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Oracle 1z0-071 Exam Syllabus Topics:

Section	Weight	Objectives
Using Subqueries to Solve Queries	15%	- Describe the types of problems that subqueries can solve - Define subqueries - Write single-row and multi-row subqueries - List the types of subqueries
Using DDL Statements to Create and Manage Tables	14%	- Describe the data types that are available for columns - Drop, rename, and truncate tables - Alter table definitions - Create a simple table - Categorize the main database objects - Review the table structure
Using Single-Row Functions to Customize Output	10%	- Manipulate strings with character function in the SELECT and WHERE clauses - Perform arithmetic with date data - Describe the differences between single-row and multiple-row functions - Manipulate numbers with the ROUND, TRUNC, and MOD functions - Manipulate dates with the date functions
Creating Other Schema Objects	10%	- Create and maintain indexes - Create synonyms for database objects - Create, maintain, and use sequences - Create, modify, and retrieve data from a view
Manipulating Data	14%	- Insert rows into a table - Update rows in a table - Delete rows from a table - Control transactions - Describe each data manipulation language (DML) statement
Controlling User Access	5%	- Differentiate system privileges from object privileges - Grant privileges on tables and on a view - Distinguish between granting privileges and roles
Using Set Operators	5%	- Describe set operators - Use a set operator to combine multiple queries into a single query - Describe the order of precedence when the set operators are used
Displaying Data from Multiple Tables Using Joins	15%	- View data that does not meet a join condition by using outer joins - Join a table to itself by using a self-join - Cross join tables - Write SELECT statements to access data from more than one table using equijoins and non-equijoins
Using Conversion Functions and Conditional Expressions	10%	- Apply the NVL, NULLIF, and COALESCE functions to data - Describe implicit and explicit data type conversion - Use the TO_CHAR, TO_NUMBER, and TO_DATE conversion functions - Use conditional IF THEN ELSE logic in a SELECT statement
Reporting Aggregated Data Using Group Functions	15%	- Describe the use of group functions - Group data by using the GROUP BY clause - Include or exclude grouped rows by using the HAVING clause - Identify the available group functions
Restricting and Sorting Data	6%	- Sort the rows that are retrieved by a query - Use DEFINE and UNDEFINE commands - Limit the rows that are retrieved by a query - Use ampersand substitution to restrict and sort output at runtime
Working with Oracle Database SQL	8%	- Create a container database - Describe the Oracle Database architecture - Start up and shut down a database - Explain the installation of Oracle Database software
Retrieving Data Using the SQL SELECT Statement	5%	- Differentiate between SQL statements and SQL*Plus commands - Execute a basic SELECT statement - List the capabilities of SQL SELECT statements

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Oracle Database SQL Sample Questions (Q424-Q429):

NEW QUESTION # 424
Examine the description of the employees table:

Examine these requirements:
1- Display the last name, date of hire and the number of years of service for each employee.
2. If the employee has been employed 5 or more years but less than 10, display -5+ years of service".
3. If the employee has been employed 10 or more years but less than 15, display "10+ years of service".
4. If the employee has been employed 15 or more years, display "15-*- years of service".
5. If none of these conditions matches, display "<5 years of service".
6. Sort the results by the hire_date column.
Which statement satisfies all the requirements?

Answer: C

NEW QUESTION # 425
Which two are SQL features? (Choose two.)

A. providing variable definition capabilities
B. providing update capabilities for data in external files
C. providing graphical capabilities
D. processing sets of data
E. providing database transaction control

Answer: D,E

Explanation:
https://docs.oracle.com/database/121/TGSQL/tgsql_sqlproc.htm#TGSQL175
https://www.tutorialspoint.com/sql/sql-transactions.htm

NEW QUESTION # 426
Which three statements are true about Structured Query Language (SQL)?

A. It requires that data be contained in hierarchical data storage.
B. It provides independence for logical data structures being manipulated from the underlying physical data storage
C. It is the only language that can be used for both relational and object-oriented databases.
D. It is used to define encapsulation and polymorphism for a relational table.
E. It guarantees atomicity, consistency, isolation, and durability (ACID) features.
F. It best supports relational databases.

Answer: A,B,E

NEW QUESTION # 427
View the exhibit and examine the structure of the EMPLOYEES table.

You want to select all employees having 100 as their MANAGER_ID manages and their manager.
You want the output in two columns: the first column should have the employee's manager's LAST_NAME and the second column should have the employee's LAST_NAME.
Which SQL statement would you execute?

A. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eWHERE m.employee_id = e.manager_id AND e.manager_id=100
B. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE e.manager_id=100;
C. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON e.employee_id = m.manager_idWHERE m.manager_id=100;
D. SELECT m.last_name "Manager", e.last_name "Employee"FROM employees m JOIN employees eON m.employee_id = e.manager_idWHERE m.manager_id=100;

Answer: B

NEW QUESTION # 428
Examine the structure of the EMPLOYEES table.

There is a parent-child relationship between EMPLOYEE_ID and 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_ID is 123.
Which query provides the correct output?

A. SELECT e.last_name, m.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.employee_id = m.manager_id)WHERE e.employee_id = 123;
B. SELECT e.last_name, e.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.employee_id)WHERE e.employee_id = 123;
C. SELECT m.last_name, e.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.manager_id = m.manager_id)WHERE e.employee_id = 123;
D. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.manager_id = m.employee_id)AND e.employee_id = 123;

Answer: C

NEW QUESTION # 429
......

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