最近更新の2023年10月テストエンジンとPDF Oracle 1z0-071テストあなたの最速Oracle合格準備を保証させる！ [Q153-Q169]

最近更新の2023年10月テストエンジンとPDF Oracle 1z0-071テストあなたの最速Oracle合格準備を保証させる！

完全版1z0-071練習テスト305別格な問題と解釈が待ってます。今すぐゲット！

質問 # 153
The CUSTOMERS table has a CUST_CREDT_LIMIT column of data type number.
Which two queries execute successfully?

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

正解：B、E

質問 # 154
Examine the structure of the BOOKS_ TRANSACTIONS table:

Examine the SQL statement:

Which statement is true about the outcome?

A. It displays details for only members A101and A102 who have borrowed before today with RM as TRANSACTION_TYPE.
B. It displays details for members who have borrowed before today with RM as TRANSACTION_TYPE and the details for members A101 or A102.
C. It displays details only for members who have borrowed before today with RM as TRANSACTION_TYPE.
D. It displays details for members who have borrowed before today's date with either RM as TRANSACTION_TYPE or MEMBER_ID as A101 and A102.

正解：C

質問 # 155
Examine the description of the CUSTOMERStable:

You need to display last names and credit limits of all customers whose last name starts with A or B in lower or upper case, and whose credit limit is below 1000.
Examine this partial query:

Which two WHEREconditions give the required result? (Choose two.)

正解：C、E

質問 # 156
Which two are SQL features?

A. processing sets of data
B. providing database transaction control
C. providing variable definition capabilities.
D. providing update capabilities for data in external files
E. providing graphical capabilities

正解：A、B

質問 # 157
View the Exhibit and examine the structure of ORDERS and ORDER_ITEMS tables.
ORDER_ID is the primary key in the ORDERS table. It is also the foreign key in the ORDER_ITEMS table wherein it is created with the ON DELETE CASCADE option.
Which DELETE statement would execute successfully?

A. DELETE order_idFROM ordersWHERE order_total < 1000;
B. DELETE orders o, order_items IWHERE o.order_id = i.order_id;
C. DELETEFROM ordersWHERE (SELECT order_idFROM order_items);
D. DELETE ordersWHERE order_total < 1000;

正解：C

質問 # 158
View the exhibit and examine the data in the PROJ_TASK_DETAILS table. (Choose the best answer.)

The PROJ_TASK_DETAILS table stores information about project tasks and the relation between them.
The BASED_ON column indicates dependencies between tasks.
Some tasks do not depend on the completion of other tasks.
You must generate a report listing all task IDs, the task ID of any task upon which it depends and the name of the employee in charge of the task upon which it depends.
Which query would give the required result?

A. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p FULL OUTER JOIN proj_task_details d
ON (p.based_on = d.task_id);
B. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p JOIN proj_task_details d
ON (p.task_id = d.task_id);
C. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p JOIN proj_task_details d
ON (p.based_on = d.task_id);
D. SELECT p.task_id, p.based_on, d.task_in_charge
FROM proj_task_details p LEFT OUTER JOIN proj_task_details d
ON (p.based_on = d.task_id);

正解：D

質問 # 159
Examine the structure of the MEMBERS table: (Choose the best answer.)

Examine the SQL statement:
SQL > SELECT city, last_name LNAME FROM MEMBERS ORDER BY 1, LNAME DESC; What would be the result execution?

A. It displays all cities in ascending order, within which the last names are further sorted in descending order.
B. It displays all cities in descending order, within which the last names are further sorted in descending order.
C. It fails because a column number and a column alias cannot be used together in the ORDER BY clause.
D. It fails because a column alias cannot be used in the ORDER BY clause.

正解：A

質問 # 160
View the exhibit and examine the structure of the EMPLOYEEStable.

You want to display all employees and their managers having 100 as the MANAGER_ID. You want the output in two columns: the first column would have the LAST_NAMEof the managers and the second column would have LAST_NAMEof the employees.
Which SQL statement would you execute?
SELECT m.last_name "Manager", e.last_name "Employee"

A. FROM employees m JOIN employees e
ON e.employee_id = m.manager_id
WHERE m.manager_id = 100;
SELECT m.last_name "Manager", e.last_name "Employee"
B. FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE m.manager_id = 100;
SELECT m.last_name "Manager", e.last_name "Employee"
C. FROM employees m JOIN employees e
ON m.employee_id = e.manager_id
WHERE e.manager_id = 100;
SELECT m.last_name "Manager", e.last_name "Employee"
D. FROM employees m JOIN employees e
WHERE m.employee_id = e.manager_id AND e.manager_id = 100

正解：C

質問 # 161
Which statement falls to execute successfully?

A. SELECT *
FROM employees e
JOIN departments d
ON e.department_id=d.department_id
WHERE d.department_id=90;
B. SELECT *
FROM employees e
JOIN departments d
ON d.departments_id=90
WHERE e.department_id=d.department_id;
C. SELECT *
FROM employees e
JOIN department d
WHERE e.department_id=d.department_id
AND d.department_id=90;
D. SELECT *
FROM employees e
JOIN departments d
ON e.department_id=d.department_id
AND d.department_id=90;

正解：B

質問 # 162
Examine the structure of the PROGRAMS table:

Which two SQL statements would execute successfully?

A. SELECT TO_DATE (NVL (SYSDATE-END_DATE, SYSDATE)) FROM programs;
B. SELECT NVL (MONTHS_BETWEEN (start_date, end_date), 'Ongoing') FROM programs;
C. SELECT NVL (TO_CHAR (MONTHS_BETWEEN (start-date, end_date)), 'Ongoing') FROM programs
D. SELECT NVL (ADD_MONTHS (END_DATE,1) SYSDATE) FROM programs;

正解：A、C

質問 # 163
View the Exhibit and examine the structure in the EMPLOYEES tables.
Evaluate the following SQL statement:
SELECT employee_id, department_id
FROM employees
WHERE department_id= 50 ORDER BY department_id
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=90
UNION
SELECT employee_id, department_id
FROM employees
WHERE department_id=10;
What would be the outcome of the above SQL statement?

A. The statement would execute successfully but it will ignore the ORDER BY clause and display the rows in random order.
B. The statement would not execute because the positional notation instead of the column name should be used with the ORDER BY clause.
C. The statement would not execute because the ORDER BY clause should appear only at the end of the SQL statement, that is, in the last SELECT statement.
D. The statement would execute successfully and display all the rows in the ascending order of DEPARTMENT_ID.

正解：C

質問 # 164
In which normal form is a table, if it has no multi-valued attributes and no partial dependencies?

A. first normal form
B. fourth normal form
C. second normal form
D. third normal form

正解：C

解説：
References:
https://blog.udemy.com/database-normal-forms/

質問 # 165
View the Exhibit and examine the structure of ORDER_ITEMSand ORDERStables.

You need to remove from the ORDER_ITEMStable those rows that have an order status of 0 or 1 in the ORDERStable.
Which two DELETEstatements are valid (Choose two.)
DELETE *

A. FROM order_items
WHERE order_id IN (SELECT order_id
FROM orders
WHERE orders_status in (0,1));
B. FROM (SELECT * FROM order_items I,orders o
WHERE i.order_id = o.order_id AND order_status IN (0,1));
DELETE FROM order_items i
C. FROM order_items
WHERE order_id IN (SELECT order_id)
FROM orders
WHERE order_status IN (0,1));
DELETE
D. WHERE order_id = (SELECT order_id FROM orders o
WHERE i.order_id = o.order_id AND order_status IN (0,1));
DELETE

正解：A、B

質問 # 166
Examine the structure of the MEMBERS table:

You want to display details of all members who reside in states starting with the letter A followed by exactly one character.
Which SQL statement must you execute?

A. SELECT * FROM MEMBERS WHERE state LIKE 'A%';
B. SELECT * FROM MEMBERS WHERE state LIKE 'A_';
C. SELECT * FROM MEMBERS WHERE state LIKE 'A_%';
D. SELECT * FROM MEMBERS WHERE state LIKE '%A_';

正解：B

質問 # 167
Examine the commands used to create DEPARTMENT_DETAILSand COURSE_DETAILS tables:

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

What is the outcome?

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

正解：D

質問 # 168
The BOOKS_TRANSACTIONS table exists in your schema in this database.
You execute this SQL statement when connected to your schema in your database instance.
SQL> SELECT * FROM books_transactions ORDER BY 3;
What is the result?

A. Only the three rows with the lowest values in the key column are displayed in the order that they are stored.
B. The first three rows in the table are displayed in the order that they are stored.
C. The execution fails unless the numeral 3 in the ORDER BY clause is replaced by a column name.
D. All table rows are displayed sorted in ascending order of the values in the third column.

正解：D

質問 # 169
......

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