In your session, the NLS_DATE_FORMAT is DD-MM-YYYY.There are 86400 seconds in a day.Examine this result:DATE -----------
Posted: Sun Jun 11, 2023 3:54 pm
In your session, the NLS_DATE_FORMAT is DD-MM-YYYY.There are 86400 seconds in a day.Examine this result:DATE ------------02-JAN-2020Which statement returns this?
A. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '4' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
B. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '3' MONTH + INTERVAL '7' DAY - INTERVAL '360' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
C. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
D. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '86410' SECOND, 'DD-MON- YYYY') AS "date" FROM DUAL; E. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '6' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
A. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '4' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
B. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '3' MONTH + INTERVAL '7' DAY - INTERVAL '360' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
C. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;
D. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '5' DAY - INTERVAL '86410' SECOND, 'DD-MON- YYYY') AS "date" FROM DUAL; E. SELECT TO_CHAR(TO_DATE('29-10-2019') + INTERVAL '2' MONTH + INTERVAL '6' DAY - INTERVAL '120' SECOND, 'DD-MON-YYYY') AS "date" FROM DUAL;