RE: subtract dates

From: Scott Canaan <srcdco_at_rit.edu>
Date: Wed, 12 May 2004 11:27:26 -0400
Message-ID: <D0A642D37DE30842AA667A9AFFE3951A628020@svits11.main.ad.rit.edu>

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_at_rit.edu)
(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_at_freelists.org
[mailto:oracle-l-bounce_at_freelists.org] On Behalf Of Huascar Espinoza Sent: Wednesday, May 12, 2004 11:16 AM
To: oracle-l_at_freelists.org
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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Received on Wed May 12 2004 - 10:29:59 CDT