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Oracle 1z1-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q125-Q130):

NEW QUESTION # 125
View the exhibit and examine the description of the PRODUCT_INFORMATION table.

Which SQL statement would retrieve from the table the number of products having LIST_PRICE as NULL?

A. SELECT COUNT (DISTINCT list_price)FROM product_informationWHERE list_price is NULL
B. SELECT COUNT (NVL(list_price, 0))FROM product_informationWHERE list_price is NULL
C. SELECT COUNT (list_price)FROM product_informationWHERE list_price i= NULL
D. SELECT COUNT (list_price)FROM product_informationWHERE list_price is NULL

Answer: B

NEW QUESTION # 126
You want to write a query that prompts for two column names and the WHERE condition each time it is executed in a session but only prompts for the table name the first time it is executed.
The variables used in your query are never undefined in your session?
Which query can be used?

A. Option A
B. Option C
C. Option D
D. Option E
E. Option B

Answer: C

NEW QUESTION # 127
Which three queries execute successfully?

A. SELECT SYSDATE / DATE '2019-01-01' - 1 FROM DUAL
B. SELECT SYSDATE - 1 - DATE'2019-01-01' EROM DUAL;
C. SELECT 1 / SYSDATE - DATE '2019-01-01' FROM DUAL;
D. SELECT SYSDATE - DATE '2019-01-01' - 1 FROM DUAL;
E. SELECT 1 - SYSDATE- DATE '2019-01-01' FROM DUAL;
F. SELECT (SYSDATE-DATE '2019-01-01') / 1 FROM DUAL;

Answer: B,D,F

Explanation:
In Oracle SQL, date arithmetic can be performed directly, subtracting one date from another to get the number of days between them:
* Option A:
* SELECT (SYSDATE - DATE '2019-01-01') / 1 FROM DUAL; This query successfully calculates the difference between the current date and a specific date.
* Option D:
* SELECT SYSDATE - DATE '2019-01-01' - 1 FROM DUAL; This query subtracts a specific date from the current date and then subtracts 1 more day, which will execute successfully.
* Option F:
* SELECT SYSDATE - 1 - DATE '2019-01-01' FROM DUAL; Similar to D, this will successfully compute the date difference minus one day.
Options B, C, and E are incorrect because:
* Option B and Option C: You cannot divide by a date or divide a date by a number.
* Option E: You cannot subtract a date from a number.

NEW QUESTION # 128
Which two statements are true regarding the COUNTfunction? (Choose two.)

A. A SELECTstatement using the COUNTfunction with a DISTINCTkeyword cannot have a WHEREclause.
B. COUNT(DISTINCT inv_amt)returns the number of rows excluding rows containing duplicates and NULLsin the INV_AMTcolumn.
C. It can only be used for NUMBERdata types.
D. COUNT(*)returns the number of rows including duplicate rows and rows containing NULLvalue in any column.
E. COUNT(inv_amt)returns the number of rows in a table including rows with NULLin the INV_AMTcolumn.

Answer: B,D

Explanation:
Explanation/Reference:

NEW QUESTION # 129
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, m.manager_id
FROM employees e LEFT OUTER JOIN employees m
on (e.employee_id = m.manager_id)
WHERE e.employee_id = 123;
B. 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;
C. 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;
D. 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;

Answer: C

NEW QUESTION # 130
......

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