Return-Path: Mailing-List: contact tomcat-user-help@jakarta.apache.org; run by ezmlm Delivered-To: mailing list tomcat-user@jakarta.apache.org Received: (qmail 26857 invoked from network); 1 Mar 2001 18:41:53 -0000 Received: from slkcpop2.slkc.uswest.net (206.81.128.2) by h31.sny.collab.net with SMTP; 1 Mar 2001 18:41:53 -0000 Received: (qmail 93786 invoked by alias); 1 Mar 2001 18:41:54 -0000 Delivered-To: fixup-tomcat-user@jakarta.apache.org@fixme Received: (qmail 93771 invoked by uid 0); 1 Mar 2001 18:41:54 -0000 Received: from unknown (HELO pbganjanette) (204.99.158.254) by slkcpop2.slkc.uswest.net with SMTP; 1 Mar 2001 18:41:54 -0000 Message-ID: <004901c0a27f$b88a6dc0$1401290a@pbganjanette> From: "Brett G. Palmer" To: Subject: Reading a file from a Context Date: Thu, 1 Mar 2001 11:44:59 -0700 MIME-Version: 1.0 Content-Type: text/plain; charset="iso-8859-1" Content-Transfer-Encoding: 7bit X-Priority: 3 X-MSMail-Priority: Normal X-Mailer: Microsoft Outlook Express 5.00.2919.6600 X-MimeOLE: Produced By Microsoft MimeOLE V5.00.3018.1300 X-Spam-Rating: h31.sny.collab.net 1.6.2 0/1000/N What is the best way to read a file from a Context in Tomcat? I know I can get the context path from the "getContextPath()" method, but most file stream APIs require a full path. There most be a clean way to accomplish this without hard coding any directory path info. Any suggestions would be appreciated? Thanks, Brett