[2024-06-24更新,250問] 無料Oracle 1z0-071試験問題集、1z0-071予想試験(ページ 10)

1z0-071 試験問題 41

Examine the data in the NEW_EMPLOYEES table:

Examine the data in the EMPLOYEES table:

You want to:
1. Update existing employee details in the EMPLOYEES table with data from the NEW EMPLOYEES
table.
2. Add new employee detail from the NEW_ EMPLOYEES able to the EMPLOYEES table.
Which statement will do this:

A. MERGE INTO employees e
USING new_employees n
WHERE e.employee_id = ne.employee_id
WHEN FOUND THEN
UPDATE SET e.name=ne.name,e.job_id =ne.job_id, e.salary=ne.salary
WHEN NOT FOUND THEN
INSERT VALUES (ne.employee_ id,ne.name,ne.job id,ne.salary) ;

B. MERGE INTO employees e
USING new employees ne
ON (e.employee_id = ne.employee_id)
WHEN FOUND THEN
UPDATE SET e.name =ne.name, e.job_id=ne.job_id, e.salary =ne.salary
WHEN NOT FOUND THEN
INSERT VALUES (ne.employee_id,ne.name,ne.job_id,ne.salary) ;

C. MERGE INTO employees e
USING new employees ne
WHERE e.employee_id = ne.employee_ id
WHEN MATCHED THEN
UPDATE SET e.name = ne.name, e.job_id = ne.job_id,e.salary =ne. salary
WHEN NOT MATCHED THEN
INSERT VALUES (ne. employee_id,ne.name, ne.job_id,ne.salary) ;

D. MERGE INTO employees e
USING new_employees n
ON (e.employee_id = ne.employee_id)
WHEN MATCHED THEN
UPDATE SET e.name = ne.name, e.job id = ne.job_id,e.salary =ne. salary
WHEN NOT MATCHED THEN
INSERT VALUES (ne. employee_id,ne.name,ne.job_id,ne.salary);

1z0-071 試験問題 42

CUSTOMERS テーブルの説明を調べます。

姓が A または B (小文字または大文字) で始まり、与信限度額が 1000 未満であるすべての顧客の姓と与信限度額を表示する必要があります。
この部分的なクエリを調べてください。
顧客から cust_last_nare、cust_credit_limit を選択します
必要な結果を与える 2 つの WHERE 条件はどれですか?

A. WHERE (UPPER(cust_last_name) like INITCAP ('A') OR UPPER(cust_last_name) like INITCAP('B')) AND ROUND(cust_credit_limit) < ROUND(1000) ;

B. WHERE UPPER(cust_last_name) BETWEEN UPPER('A%' AND 'B%') AND ROUND(cust_credit_limit) < 1000;

C. WHERE (INITCAP(cust_last_name) LIKE 'A%' または ITITCAP(cust_last_name) LIKE 'B%') AND cust_credit_limit < 1000

D. WHERE UPPER(cust_last_name) IN ('A%', 'B%') かつ cust_credit_limit < 1000:

E. WHERE (UPPER(cust_last_name) LIKE 'A%' OR UPPER(cust_last_name) LIKE 'B%') AND ROUND(cust_credit_limit) < 1000;

1z0-071 試験問題 43

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 cust credit_level !=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 IS NOT NULL
AND due_amount IS NOT NULL;

E. SELECT cust_first_name, cust_credit_limit * . 05 AS DUE AMOUNT
FROM customers
WHERE cust income_level !=NULL
AND due_amount !=NULL;

1z0-071 試験問題 44

Which three statements are true regarding single row subqueries?

A. They can be used in the where clause.

B. They must return a row to prevent errors in the SQL statement.

C. They must be placed on the left side of the comparison operator or condition.

D. They can be used in the having clause.

E. A SQL statement may have multiple single row subquery blocks.

F. They must be placed on the right side of the comparison operator or condition.

1z0-071 試験問題 45

See the Exhibit and examine the structure of the PROMOTIONS table: Exhibit:

Using the PROMOTIONS table, you need to find out the average cost for all promos in the
range $0-2000 and $2000-5000 in category A.
You issue the following SQL statements:
Exhibit:

What would be the outcome?

A. It generates an error because multiple conditions cannot be specified for the WHEN clause

B. It executes successfully and gives the required result

C. It generates an error because CASE cannot be used with group functions

D. It generates an error because NULL cannot be specified as a return value

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1z0-071 試験問題 41

1z0-071 試験問題 42

1z0-071 試験問題 43

1z0-071 試験問題 44

1z0-071 試験問題 45

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