web.xml 404 redirect to servlet, how to get the original URI?

servlets http-status-code-404 web.xml

Eric Pierce · Sep 19, 2011 · Viewed 8.6k times · Source

I'm redirecting 404 errors to a servlet via the following in my web.xml.

<error-page>
    <error-code>404</error-code>
    <location>/notFound.do</location>
</error-page>

I'd like to log where the request was trying to go, but I'm not getting it from the referrer header: request.getHeader("referer")

That shows 'null' if I just hit any old random non-existent page.

And request.getRequestURL()/request.getRequestURI() both merely shows the final landing servlet info (I.e., /notFound).

Any way to get the 'bad' page URL that was requested?

Answer

Yes, it's available as a request attribute with the name javax.servlet.forward.request_uri, which is keyed by RequestDispatcher#FORWARD_REQUEST_URI. The error page location is namely invoked by a simple RequestDispatcher#forward() call.

So, you can get it as follows in servlet:

String originalUri = request.getAttribute(RequestDispatcher.FORWARD_REQUEST_URI);

or in EL:

<p>Original URI: ${requestScope['javax.servlet.forward.request_uri']}</p>