From seema_56@yahoo.com Fri May  25 15:16:46 2001
Received: from pop.sat.net.pk ([202.133.64.2])
	by postgresql.org (8.11.3/8.11.1) with ESMTP id f4PFGgA49309;
	Fri, 25 May 2001 11:16:46 -0400 (EDT)
	(envelope-from seema_56@yahoo.com)
Received: from noor ([202.133.67.31])
	by pop.sat.net.pk (8.11.2/8.11.1) with SMTP id f4Q1EUp23758;
	Fri, 25 May 2001 20:14:35 -0500 (GMT)
Message-ID: <004601c0e52d$1af00340$0100a8c0@noor>
From: "Seema Noor" <seema_56@yahoo.com>
To: <pgsql-sql@postgresql.org>
Cc: <pgsql-novice@postgresql.org>
Subject: time difference
Date: Fri, 25 May 2001 20:12:16 +0500
MIME-Version: 1.0
Content-Type: multipart/alternative;
	boundary="----=_NextPart_000_0037_01C0E556.FE45A860"
X-Priority: 3
X-MSMail-Priority: Normal
X-Mailer: Microsoft Outlook Express 5.00.2615.200
X-MimeOLE: Produced By Microsoft MimeOLE V5.00.2615.200
X-Archive-Number: 200105/124
X-Sequence-Number: 1535

This is a multi-part message in MIME format.

------=_NextPart_000_0037_01C0E556.FE45A860
Content-Type: text/plain; charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable


i want to calculate the session time. my query is this

select now() - '2001-05-19 12:09:00+5;

it gives the difference in date and difference in time separately. if the
current time is greater than the second time then it gives simply their
difference but i want to calcutate the session time from these two times.
if now() =3D 2001-05-25 01:00:00+5
and second timestamp =3D 2001-05-24 23:00:00+5

then how can i get their time differernce EQUAL TO '3:00:00' instead of '21=
:00:00 ago' ?

and is there a way to get the total time difference including the date fact=
or involved i.e.
if now() =3D 2001-05-25 23:00:00+5
and second timestamp =3D 2001-05-24 01:00:00+5
then is there any function to get their difference equal to '46:00:00' inst=
ead of '1 22:00:00' ?

please reply me soon.

------=_NextPart_000_0037_01C0E556.FE45A860
Content-Type: text/html; charset="iso-8859-1"
Content-Transfer-Encoding: quoted-printable

<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META content=3D"text/html; charset=3Diso-8859-1" http-equiv=3DContent-Type>
<META content=3D"MSHTML 5.00.2614.3500" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3D"Comic Sans MS" size=3D2>
<DIV>i want to calculate the session time. my query is this<BR><BR>select n=
ow()=20
- '2001-05-19 12:09:00+5;<BR><BR>it gives the difference in date and differ=
ence=20
in time separately. if the<BR>current time is greater than the second time =
then=20
it gives simply their<BR>difference but i want to calcutate the session tim=
e=20
from these two times.<BR>if&nbsp;now()&nbsp;=3D 2001-05-25 01:00:00+5</DIV>
<DIV>
<DIV>and second timestamp&nbsp;=3D 2001-05-24 23:00:00+5</DIV></DIV>
<DIV>&nbsp;</DIV>
<DIV>then how can i get their time differernce EQUAL TO '3:00:00' instead o=
f=20
'21:00:00 ago' ?</DIV>
<DIV>&nbsp;</DIV>
<DIV>and is there a way to get the total time difference including the date=
=20
factor involved i.e.</DIV>
<DIV>
<DIV>if&nbsp;now()&nbsp;=3D 2001-05-25 23:00:00+5</DIV>
<DIV>and second timestamp&nbsp;=3D 2001-05-24 01:00:00+5</DIV>then is there=
 any=20
function to get their difference equal to '46:00:00' instead of '1 22:00:00=
'=20
?<BR></DIV>
<DIV>please reply me soon.</DIV></FONT></DIV></BODY></HTML>

------=_NextPart_000_0037_01C0E556.FE45A860--




From quadratini@sni.ph Fri Jun  28 03:53:24 2002
Received: from localhost.localdomain (postgresql.org [64.49.215.8])
	by localhost (Postfix) with ESMTP id 6D655476E8F
	for <pgsql-sql@postgresql.org>; Thu, 27 Jun 2002 23:53:24 -0400 (EDT)
Mailbox-Line: From quadratini@sni.ph  Thu Jun 27 23:53:24 2002
Received: from ns.sni.ph (ns.sni.ph [210.23.103.9])
	by postgresql.org (Postfix) with ESMTP id 6A1B1476ABB
	for <pgsql-sql@postgresql.org>; Thu, 27 Jun 2002 23:52:27 -0400 (EDT)
Received: (from root@localhost) by ns.sni.ph (8.11.6/8.11.6) id g5S3qVE27296
	for pgsql-sql@postgresql.org; Fri, 28 Jun 2002 11:52:31 +0800 (PHT)
Received: from sni.ph ([210.23.103.26])
	by ns.sni.ph (8.11.6/8.11.6) with ESMTP id g5S3qU227231
	for <pgsql-sql@postgresql.org>; Fri, 28 Jun 2002 11:52:30 +0800 (PHT)
Message-ID: <3D1BDEEF.50409@sni.ph>
Date: Fri, 28 Jun 2002 11:58:39 +0800
From: q u a d r a <quadratini@sni.ph>
User-Agent: Mozilla/5.0 (Windows; U; Windows NT 5.0; en-US;
	rv:0.9.4) Gecko/20011019 Netscape6/6.2
X-Accept-Language: en-us
MIME-Version: 1.0
To: pgsql-sql@postgresql.org
Subject: time difference
X-scanner: scanned by SNI 1.0.10 - (http://www.sni.ph)
Content-Type: text/plain; charset=us-ascii; format=flowed
Content-Transfer-Encoding: 7bit
X-Spam-Status: No, hits=0.0 required=5.0
	tests=none
	version=2.30
X-Spam-Level: 
X-Archive-Number: 200206/378
X-Sequence-Number: 8343


I have a query that returns
    ?column?    
-----------------
 6 days 01:56:14
(1 row)


How can I get it as total number of hours?
Instead of days and hours, I need hours only (all in all, so 24hours * 6 
days + 01:56:14)
Can you help me please? Thanks