从Spring MVC中的错误页面控制器获取导致错误(404)的URL

Question

说,我有一个带有以下web.xml条目的Spring MVC应用程序:

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

并跟随错误页面控制器:

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

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

        return "root/error.tile";
    }
}

现在用户请求不存在的URL/user/show/iamnotauser,它触发了错误页面控制器.如何从RootController的errorPage()方法获取这个不存在的'/ user/show/iamnotauser'URL以将其放入模型并显示在错误页面上？

Answer 1

诀窍是请求属性javax.servlet.forward.request_uri,它包含原始请求的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);
}

如果您仍然使用Servlet API 2.5,那么常量RequestDispatcher.FORWARD_REQUEST_URI不存在,但您可以使用request.getAttribute("javax.servlet.forward.request_uri").或升级到javax.servlet:javax.servlet-api:3.0.1