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Oracle 1z0-071 Exam Syllabus Topics:

Section	Weight	Objectives
Topic 1: Using Data Dictionary Views		- Retrieve metadata and object information
Topic 2: Using Single-Row Functions to Customize Output	20%	- Character, number, and date functions - Nested functions usage
Topic 3: Managing Views		- Updatable and read-only views - Create, modify and drop views
Topic 4: Managing Indexes, Synonyms and Sequences		- Create and use sequences - Create and use synonyms - Create and maintain indexes
Topic 5: Aggregating Data Using Group Functions	10%	- SUM, AVG, COUNT, MAX, MIN functions - Handling NULLs in group functions - GROUP BY and HAVING clauses
Topic 6: Controlling User Access		- System and object privileges - Roles, granting and revoking privileges
Topic 7: Using SET Operators		- UNION, UNION ALL, INTERSECT, MINUS operators - Ordering results with set operations
Topic 8: Manipulating Data and Managing Transactions		- INSERT, UPDATE, DELETE, MERGE statements - Transaction control: COMMIT, ROLLBACK, SAVEPOINT - Multi-table inserts
Topic 9: Restricting and Sorting Data	15%	- Substitution variables, DEFINE and VERIFY commands - WHERE clause and filtering conditions - ORDER BY clause and sorting rules - Operator precedence
Topic 10: Displaying Data from Multiple Tables	15%	- Self-joins and outer joins - Equijoins and non-equijoins - Cartesian products
Topic 11: Using Subqueries to Solve Queries		- Single-row and multi-row subqueries - Correlated subqueries for update/delete
Topic 12: Relational Database Concepts	10%	- Theoretical and physical aspects of relational databases - Relationship between database and SQL - ERD and SQL statement mapping
Topic 13: Conversion Functions and Conditional Expressions	10%	- NVL, NULLIF, COALESCE and CASE expressions - TO_CHAR, TO_NUMBER, TO_DATE functions - Implicit and explicit data type conversion
Topic 14: Retrieving Data Using SQL SELECT Statement	20%	- SELECT statement syntax and usage - Arithmetic operations and NULL handling - Column aliases, concatenation, literals and DISTINCT
Topic 15: Working with Time Zones and Date/Time Data		- TIMESTAMP, INTERVAL data types - Time zone conversion functions
Topic 16: Managing Tables and Constraints		- Constraints: primary key, foreign key, unique, check, not null - Temporary and external tables - Data types, creating/altering/dropping tables

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Oracle Database SQL Sample Questions (Q222-Q227):

NEW QUESTION # 222
Which two statements are true?

A. All conditions evaluated using CASE can also be evaluated using DECODE.
B. AIl conditions evaluated using DECODE can also be evaluated using CASE.
C. Both CASE and DECODE are functions.
D. DECODE is a function and CASE is not.
E. CASE is a function and DECODE is not.
F. Neither CASE nor DECODE is a function.

Answer: B,D

NEW QUESTION # 223
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 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;
B. 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;
C. 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;
D. 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;

Answer: B

NEW QUESTION # 224
View the exhibit and examine the structure of the CUSTOMERS table.
Which two tasks would require subqueries or joins to be executed in a single statement?

A. finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers
B. listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'.
C. listing of customers who do not have a credit limit and were born before 1980
D. finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'
E. finding the number of customers, in each city, who's marital status is 'married'.

Answer: A,B

NEW QUESTION # 225
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 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;
B. 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;
C. 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;
D. 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;

Answer: B

NEW QUESTION # 226
Examine the data in the PRODUCTS table:

Examine these queries:
1. SELECT prod name, prod list
FROM products
WHERE prod 1ist NOT IN(10,20) AND category _id=1;
2. SELECT prod name, | prod _ list
FROM products
WHERE prod list < > ANY (10,20) AND category _id= 1;
SELECT prod name, prod _ list
FROM products
WHERE prod_ list <> ALL (10, 20) AND category _ id= 1;
Which queries generate the same output?

A. 1 and 2
B. 1 and 3
C. 1, 2 and 3
D. 2 and 3

Answer: B

Explanation:
Based on the given PRODUCTS table and the SQL queries provided:
Query 1: Excludes rows where prod_list is 10 or 20 and category_id is 1.
Query 2: Includes rows where prod_list is neither 10 nor 20 and category_id is 1.
Query 3: Excludes rows where prod_list is both 10 and 20 (which is not possible for a single value) and category_id is 1.
The correct answer is A, queries 1 and 3 will produce the same result. Both queries exclude rows where prod_list is 10 or 20 and include only rows from category_id 1. The NOT IN operator excludes the values within the list, and <> ALL operator ensures that prod_list is not equal to any of the values in the list, which effectively excludes the same set of rows.
Query 2, using the <> ANY operator, is incorrect because this operator will return true if prod_list is different from any of the values in the list, which is not the logic represented by the other two queries.

NEW QUESTION # 227
......

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