Online Access Free Oracle.1z0-071.v2025-09-16.q232 Practice Test (Page 45)

1z0-071 Exam Question 216

You need to calculate the number of days from 1st January 2019 until today. Dates are stored in the default format of DD-MON-RR. Which two queries give the required output?

A.SELECT ROUND(SYSDATE - '01-JAN-2019') FROM DUAL;

B.SELECT ROUND(SYSDATE - TO_DATE('01/JANUARY/2019')) FROM DUAL;

C.SELECT TO_DATE(SYSDATE, 'DD/MONTH/YYYY") - '01/JANUARY/2019" FROM DUAL;

D.SELECT TO_CHAR(SYSDATE, *DD-MON-YYYY*) - '01-JAN-2019' FROM DUAL;

E.SELECT SYSDATE - TO_DATE('01-JANUARY-2019*) FROM DUAL;

1z0-071 Exam Question 217

Examine the structure of the INVOICE table.
NameNull?Type
--------------------------------------------------
INV_NONOT NULLNUMBER(3)
INV_DATEDATE
INV_AMTNUMBER(10,2)
Which two SQL statements would execute successfully?

A.SELECT inv_no, NVL2(inv_date, sysdate-inv_date, sysdate)FROM invoice;

B.SELECT inv_no, NVL2(inv_amt, inv_amt*.25, 'Not Available')FROM invoice;

C.SELECT inv_no, NVL2(inv_date, 'Pending', 'Incomplete')FROM invoice;

D.SELECT inv_no, NVL2(inv_amt, inv_date, 'Not Available')FROM invoice;

1z0-071 Exam Question 218

View the exhibit and examine the structure of the CUSTOMERS table.

Which two tasks would require subqueries or joins to be executed in a single statement?

A.finding the number of customers, in each city, whose credit limit is more than the average credit limit of all the customers

B.finding the number of customers, in each city, who's marital status is 'married'.

C.listing of those customers, whose credit limit is the same as the credit limit of customers residing in the city 'Tokyo'.

D.listing of customers who do not have a credit limit and were born before 1980

E.finding the average credit limit of male customers residing in 'Tokyo' or 'Sydney'

1z0-071 Exam Question 219

Which three are true about scalar subquery expressions?

A.A scalar subquery expression that returns zero rows evaluates to zoro

B.They cannot be used in the values clause of an insert statement*

C.They can be nested.

D.A scalar subquery expression that returns zero rows evaluates to null.

E.They cannot be used in group by clauses.

F.They can be used as default values for columns in a create table statement.

1z0-071 Exam Question 220

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
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);

B.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) ;

C.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) ;

D.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) ;

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1z0-071 Exam Question 216

1z0-071 Exam Question 217

1z0-071 Exam Question 218

1z0-071 Exam Question 219

1z0-071 Exam Question 220

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