Custom error page - get originally requested URL

问题:

Problem

This is a follow-up to yesterday's (unanswered) question (see here) as I try to find an alternative approach.

I added the basic

    <error-page>  
            <error-code>404</error-code>  
            <location>/404search.jsf</location>  
    </error-page>

..to my web.xml. I now need to get the URL the user entered to submit to my search function, but I only manage to get the current URL (in this case, ...404search.jsf) instead of the actual query the user entered.

Attempts

• HttpServletRequest.getRequestURL returns http://www.website.com/foldername/404search.jsf
• HttpServletRequest.getRequestURI returns /foldername/404search.jsf
• HttpServletRequest.getQueryString returns nothing at all

I want it to return /foldername/wrong-url-the-user-entered#anchor-if-there-is-any

Details...

The idea is to get the URL the user entered (such as www.site.com/product/99999-product-that-cant-be-found or www.site.com/faq/support-question-that-doesnt-exist), REGEX it to remove the hyphens and run a search query with 99999 product that cant be found or support question that doesnt exist.

Any suggestions?

回答1:

The <error-page> is under the covers served by a RequestDispatcher#forward() call. All details of the original request are available as request attribues which are keyed by the keys as identified by RequestDispatcher#FORWARD_XXX constants:

• FORWARD_CONTEXT_PATH: "javax.servlet.forward.context_path"
• FORWARD_PATH_INFO: "javax.servlet.forward.path_info"
• FORWARD_QUERY_STRING: "javax.servlet.forward.query_string"
• FORWARD_REQUEST_URI: "javax.servlet.forward.request_uri"
• FORWARD_SERVLET_PATH: "javax.servlet.forward.servlet_path"

You as starter should know that all request attributes are in EL available via the implicit EL object #{requestScope}.

So, all with all, this should do in the view:

<p>Unfortunately, the page you requested, #{requestScope['javax.servlet.forward.request_uri']} does not exist</p>

And equivalently, this should do in the bean, if necessary:

String forwardRequestURI = externalContext.getRequestMap().get(RequestDispatcher.FORWARD_REQUEST_URI);

回答2:

You might be able to get that URL with the "referer" (misspelled) request header.

http://www.w3.org/Protocols/rfc2616/rfc2616-sec14.html#sec14.36

Java usage: HttpServletRequest - how to obtain the referring URL?