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 56879 invoked from network); 18 Dec 2000 20:52:05 -0000 Received: from unknown (HELO jupiter) (firewall-user@216.52.212.10) by locus.apache.org with SMTP; 18 Dec 2000 20:52:05 -0000 Received: by jupiter; id MAA16466; Mon, 18 Dec 2000 12:52:04 -0800 (PST) Received: from unknown(10.10.20.40) by jupiter.guidance.com via smap (V5.5) id xma016442; Mon, 18 Dec 00 12:51:48 -0800 Received: from kopelovich (10-10-40-175.guidance.com [10.10.40.175]) by uranus.guidance.com (Sun Internet Mail Server sims.4.0.2000.05.17.04.13.p6) with SMTP id <0G5S00HJL7ACWQ@uranus.guidance.com> for tomcat-user@jakarta.apache.org; Mon, 18 Dec 2000 12:51:48 -0800 (PST) Date: Mon, 18 Dec 2000 12:58:04 -0800 From: MK Subject: link problem To: tomcat-user@jakarta.apache.org Message-id: <001b01c06935$376939c0$af280a0a@guidance.com> MIME-version: 1.0 X-MIMEOLE: Produced By Microsoft MimeOLE V5.00.2314.1300 X-Mailer: Microsoft Outlook Express 5.00.2314.1300 Content-type: MULTIPART/ALTERNATIVE; BOUNDARY="Boundary_(ID_Z3FYEV48O8lXPQvVO8vf6w)" X-Priority: 3 X-MSMail-priority: Normal X-Spam-Rating: locus.apache.org 1.6.2 0/1000/N This is a multi-part message in MIME format. --Boundary_(ID_Z3FYEV48O8lXPQvVO8vf6w) Content-type: text/plain; charset=iso-8859-1 Content-transfer-encoding: 7BIT I have a problem which breaks my head :-) I am using Tomcat 3.2.1. Here is what I have in server.xml I have two pages one is located: D:\GuidanceExchange\exchange\index.jsp and have a link test test.jsp is another page located at D:\GuidanceExchange\exchange\jsp\test.jsp The problem is when I execute index.jsp page the link is pointing to http:///exchange/jsp/test.jsp which is incorrect...it should point to http://127.0.0.1:8080/exchange/jsp/test.jsp. So, how do I get 127.0.0.1:8080 into my URL? Thank you very much for any help in advance, Maksim. --Boundary_(ID_Z3FYEV48O8lXPQvVO8vf6w) Content-type: text/html; charset=iso-8859-1 Content-transfer-encoding: 7BIT

I have a problem which breaks my head :-)

I am using Tomcat 3.2.1.

Here is what I have in server.xml

I have two pages one is located:

D:\GuidanceExchange\exchange\index.jsp

and have a link <a href="<%= request.getContextPath() %>"/jsp/test.jsp"> test</a>

test.jsp is another page located at

D:\GuidanceExchange\exchange\jsp\test.jsp

The problem is when I execute index.jsp page the link is pointing to http:///exchange/jsp/test.jsp which is incorrect...it should point to

http://127.0.0.1:8080/exchange/jsp/test.jsp. So, how do I get 127.0.0.1:8080 into my URL?

Thank you very much for any help in advance,

Maksim.

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