최신버전1z0-071최신덤프문제모음집덤프는Oracle Database SQL시험대비최고의자료

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오라클 1z0-071 (Oracle Database SQL) 시험은 SQL 프로그래밍 언어와 오라클 데이터베이스에 대한 전문 지식을 증명하고자 하는 IT 전문가들을 위한 자격증 시험입니다. 이 시험은 오라클 인증 협회 (OCA) 자격증 프로그램의 일부이며 오라클 인증 전문가가 되기를 원하는 전문가들에게 필수적인 요건입니다. 이 시험은 SQL 문, 데이터 조작, 데이터베이스 개체 및 데이터 제어 언어 등 다양한 주제를 다룹니다.

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1z0-071퍼펙트 덤프문제 - 1z0-071최신 업데이트 인증덤프자료

Oracle 1z0-071인증덤프는 최근 출제된 실제시험문제를 바탕으로 만들어진 공부자료입니다. Oracle 1z0-071 시험문제가 변경되면 제일 빠른 시일내에 덤프를 업데이트하여 최신버전 덤프자료를Oracle 1z0-071덤프를 구매한 분들께 보내드립니다. 시험탈락시 덤프비용 전액환불을 약속해드리기에 안심하시고 구매하셔도 됩니다.

Oracle 1Z1-071 인증 시험을 통과 한 응시자는 Oracle Database SQL 인증을받을 수 있습니다. 이 인증은 3 년 동안 유효하며, 후보자는 인증을 유지하기 위해 추천해야합니다.

최신 Oracle PL/SQL Developer Certified Associate 1z0-071 무료샘플문제 (Q171-Q176):

질문 # 171
Examine the description of the PRODUCTS table:

Which three queries use valid expressions? (Choose three.)

A. SELECT product_id, (unit_price * 0.15 / (4.75 + 552.25)) FROM products;
B. SELECT product_id, unit_price, unit_price + surcharge FROM products;
C. SELECT product_id, unit_price, 5 "Discount", unit_price + surcharge - discount FROM products;
D. SELECT product_id, expiry_date * 2 FROM products;
E. SELECT product_id, unit_price | | 5 "Discount", unit_price + surcharge - discount FROM products;
F. SELECT product_id, (expiry_date - delivery_date) * 2 FROM products;

정답：A,E,F

질문 # 172
Examine the commands used to create department details and course details:

You want to generate a list of all department IDs that do not exist in the COURSE_DETAILS table. You execute the SQL statement:

What is the outcome?

A. It fails because the ON clause condition is not valid
B. it fails because the Join type used is Incorrect
C. It executes successfully and displays the required list.
D. It executes successfully but displays an incorrect list.

정답：B

질문 # 173
Which three statements are true about performing Data Manipulation Language (DML) operations on a view In an Oracle Database?

A. Views cannot be used to query rows from an underlying table if the table has a PRIPOARY KEY and the PRIMARY KEY columns are not referenced in the defining query of the view.
B. Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains the DISTINCT keyword.
C. Views cannot be used to add on modify rows in an underlying table if the defining query of the view contains aggregating functions.
D. The WITH CHECK clause has no effect when deleting rows from the underlying table through the view.
E. Views cannot be used to add rows to an underlying table if the table has columns with NOT NULL constraints lacking default values which are not referenced in the defining query of the view.
F. Insert statements can always be done on a table through a view.

정답：B,C,E

설명：
When performing DML operations on a view, certain restrictions apply:
D . Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains the DISTINCT keyword: If the defining query of a view contains the DISTINCT keyword, it cannot be used for certain DML operations, such as INSERT and UPDATE, because the set of rows it represents is not directly modifiable.
E . Views cannot be used to add or modify rows in an underlying table if the defining query of the view contains aggregating functions: Aggregating functions create a result that summarises multiple rows and cannot be reversed to point to a single row for modification.
F . Views cannot be used to add rows to an underlying table if the table has columns with NOT NULL constraints lacking default values which are not referenced in the defining query of the view: If a column with a NOT NULL constraint is not included in the view's defining query, and it does not have a default value, it is not possible to insert through the view because it would violate the NOT NULL constraint.
Reference:
Oracle Database SQL Language Reference 12c, particularly the sections on views and the restrictions on DML operations through views.

질문 # 174
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 employees e1
WHERE salary >!
SELECT AVG (salary)
FROM employees e2
WHERE e1. Department _id = e2, department_ id
B. SELECT.
FROM
SELECT e.", AVG (salary) OVER (PARTITION BY department id) avg_ sal
FROM employees e
WHERE salary > avg_ sal;
C. SELECT"
FROM employees
WHERE salary >
( SELECT AVG
(salary) FROM
employees
GROUP BY department _ id
D. SELECT "
FROM employees
WHERE salary > ANY
SELECT AVG (salary)
EROM employees
GROUP BY department_ id);
E. SELECT
FROM employees
WHERE salary > AVG (salary) OVER (PARTITION BY department _ id);

정답：A,E

설명：
To return all rows for employees whose salary is greater than the average salary in their department, you would use either a subquery or an analytic function:
* Option B:
* SELECT ... FROM employees WHERE salary > AVG(salary) OVER (PARTITION BY department_id);
* This uses the window function AVG with PARTITION BY to calculate the average salary per department, and it compares each employee's salary to this average.
* Option C:
* SELECT ... FROM employees e1 WHERE salary > (SELECT AVG(salary) FROM
employees e2 WHERE e1.department_id = e2.department_id);
* This correlated subquery compares each employee's salary to the average salary in their department using a subquery to calculate the average salary for that department.
Options A, D, and E are incorrect because:
* Option A: The use of ANY with the subquery does not ensure comparison with the average salary of their respective department.
* Option D: This is syntactically incorrect; the subquery alias avg_sal is not accessible outside the subquery.
* Option E: The subquery does not correlate with the outer query to ensure that each employee's salary is compared to the average salary of their respective department.

질문 # 175
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, salary
FROM employees
WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = 100) UNION ALL (SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000);
B. (SELECT last_name, hire_date, salary
FROM employees
WHERE salary > 10000
UNION ALL
SELECT last_name, hire_date, salary
FROM employees
WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100)) UNION SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180;
C. SELECT last_name, hire_date, salary
FROM employees
WHERE salary > 10000
UNION ALL
SELECT last_name, hire_date, salary
FROM employees
WHERE manager_ID = (SELECT employee_id FROM employees WHERE employee_id = 100) INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE- 180;
D. SELECT last_name, hire_date, salary
FROM employees
WHERE manager_id = (SELECT employee_id FROM employees WHERE employee_id = '100') UNION SELECT last_name, hire_date, salary FROM employees WHERE hire_date > SYSDATE -180 INTERSECT SELECT last_name, hire_date, salary FROM employees WHERE salary > 10000;

정답：D

질문 # 176
......

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