Java HttpServletRequest get URL in browsers URL bar

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.

Solution 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.

Solution 2

Just did a slight tidy of the solution by Ballsacian1

String currentURL = null;
if( request.getAttribute("javax.servlet.forward.request_uri") != null ){
    currentURL = (String)request.getAttribute("javax.servlet.forward.request_uri");
}
if( currentURL != null && request.getAttribute("javax.servlet.include.query_string") != null ){
    currentURL += "?" + request.getQueryString();
}

The null checks are going to run a lot more efficiently than String comparisons.

Solution 3

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;

Solution 4

${requestScope['javax.servlet.forward.query_string']} -- if you access it form jsp, using Expression Language

Solution 5

To get the HTTP requested path without know the state of the internal flow of the request, use this method:

public String getUri(HttpServletRequest request) {
    String r = (String) request.getAttribute("javax.servlet.forward.request_uri");
    return r == null ? request.getRequestURI() : r;
}

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