Your servlets belong to 2 different contexts, moreover - jvms ... sessions are 
defined per context ... 
hth

*********************************************************************
* Boris Niyazov                Ph:  212-854-4094  Fax: 212-854-1749 *
* Systems Manager              Email: [EMAIL PROTECTED]  * 
* Columbia Law School          URL: http://www.law.columbia.edu     *
*********************************************************************                  
             


>
>Hello,
>
>I've read on this list that Tomcat 3.2 supports
>serializing sessions attributes. I wanted to try it,
>but could not find more information on the subject
>here. So I tried to do it according to the servlet 2.2
>specification. As it is not working, I try to sum-up
>here what I've done:
>
>I have installed 2 Tomcat 3.2 on the same machine
>(taking care that the 2nd Tomcat starts on port 8081),
>and for each of them, I have built a very simple
>"session" webapp :
>- session/
>  - WEB-APP/
>    web.xml
>    - classes/
>      SessionServlet.class
>
>The web.xml looks like:
><?xml version="1.0" encoding="ISO-8859-1"?>
><!DOCTYPE web-app
>    PUBLIC "-//Sun Microsystems, Inc.//DTD Web
>Application 2.2//EN"
>    "http://java.sun.com/j2ee/dtds/web-app_2_2.dtd";>
>
>    <web-app>
>       <distributable>
>       </distributable>
>        <servlet>
>            <servlet-name>session</servlet-name>
>           
><servlet-class>SessionServlet</servlet-class>
>        </servlet>
>        <servlet-mapping>
>            <servlet-name>session</servlet-name>
>            <url-pattern>/session</url-pattern>
>        </servlet-mapping>
>    </web-app>
>
>
>For the firt Tomcat, I compiled:
>import java.io.*;
>import javax.servlet.*;
>import javax.servlet.http.*;
>public class SessionServlet extends HttpServlet {
>    public void doGet (HttpServletRequest req,
>HttpServletResponse res)
>      throws ServletException, IOException
>      {
>         ServletOutputStream out = res.getOutputStream();
>         res.setContentType("text/plain");
>         out.println("SessionServlet 1 output");
>
>         ServletContext servletContext =
>this.getServletContext();
>         servletContext.setAttribute("foo","abcdef..foo");
>         out.println("attribute foo, with value
>"abcde...foo", added in ServletContext");
>
>         HttpSession httpSession = req.getSession();
>         httpSession.setAttribute("bar","abcde...bar");
>         out.println("attribute bar, with value
>"abcde...bar", added in HttpSession");
>
>         out.close();
>      }
>}
>
>For the second Tomcat, I compiled:
>import java.io.*;
>import javax.servlet.*;
>import javax.servlet.http.*;
>public class SessionServlet extends HttpServlet {
>    public void doGet (HttpServletRequest req,
>HttpServletResponse res)
>      throws ServletException, IOException
>      {
>         ServletOutputStream out = res.getOutputStream();
>
>         res.setContentType("text/plain");
>         out.println("SessionServlet 2 Output");
>
>         ServletContext servletContext =
>this.getServletContext();
>         String foo = (String)
>servletContext.getAttribute("foo");
>         out.println("foo="+ foo +" (get from
>ServletContext)");
>
>         HttpSession httpSession = req.getSession();
>         String bar = (String)
>httpSession.getAttribute("bar");
>         out.println("bar="+ bar +" (get from
>HttpSession)");
>
>         out.close();
>      }
>}
>
>Requesting respectively the first servlet, and the
>second serlvet did not give the expected results:
>
>http://127.0.0.1:8080/session/servlet/SessionServlet
>SessionServlet 1 output 
>attribute foo, with value "abcde...foo", added in
>ServletContext 
>attribute bar, with value "abcde...bar", added in
>ServletContext
>
>http://127.0.0.1:8081/session/servlet/SessionServlet
>SessionServlet 2 Output 
>foo=null (get from ServletContext) 
>bar=null (get from HttpSession) 
>
>
>So did I made a mistake, did I forget something, or is
>it simply not possible to do it ? 
>Thanks for any help !
>
>Xavier Marjou
>
>__________________________________________________
>Do You Yahoo!?
>Spot the hottest trends in music, movies, and more.
>http://buzz.yahoo.com/


Reply via email to