2


Is it possible to change the Servlet request object while forwarding the request from one servlet to another? or a work around for achieving this?

I have 2 servlet's, Servlet1 and Servlet2 like for e.g. - 

public class Servlet1 extends HttpServlet {
    doPost(HttpServletRequest rq, HttpServletResponse rs) {
       // do something meaningful, call other different web-apps/servlets

       InputStream is1 = rq.getInputStream();
       RequestDispatcher rd = getServletContext().getRequestDispatcher("/Servlet2");
       rd.forward(rq, rs);
    }
}

If I print the is1 it is something like - 

-----Part2_324<?xml version="1.0" encoding="utf-8"?><Head><Body><Text>This is the first File</Text></Body></Head>-------Part2_65623

I dont care about this o/p, when later the request is to be forwarded to the Servlet2.

I have an xml file file2.xml, contents are - 

<?xml version="1.0" encoding="utf-8"?><Head><Body><Top>Start</Top><Middle>Process</Middle><Bottom>End</Bottom></Body></Head>

I would like this to be as the request content for the Servlet2, as it cannot process the contents of is1, it is meant to be processing content of file2.xml.
There are pretty much no Attributes/Parameters set.

Is it possible to achieve this? I hope, the question is clear.
Thank you

Improve this question

1 Answer 1

Reset to default
8

The servlet spec forbids the substitution of one request for another when forwarding.

However, it does permit the forwarding of an HttpServletRequestWrapper, where the wrapper is wrapping the original request. So try and implement your logic as a subclass of HttpServletRequestWrapper, wrapping the original request, and overriding the various methods of HttpServletRequest as appropriate.

Improve this answer
1
  • That is it, thats pretty much what I needed. Thanks a lot.
    – Swift-Tuttle
    Commented Feb 8, 2011 at 12:06

Not the answer you're looking for? Browse other questions tagged or ask your own question.