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From Michael Labhard <m.labh...@comcast.net>
Subject Re: How to "start" a web app?
Date Thu, 08 Jul 2004 17:03:46 GMT
Mike:

Thanks for the response.  This application is slightly different but has the 
identical problem.  Yes I have a web.xml.  The directory structure for the 
app looks like this:

=======================================
/opt/tomcat/webapps/springapps/:
niagra2

/opt/tomcat/webapps/springapps/niagra2:
WEB-INF  index.jsp

/opt/tomcat/webapps/springapps/niagra2/WEB-INF:
lib  web.xml
=======================================

The web.xml file contains this:

=======================================
<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE web-app PUBLIC '-//Sun Microsystems, Inc.//DTD Web Application 
2.3//EN' 'http://java.sun.com/dtd/web-app_2_3.dtd'>

<web-app>

</web-app>
=======================================

Tomcat was stop/started.  Still when I try

http://localhost:8080/springapps/niagra2/index.jsp

I get

" The requested resource (/springapps/niagra2/index.jsp) is not available."

Any other suggestions?

-- Michael

On Thursday 08 July 2004 09:48 am, Mike Curwen wrote:
> in order to be considered a webapp, you need to have an empty web.xml
> file in the WEB-INF folder under springapps.  Do you have this?
>
> empty means web.xml contains:
>
> <web-app>
> </web-app>
>
> You will also need to restart Tomcat for it to pick this up.

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