using this command:
SQL > CREATE INDEX upper_name_idx
ON product_information(UPPER(product_name));
No other indexes exist on the PRODUCT_INFORMATION table.
Which query would use the UPPER_NAME_IDX index?

A. SELECT product_id
FROM product_information
WHERE UPPER(product_name) IN ('LASERPRO', 'CABLE');
B. SELECT product_id, UPPER(product_name)
FROM product_information
WHERE UPPER(product_name) = 'LASERPRO' OR list_price > 1000;
C. SELECT UPPER(product_name)
FROM product_information;
D. SELECT UPPER(product_name)
FROM product_information
WHERE product_id = 2254;

Answer: A

NEW QUESTION # 420
......

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