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5462

answers:

3

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.

+7  A: 

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.

kdgregory
I tried out.println(request.getAttribute("request_uri")); out.println(request.getAttribute("context_path")); out.println(request.getAttribute("servlet_path")); out.println(request.getAttribute("path_info")); out.println(request.getAttribute("query_string"));All returned null. Does this mean I'm screwed?
Ballsacian1
The actual attribute name is "javax.servlet.include.request_uri", not "request_uri"
kdgregory
Thanks that works. Need to add a little more logic in my navigation, but hey saves me a lot of trouble. Thanks.
Ballsacian1
+1  A: 
String activePage = "";
    // using getAttribute allows us to get the orginal url out of the page when a forward has taken place.
    String queryString = "?"+request.getAttribute("javax.servlet.forward.query_string");
    String requestURI = ""+request.getAttribute("javax.servlet.forward.request_uri");
    if(requestURI == "null") {
     // using getAttribute allows us to get the orginal url out of the page when a include has taken place.
     queryString = "?"+request.getAttribute("javax.servlet.include.query_string");
     requestURI = ""+request.getAttribute("javax.servlet.include.request_uri");
    }
    if(requestURI == "null") {
     queryString = "?"+request.getQueryString();
     requestURI = request.getRequestURI();
    }
    if(queryString.equals("?null")) queryString = "";
    activePage = requestURI+queryString;
Ballsacian1
A: 

how to get url parameters i m getting null pointer exception for the lines

String ename=request.getAttribute("javax.servlet.forword.q").toString(); System.out.println("----------------"+ename);

smita