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Oracle 1z0-071 Exam Syllabus Topics:

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

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Oracle Database SQL Sample Questions (Q248-Q253):

NEW QUESTION # 248
Examine the description of the CUSTONERS table:

CUSTNO is the PRIMARY KEY.
You must determine if any customers' details have been entered more than once using a different CUSTNO, by listing all duplicate names.
Which two methods can you use to get the required result?

A. LEFT OUTER JOIN with self join
B. self Join
C. RIGHT OUTER JOIN with self join
D. PULL OUTER JOIN with self join
E. subquery

Answer: B,E

NEW QUESTION # 249
You need to display the first names of all customers from the CUSTOMERStable that contain the character 'e' and have the character 'a' in the second last position.
Which query would give the required output?
SELECT cust_first_name

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

Answer: A

NEW QUESTION # 250
View the Exhibit and examine the details of the PRODUCT_INFORMATION table.

Evaluate this SQL statement:
SELECT TO_CHAR (list_price, '$9,999')
From product_information;
Which two statements are true regarding the output? (Choose two.)

A. A row whose LIST_PRICE column contains value 11235.90 would be displayed as #######.
B. A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,123.
C. A row whose LIST_PRICE column contains value 11235.90 would be displayed as $1,123.
D. A row whose LIST_PRICE column contains value 1123.90 would be displayed as $1,124.

Answer: A,D

NEW QUESTION # 251
The CUSTOMERS table has a CUST_CREDT_LIMIT column of data type number.
Which two queries execute successfully?

A. SLECT NVL(TO_CHAR(cust_credit_limit * .15),'Not available') from customers;
B. SELECT NVL2(cust_credit_limit * .15,'Not Available') FROM customers;
C. SELECT TO_CHAR(NVL(cust_credit_limit * .15,'Not Available')) FROM customers;
D. SELECT NVL(cust_credit_limit * .15,'Not Available') FROM customers;
E. SELECT NVL2(cust_credit_limit,TO_CHAR(cust_credit_limit * .15),'NOT Available') FROM customers;

Answer: C,E

Explanation:
A). True - The TO_CHAR function is used correctly here to convert the numeric value to a string, and NVL handles the case where cust_credit_limit might be NULL. The expression inside NVL computes 15% of the credit limit or displays 'Not Available' if the credit limit is NULL. The syntax is correct.
B). False - The NVL2 function requires three parameters: the expression to check for NULL, the value to return if it's not NULL, and the value to return if it is NULL. The given usage lacks the required parameters and syntax.
C). False - The NVL function expects both parameters to be of the same data type. Since the second parameter
'Not Available' is a string, it causes a data type conflict with the numerical result of the first parameter.
D). False - The keyword SELECT is misspelled as SLECT, making the syntax incorrect.
E). True - This query uses NVL2 correctly by checking if cust_credit_limit is not NULL, then applying TO_CHAR to compute 15% of it and converting it to string, or returning 'NOT Available' if it is NULL. The syntax and function usage are correct.

NEW QUESTION # 252
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 + NVL (commission_pct, 0)*salary)*12 "Total"FROM EMPLOYEES;
C. SELECT first_name, salary, salary*12+salary*commission_pct "Total"FROM EMPLOYEES;
D. SELECT first_name, salary, salary*12+(salary*NVL2 (commission_pct,
salary,salary+commission_pct))"Total"FROM EMPLOYEES;

Answer: D

NEW QUESTION # 253
......

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