How do I capture a request and dispatch the exact request (with all the parameters received) to another URL?

David Garcia

As far as i know it depends on the location of the next target url.

• If the next servlet url is in the same host, then you can use the forward method.
  Here is an example code about using forward:

     RequestDispatcher rd = null;
     String targetURL = "target_servlet_name";
     ServletContext ctx = this.getServletContext();
     rd = ctx.getRequestDispatcher(targetURL);
     rd.forward(request, response);
  




• The other possibility is that the target servlet location is in a different host. Then you will need to use Streams,Sockets... and the code could be something like this:

      URL url=null;
      URLConnection conn=null;
      try{
        url = new URL
  (protocol,host,port,"/"+servlet+"?"+requestParams);
        // the requestParam contains all the attributes (name=value pairs) you want to send taken from the incomming request
  
        conn = url.openConnection();
        conn.setUseCaches(false);
        conn.setDoOutput(false);
      } catch(MalformedURLException e) {
         // whatever you want
      }
      BufferedReader inStream=new BufferedReader(new InputStreamReader(conn.getInputStream()));
      String respuesta=inStream.readLine();
      inStream.close();
  

You can read more about the first approach in the Java Developer Journal (Number of May 2000 page 102).

Test Conditions
I have test this approach over webLogic 451 with the Service Pack 8 or higher. I can asure you that with Service Pack 7 or lower, the forward approach doesn't work.

[That's all well and good, but you've dodged an important part of the answer: How do we grab the values of the parameters? I have source code that does this, but unfortunately my hard drive just crashed -- can someone else fill this in? -Alex]