RE: subtract dates
- From: "Scott Canaan" <srcdco@xxxxxxx>
- To: <oracle-l@xxxxxxxxxxxxx>
- Date: Wed, 12 May 2004 11:27:26 -0400
I use this code to convert the difference between to dates into date
format:
decode(trunc((a.event_time - b.event_time) * 24),0,
to_char(trunc((a.event_time - b.event_time) * 1440)),
to_char(trunc((a.event_time - b.event_time) * 24)) ||
':' ||
to_char(trunc((a.event_time - b.event_time) * 1440) -
trunc((a.event_time - b.event_time) * 24) *
60,'FM00'))
|| ':' ||
to_char(round(((a.event_time - b.event_time) * 1440 -
trunc((a.event_time - b.event_time) * 1440)) *
60),'FM00
')
It displays the time in the format of hours:minutes:seconds.
Scott Canaan (srcdco@xxxxxxx)
(585) 475-7886
"Life is like a sewer, what you get out of it depends on what you put
into it." - Tom Lehrer.
-----Original Message-----
From: oracle-l-bounce@xxxxxxxxxxxxx
[mailto:oracle-l-bounce@xxxxxxxxxxxxx] On Behalf Of Huascar Espinoza
Sent: Wednesday, May 12, 2004 11:16 AM
To: oracle-l@xxxxxxxxxxxxx
Subject: subtract dates
Hi guys: =3D20
=3D20
How can I subtract two dates ('DD/MM/YY HH:MI:SS') to obtain the result
=3D
in hours? MONTHS_BETWEEN don't obtain a precise value. =3D20
=3D20
Thank you,
Hu=3DE1scar
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