Question

So I’m trying to grab the current URL of the page using Java’s request object. I’ve been using request.getRequestURI() to preform this, but I noticed that when a java class reroutes me to a different page off a servlet request getRequestURI gives that that address as opposed to the orginal URL that was typed in the browser and which still shows in the browser.

Ex: \AdvancedSearch:
getRequestURI() returns “\subdir\search\search.jsp”

I’m looking for a way to grab what the browser sees as the URL and not what that page knows is only a servlet wrapper.

Answer 1

If your current request is coming from an “inside the app-server” forward or include, the app-server is expected to preserve request information as request attributes. The specific attributes, and what they contain, depends on whether you’re doing a forward or an include.

For <jsp:include>, the original parent URL will be returned by request.getRequestURL(), and information about the included page will be found in the following request attributes:

     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

For <jsp:forward>, the new URL will be returned by request.getRequestURL(), and the original request’s information will be found in the following request attributes:

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

These are set out in section 8.3 and 8.4 of the Servlet 2.4 specification.

However, be aware that this information is only preserved for internally-dispatched requests. If you have a front-end web-server, or dispatch outside of the current container, these values will be null. In other words, you may have no way to find the original request URL.