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From Roman Mikhailov <ro...@tugboatmedia.com>
Subject <context-param><param-value> question
Date Fri, 04 Jan 2002 18:58:10 GMT
Hi guys,
the following piece of code from the web.xml file always gives me
"Apache Tomcat/4.0.1-HTTP Status 404 - /chat/servlet/ChatAdminServlet" 
error
when I run it on Linux and (?!) works fine on Windows
How can I fix it ? Can anyone give me advice how can I make it work on 
Linux?
Thanks in advance

Roman Mikhailov

***********************

<context-param>
     <param-name>ADMIN_PATH</param-name>

<param-value>/chat/servlet/ChatAdminServlet</param-value>
   </context-param>

   <context-param>
     <param-name>LISTROOMS_PATH</param-name>

<param-value>/chat/servlet/ListRoomsServlet</param-value>
   </context-param>

   <context-param>
     <param-name>CHATROOM_PATH</param-name>

<param-value>/chat/servlet/ChatRoomServlet</param-value>
   </context-param>


******************
  ChatAdminServlet.class



import javax.servlet.*;
import javax.servlet.http.*;

import java.io.*;
import java.util.*;

public class ChatAdminServlet extends HttpServlet {

   String chatRoomPath;
   String listRoomsPath;
   String chatAdminPath;

   public void init() {
     ServletContext context = getServletContext();


------->>> I guess it breaks somewhere here

     chatRoomPath = context.getInitParameter("CHATROOM_PATH");
     listRoomsPath = context.getInitParameter("LISTROOMS_PATH");
     chatAdminPath = context.getInitParameter("ADMIN_PATH");
   }


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