Always resolve the URL yourself before letting the tomcat do it. For every sub
application and context's path is equals to <root-context> + sub application's
context.
[EMAIL PROTECTED] wrote:
Thanks. Yes, of course on the matter of the package name. I think I've
grasped the points I was struggling with. I have saved the spec to my local
disk, and when I'm comfortable enough with the general concepts, I'll be able
to read it and understand it.
--
Charles Knell
[EMAIL PROTECTED] - email
-----Original Message-----
From: David Smith
Sent: Mon, 20 Mar 2006 10:28:29 -0500
To: Tomcat Users List
Subject: Re: What, exactly, is meant by "full path" when construction web.xml
entries
More like:
package com.kilonovember ;
// imports here
public class Monkey extends HttpServlet{
// methods and programming
}
Note I didn't include the class name in the package name. This creates
a class with the full name of com.kilonovember.Monkey.
--David
[EMAIL PROTECTED] wrote:
>Thanks for the sanity check. That did what I expected, so to expand on this,
>if I were creating a "real" servlet whose source code opened like this:
>
>package com.kilonovember.Monkey;
>import javax.servlet.*;
>import javax.servlet.http.*;
>import java.io.*;
>
>public class Monkey extends HttpServlet{
>
>.. programming stuff goes here ...
>
>}
>
>I would create this directory structure under "C:\Apache Software
>Foundation\Tomcat 5.5\webapps"?
>
>com/kilonovember
>
>and place my compiled Monkey.class in kilonovember?
>
>Next I would edit the web.xml file and add these elements?
>
>
> monkey
> com.kilonovember.Monkey
>
>
>
> monkey
> /monkey
>
>
>Having done that, I would re-start Tomcat and, providing my Monkey.class
>compiled correctly, I should expect to be able to type
>"http://localhost:8080/monkey"; into my browser's address window and see the
>output of Monkey.class. Is that correct?
>
>
>Thanks to everyone who responded.
>
>
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--
Manivannan.P
http://geocities.com/manivannan57
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