Subtracting time in oracle sql - sql

Question

I am trying to subtract time from each other based on a case condition
but the error is am getting is non-numeric character found where numeric character was expected
HERE IS THE CODE
select
case
when to_date('01-JAN-2019 05:00 AM', 'HH:MI:SS AM') <
to_date('01-JAN-2019 05:00 PM', 'HH:MI:SS AM')
then round((to_date('01-JAN-2019 05:00 AM', 'HH:MI:SS AM') -
to_date('01-JAN-2019 09:00 AM', 'HH:MI:SS AM'))*24,2)
else 0
end late
from dual

Answer 1

One option would be using ISO 8601 standard timestamp format as
select case
when timestamp'2019-01-01 05:00:00' < timestamp'2019-01-01 17:00:00'
then
timestamp'2019-01-01 05:00:00' - timestamp'2019-01-01 09:00:00'
else interval '0' second
end as late
from dual
If you need numeric value for hour only, then consider :
select case
when timestamp'2019-01-01 05:00:00' < timestamp'2019-01-01 17:00:00'
then
extract( hour from timestamp'2019-01-01 05:00:00'
- timestamp'2019-01-01 09:00:00' )
else 0
end as late
from dual
Demo

Answer 2

You have to match date format and format mask:
SQL> select
2 round(
3 (to_date('01-JAN-2019 05:00 AM', 'dd-mon-yyyy HH:MI AM') -
4 to_date('01-JAN-2019 09:00 AM', 'dd-mon-yyyy HH:MI AM')
5 ) * 24, 2) result
6 from dual;
RESULT
----------
-4
SQL>

Answer 3

First conversion format and real string format must match. Next, convert date to timestamp before substracting, so the result is interval. Finally, extract parts needed from the interval
select
extract( day from diff ) Days,
extract( hour from diff ) Hours,
extract( minute from diff ) Minutes
from (
select CAST(to_date('01-JAN-2019 09:00 AM', 'DD-MON-YYYY HH:MI AM') as timestamp)
- CAST(to_date('01-JAN-2019 05:00 AM', 'DD-MON-YYYY HH:MI AM') as timestamp) diff
from dual
)