Received: from localhost.localdomain (postgresql.org [64.49.215.8]) by localhost (Postfix) with ESMTP id 6D655476E8F for ; 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 ; 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 ; 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 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