Getting an URL which led to error (404) from error-page controller in spring MVC

Question

Say, i have a Spring MVC application with the following web.xml entry:

<error-page>
    <error-code>404</error-code>
    <location>/error/404</location>
</error-page>

and following error-page controller:

@RequestMapping({"","/"})
@Controller
public class RootController {

    @RequestMapping("error/{errorId}")
    public String errorPage(@PathVariable Integer errorId, Model model) {
        model.addAttribute("errorId",errorId);

        return "root/error.tile";
    }
}

Now user requested non-existent URL /user/show/iamnotauser which triggered error page controller. How do i get this non-existent '/user/show/iamnotauser' URL from errorPage() method of RootController to put it into model and display on error page ?

Ralph · Accepted Answer · 2012-12-16 13:22:49Z

15

The trick is request attribute javax.servlet.forward.request_uri, it contains the original requested uri.

@RequestMapping("error/{errorId}")
public ModelAndView resourceNotFound(@PathVariable Integer errorId,
                                                   HttpServletRequest request) {
    //request.getAttribute("javax.servlet.forward.request_uri");
    String origialUri = (String) request.getAttribute(
                                               RequestDispatcher.FORWARD_REQUEST_URI);

    return new ModelAndView("root/error.jspx", "originalUri", origialUri);
}

If you still use Servlet API 2.5, then the constant RequestDispatcher.FORWARD_REQUEST_URI does not exist, but you can use request.getAttribute("javax.servlet.forward.request_uri"). or upgrad to javax.servlet:javax.servlet-api:3.0.1

answered Dec 16, 2012 at 13:22

Ralph

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1 Comment

Jens Schauder Over a year ago

We are on a server supporting Servlet API 3.0.x but using 2.5 as of specification in the web.xml. With this setting the approaches above didn't yield anything usefull, but request.getAttribute("javax.servlet.error.request_uri"); did the trick

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