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Oracle 1z0-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q10-Q15):

NEW QUESTION # 10
Which two statements are true about the COUNT function?

A. A SELECT statement using the COUNT function with a DISTINCT keyword cannot have a WHERE clause.
B. COUNT(*) returns the number of rows in a table including duplicate rows and rows containing NULLs in any column.
C. It can only be used for NUMBER data types.
D. COUNT(inv_amt) returns the number of rows in a table including rows with NULL in the INV_AMT column.
E. COUNT (DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULLs in the INV_AMT column

Answer: B,E

Explanation:
The COUNT function is one of the most commonly used aggregate functions in SQL for determining the number of items in a group. Here's why the correct answers are B and C:
* A: Incorrect. COUNT can be used with any data type, not just NUMBER. It counts rows without regard to data type.
* B: Correct. COUNT(DISTINCT inv_amt) counts the number of unique non-null values in the column inv_amt. It excludes duplicates and ignores NULL values.
* C: Correct. COUNT(*) counts all rows in a table, including those with duplicates and those with NULLs in any column. It is a total row count regardless of content.
* D: Incorrect. You can use a WHERE clause with COUNT. The WHERE clause filters rows before counting, which is standard in SQL.
* E: Incorrect. COUNT(inv_amt) counts the rows where inv_amt is not NULL. Rows with NULL in the inv_amt column are not counted.

NEW QUESTION # 11
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. SELECT cust_first_name
FROM customers
WHERE INSTR(cust_first_name, `e')IS NOT NULL AND
SUBSTR(cust_first_name, 1, -2)='a';
B. SELECT cust_first_name
FROM customers
WHERE INSTR(cust_first_name, `e')<>0 AND
SUBSTR(cust_first_name, -2, 1)='a';
C. SELECT cust_first_name
FROM customers
WHERE INSTR(cust_first_name, `e')<>'' AND
SUBSTR(cust_first_name, -2, 1)='a';
D. SELECT cust_first_name
FROM customers
WHERE INSTR(cust_first_name, `e')<>0 AND
SUBSTR(cust_first_name, LENGTH(cust_first_name), -2)='a';

Answer: B

NEW QUESTION # 12
Examine these SQL statements which execute successfully:

Which two statements are true after execution? (Choose two.)

A. The foreign key constraint will be enabled and IMMEDIATE.
B. The foreign key constraint will be enabled and DEFERRED.
C. The primary key constraint will be enabled and DEFERRED.
D. The foreign key constraint will be disabled.
E. The primary key constraint will be enabled and IMMEDIATE.

Answer: A,C

Explanation:
Explanation/Reference:

NEW QUESTION # 13
Examine these requirements:
1. Display book titles for books purchased before January 17, 2007 costing less than 500 or more than 1000.
2. Sort the titles by date of purchase, starting with the most recently purchased book.
Which two queries can be used?

A. SELECT book_title FROM books WHERE (price BETWEEN 500 AND 1000) AND
(purchase_date<'17-JAN-2007') ORDER BY purchase_date;
B. SELECT book_title FROM books WHERE (price< 500 OR >1000) AND (purchase date<
'17-JAN-2007') ORDER BY purchase date DESC;
C. SELECT book_title FROM books WHERE (price IN (500, 1000)) AND (purchase date <
'17-JAN-2007') ORDER BY purchase_date ASC;
D. SELECT book_title FROM books WHERE (price NOT BETWEEN 500 AND 1000) AND (purchase_date< '17-JAN-2007') ORDER BY purchase_date DESC;

Answer: B,D

NEW QUESTION # 14
Examine the structure of the EMPLOYEES table.
You must display the details of employees who have manager with MANAGER_ID 100, who were hired in the past 6 months and who have salaries greater than 10000.
Which query would retrieve the required result?

A. SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100)INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;
B. SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = '100')UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000;
C. SELECT last_name, hire_date, salaryFROM employeesWHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100)UNION ALL(SELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000);
D. (SELECT last_name, hire_date, salaryFROM employeesWHERE salary > 10000UNION ALLSELECT last_name, hire_date, salaryFROM employeesWHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100))UNIONSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;

Answer: B

NEW QUESTION # 15
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