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Question 68/75

Examine the structure of the EMP table:
Name Null? Type
------------------ -------------- ---------------------
EMPNO NOT NULL NUMBER (4)
ENAME VARCHAR2 (10)
SAL NUMBER (7, 2)
Examine this code:

Which code should be inserted to display the collection contents?

A. FOR indx IN -1 .. 1_list.LIMIT LOOP
DBMS_OUTPUT.PUT_LINE (indx | | ‘ ’ | | 1_list (indx) );
END LOOP;

B. 1_indx := 1_list.FIRST;
WHILE (1_indx IS NOT NULL) LOOP
DBMS_OUTPUT.PUT_LINE (1_indx || ‘ ’ || 1_list (1_indx));
1_indx := 1_emp.NEXT (1_indx);
END LOOP;

C. FOR indx IN 1_list. COUNT .. -1 LOOP
DBMS_OUTPUT.PUT_LINE (indx | | ‘ ’ | | 1_list (indx));
END LOOP;

D. FOR indx IN 1_list.FIRST . . 1_list.LAST LOOP
DBMS_OUTPUT.PUT_LINE (indx | | ‘ ’ | | 1_list (indx));
END LOOP;

Recent Comments (The most recent comments are at the top.)

Sshumeev - Apr 08, 2019

How about l_list (1) = 800 ?

Sshumeev - Feb 04, 2019

Correct B and D

aa - Jan 27, 2019

Answer B

SShumeev - Nov 13, 2018

B is correct to
indx := l_list.FIRST;
WHILE (indx IS NOT NULL) LOOP
DBMS_OUTPUT.PUT_LINE (indx || ' ' || l_list (indx));
indx := l_list.NEXT (indx);
END LOOP;

Milan Babic - Oct 30, 2018

Correct answer is D.

Question 68/75

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