1z0-071시험대비덤프데모다운완벽한시험최신버전덤프

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Oracle 1z0-071 Exam Syllabus Topics:

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

>> 1z0-071시험대비 덤프데모 다운 <<

1z0-071퍼펙트 덤프데모 다운로드, 1z0-071최신버전 인기 덤프자료

지금 같은 세대에 많은 분들이 IT업계에 관심을 가지고 있습니다. 이렇게 인재가 많은 사회에서 IT관련인사들은 아직도 적은 편입니다. 면접 시에도 IT인증 자격증유무를 많이들 봅니다. 때문에 IT자격증이 많은 인기를 누리고 있습니다.이런 살아가기 힘든 사회에서 이런 자격증들 또한 취득하기가 넘 어렵습니다.Oracle 1z0-071인증시험 또한 아주 어려운 시험입니다. 많은 분들이 응시하지만 통과하는 분들은 아주 적습니다.

최신 Oracle PL/SQL Developer Certified Associate 1z0-071 무료샘플문제 (Q346-Q351):

질문 # 346
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 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;
B. (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;
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)INTERSECTSELECT last_name, hire_date, salaryFROM employeesWHERE hire_date > SYSDATE-180;

정답：A

질문 # 347
Which three actions can you perform only with system privileges?

A. Use the WITH GRANT OPTION clause.
B. Truncate a table in another schema.
C. Access flat files via a database, which are stored in an operating system directory.
D. Query any table in a database.
E. Log in to a database.
F. Execute a procedure in another schema.

정답：A,D,E

질문 # 348
Examine the data in the ORDERS table:

Examine the data in the INVOICES table:

Examine this query:
SELECT order_ id, order_ date FROM orders
INTERSECT
SELECT order_ 1d, order_ date FROM invoices;
Which two rows will it return?

A. 1 <null>
B. 2 <null>
C. 4 01-FEB-2019
D. 3 01-JAN-2019
E. 3 <null>
F. 5 01-MAR-2019

정답：A,C

설명：
The INTERSECT operator in SQL returns the results that are common to both of the SELECT statements. It functions similarly to a set intersection in mathematics. When comparing rows for the INTERSECT operation, Oracle Database uses all the expressions in the SELECT lists to derive the result set. NULL values are considered equal for the INTERSECT operator.
Evaluating the provided data from the ORDERS and INVOICES tables, let's see which rows have the same ORDER_ID and ORDER_DATE in both tables:
A: Order ID 3 has a NULL order date in the ORDERS table and does not match with any row in the INVOICES table, so it will not be returned.
B: Order ID 2 has a NULL order date in the ORDERS table but has a non-NULL order date in the INVOICES table, so it will not be returned.
C: Order ID 1 has a NULL order date in both tables, but INTERSECT considers NULLs as equal, so this will be returned.
D: Order ID 5 has a date of 01-MAR-2019 in the ORDERS table and 01-APR-2019 in the INVOICES table, so it will not be returned since the dates do not match.
E: Order ID 4 has a date of 01-FEB-2019 in both tables, so this row will be returned as it matches in both.
F: Order ID 3 has a NULL order date in the ORDERS table but has 01-JAN-2019 in the INVOICES table, so it will not be returned.
Based on this analysis, the query will return:
Order ID 1 with a NULL order date.
Order ID 4 with an order date of 01-FEB-2019.
So the correct answer is:

질문 # 349
View the exhibit and examine the description of the EMPLOYEES table. (Choose two.)

You executed this SQL statement:
SELECT first_name, department_id, salary
FROM employees
ORDER BY department_id, first_name, salary desc;
Which two statements are true regarding the result? (Choose two.)

A. The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.
B. The values in the FIRST_NAME column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID column.
C. The values in the SALARY column would be returned in descending order for all employees having the same value in the DEPARTMENT_ID and FIRST_NAME column.
D. The values in the FIRST_NAME column would be returned in ascending order for all employees having the same value in the DEPARTMENT_ID column.
E. The values in all columns would be returned in descending order.

정답：C,D

질문 # 350
Which two queries execute successfully?

A. SELECT NULLIF(NULL, 100) FROM DUAL
B. SELECT COALESCE(100, NULL, 200) FROM DUAL
C. SELECT NULLIF(100, 100) FROM DUAL
D. SELECT CO ALESCE(100, 'A' ) FROM DUAL
E. SELECT NULLIF(100, 'A') FROM DUAL

정답：A,B,C,D

설명：
The NULLIF and COALESCE functions in Oracle SQL serve specific purposes:
* Option A: Executes successfully. NULLIF returns NULL if the two arguments are equal; otherwise, it
* returns the first argument.
* Option B: Also correct. COALESCE returns the first non-null value among its arguments.
* Option C: This will not execute successfully because NULLIF requires both arguments to be of comparable data types. Comparing an integer to a character ('100' to 'A') is invalid.
* Option D: Executes successfully. NULLIF returns NULL because it compares NULL to a number, which is valid (though always yields NULL).
* Option E: Executes successfully. COALESCE accepts any data type as it returns the first non-null value, irrespective of type consistency among the arguments.

질문 # 351
......

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