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  • [2022年最新] 完璧1z1-071問題集問題と解答で一年無料最速更新 [Q58-Q83]

[2022年最新] 完璧1z1-071問題集問題と解答で一年無料最速更新 [Q58-Q83]

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[2022年最新] 完璧1z1-071問題集問題と解答で一年無料最速更新

更新されたのは2022年リアルな無敵1z1-071問題集で100% 無料1z1-071試験問題集

質問 58
Examine the structure of the EMPLOYEES table.

There is a parent-child relationship between EMPLOYEE_ID and MANAGER_ID.
You want to display the last names and manager IDs of employees who work for the same manager as the employee whose EMPLOYEE_ID is 123.
Which query provides the correct output?

  • A. SELECT m.last_name, e.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.manager_id = m.manager_id)WHERE e.employee_id = 123;
  • B. SELECT e.last_name, m.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.manager_id = m.employee_id)AND e.employee_id = 123;
  • C. SELECT e.last_name, m.manager_idFROM employees e LEFT OUTER JOIN employees mon (e.employee_id = m.manager_id)WHERE e.employee_id = 123;
  • D. SELECT e.last_name, e.manager_idFROM employees e RIGHT OUTER JOIN employees mon (e.employee_id = m.employee_id)WHERE e.employee_id = 123;

正解: A

 

質問 59
Evaluate the following SQL statement:
SQL> select cust_id, cust_last_name "Last name"
FROM customers
WHERE country_id = 10
UNION
SELECT cust_id CUST_NO, cust_last_name
FROM customers
WHERE country_id = 30
Identify three ORDER BY clauses either one of which can complete the query. (Choose three.)

  • A. ORDER BY "Last name"
  • B. ORDER BY CUST_NO
  • C. ORDER BY 2, cust_id
  • D. ORDER BY 2,1
  • E. ORDER BY "CUST_NO"
    Using the ORDER BY Clause in Set Operations
    -The ORDER BY clause can appear only once at the end of the compound query.
    -Component queries cannot have individual ORDER BY clauses.
    -The ORDER BY clause recognizes only the columns of the first SELECT query.
    -By default, the first column of the first SELECT query is used to sort the output in an ascending order.

正解: A,C,D

 

質問 60
Examine the structure of the EMPLOYEES table.

You must display the maximum and minimum salaries of employees hired 1 year ago.
Which two statements would provide the correct output? (Choose two.)

  • A. SELECT minsal, maxsalFROM (SELECT MIN(salary) minsal, MAX(salary) maxsalFROM employeesWHERE hire_date < SYSDATE-365)GROUP BY maxsal, minsal;
  • B. SELECT minsal, maxsalFROM (SELECT MIN(salary) minsal, MAX(salary) maxsalFROM employeesWHERE hire_date < SYSDATE-365GROUP BY MIN(salary), MAX(salary));
  • C. SELECT MIN(Salary) minsal, MAX(salary) maxsalFROM employeesWHERE hire_date < SYSDATE-365GROUP BY MIN(salary), MAX(salary);
  • D. SELECT MIN(Salary), MAX(salary)FROM (SELECT salary FROMemployeesWHERE hire_date < SYSDATE-365);

正解: A,D

 

質問 61
Examine the description of the transactions table:

Which two SQL statements execute successfully?

  • A. SELECT customer id AS CUSTOMER-ID, transaction_date AS TRANS_DATE, amount+100 "DUES AMOUNT" FROM transactions;
  • B. SELECT customer_id AS "CUSTOMER-ID", transaction_date AS DATE, amount+100 "DUES" from transactions;
  • C. SELECT customer_id AS "CUSTOMER-ID", transaction_date AS "DATE", amount+100 DUES FROM transactions;
  • D. SELECT customer_id AS 'CUSTOMER-ID',transaction_date AS DATE, amount+100 'DUES' from transactions;
  • E. SELECT customer_id CUSTID, transaction_date TRANS_DATE,amount+100 DUES FROM transactions;

正解: C,E

 

質問 62
View the Exhibit and examine the structure of the EMPLOYEES and JOB_HISTORY tables. (Choose all that apply.) Examine this query which must select the employee IDs of all the employees who have held the job SA_MAN at any time during their employment.
SELECT EMPLOYEE_ID
FROM EMPLOYEES
WHERE JOB_ID = 'SA_MAN'
-------------------------------------
SELECT EMPLOYEE_ID
FROM JOB_HISTORY
WHERE JOB_ID = 'SA_MAN';
Choose two correct SET operators which would cause the query to return the desired result.

  • A. UNION
  • B. INTERSECT
  • C. UNION ALL
  • D. MINUS

正解: A,C

 

質問 63
View the Exhibit and examine the structure of the ORDERS table.

You must select ORDER_ID and ORDER_DATE for all orders that were placed after the last order placed by CUSTOMER_ID 101.
Which query would give you the desired result?

  • A. SELECT order_id, order_date FROM orders
    WHERE order_date >
    ANY
    (SELECT order_date FROM orders WHERE customer_id = 101);
  • B. SELECT order_id, order_date FROM orders
    WHERE order_date > IN
    (SELECT order_date FROM orders WHERE customer_id = 101);
  • C. SELECT order_id, order_date FROM orders
    WHERE order_date > ALL
    (SELECT MAX(order_date) FROM orders ) AND customer_id = 101;
  • D. SELECT order_id, order_date FROM orders
    WHERE order_date > ALL
    (SELECT order_date FROM orders WHERE customer_id = 101);

正解: D

 

質問 64
Examine the structure of the DEPARTMENTS table.

You execute the following command:

Which two statements are true?

  • A. A new column, COUNTRY, can be added to the DEPARTMENTS table after executing the command.
  • B. Unique key constraints defined on the COUNTRY column are removed.
  • C. Views created in the DEPARTMENTS table that include the COUNTRY column are automatically modified and remain valid.
  • D. Indexes created on the COUNTRY column exist until the DROP UNUSED COLUMNS command is executed.
  • E. Synonyms existing of the DEPARTMENTS table would have to be re-created.

正解: A,B

 

質問 65
Which two statements are true regarding the COUNT function? (Choose two.)

  • A. COUNT(cust_id) returns the number of rows including rows with duplicate customer IDs and NULL value in the CUST_ID column
  • B. The COUNT function can be used only for CHAR, VARCHAR2 and NUMBER data types
  • C. COUNT(*) returns the number of rows including duplicate rows and rows containing NULL value in any of the columns
  • D. COUNT(DISTINCT inv_amt) returns the number of rows excluding rows containing duplicates and NULL values in the INV_AMT column
  • E. A SELECT statement using COUNT function with a DISTINCT keyword cannot have a WHERE clause

正解: C,D

 

質問 66
Examine thestructure of the BOOKS_TRANSACTIONS table:

You want to display the member IDs, due date, and late fee as $2 for all transactions.
Which SQL statement must you execute? A)

B)

C)

D)

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

正解: C

 

質問 67
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 JOIN proj_task_details d
    ON (p.task_id = d.task_id);
  • B. 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);
  • 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 FULL OUTER JOIN proj_task_details d
    ON (p.based_on = d.task_id);

正解: B

解説:
Explanation/Reference:

 

質問 68
Which two statements are true regarding the execution of the correlated subqueries?
(Choose two.)

  • A. The nested query executes first and then the outer query executes.
  • B. Each row returned by the outer query is evaluated for the results returned by the inner query.
  • C. The outer query executes only once for the result returned by the inner query.
  • D. The nested query executes after the outer query returns the row.

正解: B,D

 

質問 69
View the Exhibit and examine the structure of the ORDERS table. (Choose the best answer.)

You must select ORDER_ID and ORDER_DATE for all orders that were placed after the last order placed by CUSTOMER_ID 101.
Which query would give you the desired result?

  • A. SELECT order_id, order_date FROM ordersWHERE order_date > ALL(SELECT order_date FROM orders WHERE customer_id = 101);
  • B. SELECT order_id, order_date FROM ordersWHERE order_date > ALL(SELECT MAX(order_date) FROM orders ) AND customer_id = 101;
  • C. SELECT order_id, order_date FROM ordersWHERE order_date > IN(SELECT
    order_date FROM orders WHERE customer_id = 101);
  • D. SELECT order_id, order_date FROM ordersWHERE order_date > ANY(SELECT order_date FROM orders WHERE customer_id = 101);

正解: A

 

質問 70
Which three statements are true about sequences in a single instance Oracle database? (Choose three.)

  • A. A sequence can only be dropped by a DBA
  • B. A sequence number that was allocated can be rolled back if a transaction fails
  • C. Sequences can always have gaps
  • D. A sequence can issue duplicate values
  • E. Two or more tables cannot have keys generated from the same sequence
  • F. A sequence's unallocated cached value are lost if the instance shuts down

正解: D,E,F

 

質問 71
Examine the description of the EMP_DETAILS table given below:

Which two statements are true regarding SQL statements that can be executed on the EMP_DETAIL TABLE?

  • A. An EMP_IMAGE column can be included in the GROUP BY clause.
  • B. You can alter the table to include the NOT NULL constraint on the EMP_IMAGE column.
  • C. An EMP_IMAGE column cannot be included in the ORDER BY clause.
  • D. You cannot add a new column to the table with LONG as the data type.

正解: C,D

 

質問 72
Examine the data in the CUST_NAME column of the CUSTOMERS table:

You want to display the CUST_NAME values where the last name starts with Mc or MC.
Which two WHERE clauses give the required result?

  • A. WHERE UPPER (SUBSTR (cust_name, INSTR(cust_name, '')+1)) LIKE UPPER ('MC%')
  • B. WHERE INITCAP (SUBSTR (cust_name, INSTR(cust_name, '')+1)) IN ('MC%', 'Mc%)
  • C. WHERE SUBSTR (cust_name, INSTR (cust_name, '') +1) LIKE 'Mc%' OR 'MC%'
  • D. WHERE INITCAP (SUBSTR (cust_name, INSTR(cust_name, '')+1)) LIKE 'Mc%'
  • E. WHERE SUBSTR (cust_name, INSTR (cust_name, '') +1) LIKE 'Mc%'

正解: E

 

質問 73
You issue the following command to drop the PRODUCTStable:
SQL > DROP TABLE products;
Which three statements are true about the implication of this command? (Choose three.)

  • A. All data along with the table structure is deleted.
  • B. All indexes on the table remain but they are invalidated.
  • C. A pending transaction in the session is committed.
  • D. All data in the table is deleted but the table structure remains.
  • E. All views and synonyms on the table remain but they are invalidated.

正解: A,C,E

解説:
Explanation

 

質問 74
View the Exhibit and examine the data in the PRODUCT_INFORMATIONtable.

Which two tasks would require subqueries? (Choose two.)

  • A. displaying the number of products whose list prices are more than the average list price
  • B. displaying all supplier IDs whose average list price is more than 500
  • C. displaying all the products whose minimum list prices are more than average list price of products having the status orderable
  • D. displaying the minimum list price for each product status
  • E. displaying the total number of products supplied by supplier 102071 and having product status OBSOLETE

正解: A,C

 

質問 75
View the Exhibit and examine the structure of the CUSTOMERS table.

You want to generate a report showing the last names and credit limits of all customers whose last names start with A, B, or C, and credit limit is below 10,000.
Evaluate the following two queries:
SQL> SELECT cust_last_name, cust_credit_limit FROM customers
WHERE (UPPER(cust_last_name) LIKE 'A%' OR
UPPER (cust_last_name) LIKE 'B%' OR UPPER (cust_last_name) LIKE 'C%')
AND cust_credit_limit < 10000;
SQL>SELECT cust_last_name, cust_credit_limit FROM customers
WHERE UPPER (cust_last_name) BETWEEN 'A' AND 'C'
AND cust_credit_limit < 10000;
Which statement is true regarding the execution of the above queries?

  • A. Both execute successfully and give the same result
  • B. Both execute successfully but do not give the required result
  • C. Only the second query gives the correct result
  • D. Only the first query gives the correct result

正解: D

 

質問 76
Examine the structure of the EMPLOYEES table. (Choose two.)

You must display the maximum and minimum salaries of employees hired 1 year ago.
Which two statements would provide the correct output?

  • A. SELECT minsal, maxsal
    FROM (SELECT MIN(salary) minsal, MAX(salary) maxsal
    FROM employees
    WHERE hire_date < SYSDATE-365)
    GROUP BY maxsal, minsal;
  • B. SELECT MIN(Salary), MAX(salary)
    FROM (SELECT salary FROM
    employees
    WHERE hire_date < SYSDATE-365);
  • C. SELECT minsal, maxsal
    FROM (SELECT MIN(salary) minsal, MAX(salary) maxsal
    FROM employees
    WHERE hire_date < SYSDATE-365
    GROUP BY MIN(salary), MAX(salary);
  • D. SELECT MIN(Salary) minsal, MAX(salary) maxsal
    FROM employees
    WHERE hire_date < SYSDATE-365
    GROUP BY MIN(salary), MAX(salary);

正解: A,B

 

質問 77
View the Exhibit and examine the structure of the CUSTOMERS and CUST_HISTORY tables.

The CUSTOMERS table contains the current location of all currently active customers.
The CUST_HISTORY table stores historical details relating to any changes in the location of all current as well as previous customers who are no longer active with the company.
You need to find those customers who have never changed their address.
Which SET operator would you use to get the required output?

  • A. INTERSECT
  • B. UNION ALL
  • C. UNION
  • D. MINUS

正解: D

 

質問 78
Which three statements are true? (Choose three.)

  • A. The data dictionary is created and maintained by the database administrator.
  • B. Data dictionary views consist of joins of dictionary base tables and user-defined tables.
  • C. The usernames of all users including database administrators are stored in the data dictionary.
  • D. The USER_CONS_COLUMNS view should be queried to find the names of columns to which constraints apply.
  • E. Views with the same name but different prefixes, such as DBA, ALL and USER, reference the same base tables from the data dictionary.
  • F. Both USER_OBJECTS and CAT views provide the same information about all objects that are owned by the user.

正解: C,D,E

解説:
References:
https://docs.oracle.com/cd/B10501_01/server.920/a96524/c05dicti.htm

 

質問 79
Examine the description of the ORDERS table:

Which three statements execute successfully?

  • A. SELECT * FROM orders ORDER BY order_ id
    UNION
    SELECT * FROM invoices;
  • B. (SELECT * FROM orders
    UNION ALL
    SELECT* FROM invoices) ORDER BY order _id;
  • C. SELECT * FROM orders ORDER BY order_ id
    INTERSEOT
    SELECT * FROM invoices ORDER BY invoice_ id;
  • D. SELECT * FROM orders
    MINUS
    SELECT * FROM INVOICES ORDER BY 1
  • E. SELECT order_ id, invoice_ data order_ date FROM orders
    MINUS
    SELECT invoice_ id, invoice_ data FROM invoices ORDER BY invoice_ id;
  • F. SELECE order _id, order _ date FRON orders
    LNTERSECT
    SELECT invoice_ id, invoice_ id, order_ date FROM orders
  • G. SELECT order_ id, order_ data FROM orders
    UNION ALL
    SELECT invoice_ id, invoice_ data FROM invoices ORDER BY order_ id;

正解: B,D,G

 

質問 80
BOOK_SEQ is an existing sequence in your schema.
Which two CREATE TABLE commands are valid?

  • A. CREATE TABLE bookings (
    bk_id NUMBER(4) DEFAULT book_seq.CURRVAL,
    start_date DATE DEFAULT SYSDATE,
    end_date DATE DEFAULT start date);
  • B. CREATE TABLE bookings (
    bk_id NUMBER(4) DEFAULT book_seq.NEXTVAL PRIMARY KEY,
    start_date DATE DEFAULT SYSDATE,
    end_date DATE DEFAULT SYSDATE NOT NULL);
  • C. CREATE TABLE bookings (
    bk_id NUMBER(4) NOT NULL DEFAULT book_seq.CURRVAL,
    start_date DATE NOT NULL,
    end_date DATE DEFAULT SYSDATE);
  • D. CREATE TABLE bookings (
    bk_id NUMBER(4) NOT NULL PRIMARY KEY,
    start_date DATE NOT NULL,
    end_date DATE DEFAULT SYSDATE);
  • E. CREATE TABLE bookings ( bk_id NUMBER(4),
    start_date DATE DEFAULT SYSDATE,
    end_date DATE DEFAULT (end_date >= start_date));

正解: B,D

 

質問 81
Examine the description or the BOOKS_TRANSACTIONS table:

FOR customers whose income level has a value, you want to display the first name and due amount as 5% of their credit limit. Customers whose due amount is null should not be displayed.
Which query should be used?

  • A. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
    FROM customers
    WHERE cust income_level <> NULL
    AND due_amount <> NULL;
  • B. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
    FROM customers
    WHERE cust_income_level IS NOT NULL
    AND cust_credit_limit IS NOT NULL;
  • C. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
    FROM customers
    WHERE cust income_level !=NULL
    AND due_amount !=NULL;
  • D. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
    FROM customers
    WHERE cust income_level !=NULL
    AND cust credit_level !=NULL;
  • E. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
    FROM customers
    WHERE cust income_level IS NOT NULL
    AND due_amount IS NOT NULL;

正解: B

 

質問 82
Which two queries execute successfully?

  • A. SELECT * FROM products
    UNION
    SELECT * FROM new_products;
  • B. SELECT k FROM products
    MINUS
    SELECT prod_id FROM new_products;
  • C. SELECT prod_id FROM products
    UNION ALL
    SELECT prod_id, prod_name FROM new_products;
  • D. SELECT prod_id, prod_name FROM products
    INTERSECT
    SELECT 100, prod_name FROM newproducts;
  • E. SELECT prod_id, exp_date FROM products
    UNION ALL
    SELECT prod_id, NULL FROM new_products;

正解: A,E

 

質問 83
......

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