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Oracle 1z1-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q93-Q98):

NEW QUESTION # 93
Which three statements are true regarding group functions? (Choose three.)

A. They can be used on columns or expressions.
B. They can be used with a SQL statement that has a GROUP BY clause.
C. They can be passed as an argument to another group function.
D. They can be used only on one column in the SELECT clause.
E. They can be used together with the single-row functions in the SELECT clause.

Answer: A,C,E

Explanation:
References:
https://www.safaribooksonline.com/library/view/mastering-oracle-sql/0596006322/ch04.html

NEW QUESTION # 94
Examine the description or the CUSTOMERS 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_AMONT FROM customers
WHERE cust_income_level <> NULL AND due_amount <> NULL;
B. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers
WHERE cust_income_level IS NOT NULL AND cust_credit_limit IS NOT NULL;
C. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMOUNT FROM customers
WHERE cust_income_level != NULL AND cust_credit_level != NULL;
D. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers
WHERE cust_income_level IS NOT NULL AND due_amount IS NOT NULL;
E. SELECT cust_first_name, cust_credit_limit * .05 AS DUE_AMONT FROM customers
WHERE cust_income_level != NULL AND due_amount != NULL;

Answer: B

NEW QUESTION # 95
View the exhibit and examine the structures of the EMPLOYEESand DEPARTMENTStables.

You want to update EMPLOYEEStable as follows:
Update only those employees who work in Boston or Seattle (locations 2900 and 2700).

Set department_idfor these employees to the department_idcorresponding to London

(location_id 2100).
Set the employees' salary in location_id2100 to 1.1 times the average salary of their department.

Set the employees' commission in location_id2100 to 1.5 times the average commission of their

department.
You issue the following command:

What is outcome?

A. It generates an error because a subquery cannot have a join condition in a UPDATEstatement.
B. It executes successfully but does not give the desired update
C. It executes successfully and gives the desired update
D. It generates an error because multiple columns (SALARY, COMMISSION)cannot be specified together in an UPDATEstatement.

Answer: B

NEW QUESTION # 96
Examine the ORDER _ITEms table:

Which two queries return rows where QUANTITY is a multiple of ten?

A. SELECT FROM order_ items WHERE quantity / 10 = TRUNC (quantity);
B. SELECT * FROM order_ items WHERE MOD (quantity, 10) = 0;
C. SELECT" FROM order_ _items WHERE quantity = ROUND (quantity, 1);
D. SELECT" FROM order_ items WHERE FLOOR (quantity / 10) = TRUNC (quantity / 10);
E. SELECT * FROM order_ items WHERE quantity = TRUNC (quantity, -1);

Answer: B,E

NEW QUESTION # 97
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.

The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?

A. INTERSECT
B. UNION ALL
C. MINUS
D. UNION

Answer: C

NEW QUESTION # 98
......

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