Question

I want to include the output of another servlet into the current response.

I can call

request.getRequestDispatcher("/new/url").include(request, response);

and it does go to the right servlet, but request.getPathInfo(), request.getServletPath() and friends do not get set to the new URL (they still have the old values), so the servlet does not work.

Is that supposed to be like that? And what can I do to make this work?

Answer 1

Yes, it’s supposed to be like that (from javadoc, note that Adeel Ansari cited javadoc for forward() method instead of include()):

The ServletResponse object has its path elements and parameters remain unchanged from the caller’s.

However, included servlet can access its own path info as request attributes with the following names:

javax.servlet.include.request_uri
javax.servlet.include.context_path
javax.servlet.include.servlet_path
javax.servlet.include.path_info
javax.servlet.include.query_string