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Oracle 1z1-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q340-Q345):

NEW QUESTION # 340
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 B
B. Option C
C. Option A
D. Option D

Answer: C

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 # 341
Examine the description of the ENPLYEES table:

Which two queries return all rows for employees whose salary is greater than the average salary in their department?

A. SELECT.
FROM
SELECT e.", AVG (salary) OVER (PARTITION BY department id) avg_ sal
FROM employees e
WHERE salary > avg_ sal;
B. SELECT "
FROM employees
WHERE salary > ANY
SELECT AVG (salary)
EROM employees
GROUP BY department_ id);
C. SELECT
FROM employees
WHERE salary > AVG (salary) OVER (PARTITION BY department _ id);
D. SELECT"
FROM employees
WHERE salary >
( SELECT AVG
(salary) FROM
employees
E. SELECT"
FROM employees e1
WHERE salary >!
SELECT AVG (salary)
FROM employees e2
WHERE e1. Department _id = e2, department_ id

Answer: A,E

Explanation:
GROUP BY department _ id

NEW QUESTION # 342
Examine the structure of the MEMBERStable:
Name Null? Type
- --------------- ----------------- ---------------------------
MEMBER_ID NOT NULL VARCHAR2 (6)
FIRST_NAME VARCHAR2 (50)
LAST_NAME NOT NULL VARCHAR2 (50)
ADDRESS VARCHAR2 (50)
CITY VARCHAR2 (25)
STATE VARCHAR2 (3)
You want to display details of all members who reside in states starting with the letter A followed by exactly one character.
Which SQL statement must you execute?

A. SELECT * FROM MEMBERS WHERE state LIKE 'A%';
B. SELECT * FROM MEMBERS WHERE state LIKE 'A_%';
C. SELECT * FROM MEMBERS WHERE state LIKE '%A_';
D. SELECT * FROM MEMBERS WHERE state LIKE 'A_';

Answer: D

NEW QUESTION # 343
The EMPLOYEEStable contains columns EMP_ID of data type NUMBERand HIRE_DATE of data type DATE.
You want to display the date of the first Monday after the completion of six months since hiring.
The NLS_TERRITORY parameter is set to AMERICAin the session and, therefore, Sunday is the first day of the week.
Which query can be used?

A. SELECT emp_id, NEXT_DAY(ADD_MONTHS(hire_date, 6), 1) FROM employees;
B. SELECT emp_id, NEXT_DAY (MONTHS_BETWEEN (hire_date, SYSDATE), 6) FROM employees;
C. SELECT emp_id, ADD_MONTHS(hire_date, 6), NEXT_DAY(`MONDAY') FROM employees;
D. SELECT emp_id, NEXT_DAY(ADD_MONTHS(hire_date, 6), `MONDAY') FROM employees;

Answer: D

Explanation:
http://www.dba-oracle.com/t_add_months.htm

NEW QUESTION # 344
View the Exhibit and examine the structure of the CUSTOMERS table.

Evaluate the following SQL statement:

Which statement is true regarding the outcome of the above query?

A. It returns an error because WHERE and HAVING clauses cannot be used in the same SELECT statement.
B. It executes successfully.
C. It returns an error because WHERE and HAVING clauses cannot be used to apply conditions on the same column.
D. It returns an error because the BETWEEN operator cannot be used in the HAVING clause.

Answer: B

NEW QUESTION # 345
......

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