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Oracle 1z0-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q45-Q50):

NEW QUESTION # 45
You need to display the first names of all customers from the customers table that contain the character 'e' and have the character 'a' in the second last position.
Which query would give the required output?

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

Answer: B

Explanation:
The SUBSTR(string, start position, number of characters) function accepts three parameters and returns a string consisting of the number of characters extracted from the source string, beginning at the specified start position:
substr('http://www.domain.com', 12, 6) = domain
The position at which the first character of the returned string begins. When position is 0 (zero), then it is treated as 1. When position is positive, then the function counts from the beginning of string to find the first character.
When position is negative, then the function counts backward from the end of string.
substring_length
The length of the returned string. SUBSTR calculates lengths using characters as defined by the input character set. SUBSTRB uses bytes instead of characters. SUBSTRC uses Unicode complete characters.
SUBSTR2 uses UCS2 code points. SUBSTR4 uses UCS4 code points. When you do not specify a value for this argument, then the function The INSTR(source string, search item, [start position], [nth occurrence of search item]) function returns a number that represents the position in the source string, beginning from the given start position, where the nth occurrence of the search item begins:
instr('http://www.domain.com', '.', 1, 2) = 18

NEW QUESTION # 46
View the Exhibit and examine the description of the EMPLOYEES table.

You want to calculate the total renumeration for each employee. Total renumeration is the sum of the annual salary and the percentage commission earned for a year. Only a few employees earn commission.
Which SQL statement would you execute to get the desired output?

A. SELECT first_name, salary*12 + NVL(salary,0)*commission_pct, "Total"FROM EMPLOYEES;
B. SELECT first_name, salary, salary*12+(salary*NVL2 (commission_pct,
salary,salary+commission_pct))"Total"FROM EMPLOYEES;
C. SELECT first_name, salary, salary*12+salary*commission_pct "Total"FROM EMPLOYEES;
D. SELECT first_name, salary (salary + NVL (commission_pct, 0)*salary)*12 "Total"FROM EMPLOYEES;

Answer: B

NEW QUESTION # 47
View the Exhibit and examine the structure of ORDERS and CUSTOMERS tables. (Choose the best answer.)

You executed this UPDATE statement:

Which statement is true regarding the execution?

A. It would execute and restrict modifications to the columns specified in the SELECT statement.
B. It would not execute because a subquery cannot be used in the WHERE clause of an UPDATE statement.
C. It would not execute because a SELECT statement cannot be used in place of a table name.
D. It would not execute because two tables cannot be referenced in a single UPDATE statement.

Answer: A

NEW QUESTION # 48
Examine the description of the CUSTOMERS table:

Which two SELECT statements will return these results:
CUSTOMER_ NAME
--------------------
Mandy
Mary

A. SELECT customer_ name FROM customers WHERE customer_ name LIKE ' % a % ' ;
B. SELECT customer_ name FROM customers WHERE customer_ name='*Ma*';
C. SELECT customer_ name FROM customers WHERE UPPER (customer name) LIKE 'MA&';
D. SELECT customer_ name FROM customers WHERE customer_ name KIKE .*Ma*';
E. SELECT customer_ name FROM customers WHERE UPPER (customer_ name ) LIKE 'MA*. ;
F. SELECT customer_ name FROM customers WHERE customer name LIKE 'Ma%' ;
G. SELECT customer_ name FROM customers WHERE customer name LIKE 'Ma*';

Answer: A,F

NEW QUESTION # 49
View the Exhibit and examine the structure in the DEPARTMENTS tables. (Choose two.)

Examine this SQL statement:
SELECT department_id "DEPT_ID", department_name, 'b' FROM
departments
WHERE departments_id=90
UNION
SELECT department_id, department_name DEPT_NAME, 'a' FROM
departments
WHERE department_id=10
Which two ORDER BY clauses can be used to sort output?

A. ORDER BY 3;
B. ORDER BY 'b';
C. ORDER BY DEPT_NAME;
D. ORDER BY DEPT_ID;

Answer: A,D

NEW QUESTION # 50
......

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