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
returnshttp://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?
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?
The
<error-page>
is under the covers served by aRequestDispatcher#forward()
call. All details of the original request are available as request attribues which are keyed by the keys as identified byRequestDispatcher#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:
And equivalently, this should do in the bean, if necessary: