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

Created May 4, 2012

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]