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Oracle 1z0-071 Exam Syllabus Topics:

Section	Weight	Objectives
Aggregating Data Using Group Functions	10%	- GROUP BY and HAVING clauses - SUM, AVG, COUNT, MAX, MIN functions - Handling NULLs in group functions
Manipulating Data and Managing Transactions		- Transaction control: COMMIT, ROLLBACK, SAVEPOINT - Multi-table inserts - INSERT, UPDATE, DELETE, MERGE statements
Displaying Data from Multiple Tables	15%	- Cartesian products - Equijoins and non-equijoins - Self-joins and outer joins
Using Data Dictionary Views		- Retrieve metadata and object information
Working with Time Zones and Date/Time Data		- Time zone conversion functions - TIMESTAMP, INTERVAL data types
Restricting and Sorting Data	15%	- Substitution variables, DEFINE and VERIFY commands - Operator precedence - ORDER BY clause and sorting rules - WHERE clause and filtering conditions
Controlling User Access		- Roles, granting and revoking privileges - System and object privileges
Using SET Operators		- Ordering results with set operations - UNION, UNION ALL, INTERSECT, MINUS operators
Managing Views		- Updatable and read-only views - Create, modify and drop views
Retrieving Data Using SQL SELECT Statement	20%	- Column aliases, concatenation, literals and DISTINCT - SELECT statement syntax and usage - Arithmetic operations and NULL handling
Using Single-Row Functions to Customize Output	20%	- Character, number, and date functions - Nested functions usage
Managing Tables and Constraints		- Temporary and external tables - Constraints: primary key, foreign key, unique, check, not null - Data types, creating/altering/dropping tables
Using Subqueries to Solve Queries		- Single-row and multi-row subqueries - Correlated subqueries for update/delete
Conversion Functions and Conditional Expressions	10%	- TO_CHAR, TO_NUMBER, TO_DATE functions - NVL, NULLIF, COALESCE and CASE expressions - Implicit and explicit data type conversion
Managing Indexes, Synonyms and Sequences		- Create and use synonyms - Create and maintain indexes - Create and use sequences
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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최신 Oracle PL/SQL Developer Certified Associate 1z0-071 무료샘플문제 (Q128-Q133):

질문 # 128
Which three actions can you perform by using the ALTER TABLE command?

A. Drop pseudo columns from a table.
B. Enable or disable constraints on a table.
C. Rename a table
D. Lock a set of rows in a table CE Rename a table.
E. Restrict all DML statements on a table.
F. Drop all columns simultaneously from a table.

정답：B,C,D

질문 # 129
Examine the structure of the SALES table.

Examine this statement:

Which two statements are true about the SALES1 table? (Choose two.)

A. It has PRIMARY KEY and UNIQUE constraints on the selected columns which had those constraints in the SALES table.
B. It is created with no rows.
C. It will have NOT NULL constraints on the selected columns which had those constraints in the SALES table.
D. It will not be created because of the invalid WHERE clause.
E. It will not be created because the column-specified names in the SELECT and CREATE TABLE clauses do not match.

정답：B,C

설명：
Explanation/Reference:

질문 # 130
Evaluate the following statement: INSERT ALL WHEN order_total < 10000 THEN INTO small_orders WHEN order_total > 10000 AND order_total < 20000 THEN INTO medium_orders WHEN order_total > 2000000 THEN INTO large_orders SELECT order_id, order_total, customer_id FROM orders;
Which statement is true regarding the evaluation of rows returned by the subquery in the INSERT statement?

A. They are evaluated by the first WHEN clause. If the condition is false, then the row would be evaluated by the subsequent WHEN clauses.
B. They are evaluated by the first WHEN clause. If the condition is true, then the row would be evaluated by the subsequent WHEN clauses.
C. The INSERT statement would give an error because the ELSE clause is not present for support in case none of the WHEN clauses are true.
D. They are evaluated by all the three WHEN clauses regardless of the results of the evaluation of any other WHEN clause.

정답：D

질문 # 131
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 TO_CHAR(NVL(cust_credit_limit * .15,'Not Available')) FROM customers;
C. SELECT NVL(cust_credit_limit * .15,'Not Available') FROM customers;
D. SELECT NVL2(cust_credit_limit * .15,'Not Available') FROM customers;
E. SELECT NVL2(cust_credit_limit,TO_CHAR(cust_credit_limit * .15),'NOT Available') FROM customers;
F. 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.

정답：B,E

설명：
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.

질문 # 132
Which CREATE TABLE statement is valid?

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

정답：B

질문 # 133
......

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