如何让Spring 3.0控制器触发404?

我有一个控制器@RequestMapping(值= "/**",方法= RequestMethod.GET)和一些访问控制器的url,我希望容器提出一个404。


当前回答

这有点晚了,但如果你正在使用Spring Data REST,那么已经有org.springframework.data.rest.webmvc.ResourceNotFoundException 它还使用@ResponseStatus注释。不再需要创建自定义运行时异常。

其他回答

如果你的控制器方法是用于文件处理,那么ResponseEntity是非常方便的:

@Controller
public class SomeController {
    @RequestMapping.....
    public ResponseEntity handleCall() {
        if (isFound()) {
            return new ResponseEntity(...);
        }
        else {
            return new ResponseEntity(404);
        }
    }
}

如果你想从控制器返回404状态,你只需要这样做

@RequestMapping(value = "/something", method = RequestMethod.POST)
@ResponseBody
public HttpStatus doSomething(@RequestBody String employeeId) {
    try {
        return HttpStatus.OK;
    } 
    catch (Exception ex) { 
         return HttpStatus.NOT_FOUND;
    }
}

通过这样做,当您想从控制器返回404时,您将收到一个404错误。

使用setting配置web.xml

<error-page>
    <error-code>500</error-code>
    <location>/error/500</location>
</error-page>

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

创建新控制器

   /**
     * Error Controller. handles the calls for 404, 500 and 401 HTTP Status codes.
     */
    @Controller
    @RequestMapping(value = ErrorController.ERROR_URL, produces = MediaType.APPLICATION_XHTML_XML_VALUE)
    public class ErrorController {


        /**
         * The constant ERROR_URL.
         */
        public static final String ERROR_URL = "/error";


        /**
         * The constant TILE_ERROR.
         */
        public static final String TILE_ERROR = "error.page";


        /**
         * Page Not Found.
         *
         * @return Home Page
         */
        @RequestMapping(value = "/404", produces = MediaType.APPLICATION_XHTML_XML_VALUE)
        public ModelAndView notFound() {

            ModelAndView model = new ModelAndView(TILE_ERROR);
            model.addObject("message", "The page you requested could not be found. This location may not be current.");

            return model;
        }

        /**
         * Error page.
         *
         * @return the model and view
         */
        @RequestMapping(value = "/500", produces = MediaType.APPLICATION_XHTML_XML_VALUE)
        public ModelAndView errorPage() {
            ModelAndView model = new ModelAndView(TILE_ERROR);
            model.addObject("message", "The page you requested could not be found. This location may not be current, due to the recent site redesign.");

            return model;
        }
}

我想提一下,Spring默认提供了404异常(不仅是)。有关详细信息,请参阅Spring文档。所以如果你不需要自己的异常,你可以简单地这样做:

 @RequestMapping(value = "/**", method = RequestMethod.GET)
 public ModelAndView show() throws NoSuchRequestHandlingMethodException {
    if(something == null)
         throw new NoSuchRequestHandlingMethodException("show", YourClass.class);

    ...

  }

从Spring 5.0开始,你不需要创建额外的异常:

throw new ResponseStatusException(NOT_FOUND, "Unable to find resource");

此外,你可以用一个内置异常覆盖多个场景,你有更多的控制。

看到更多:

ResponseStatusException (javadoc) https://www.baeldung.com/spring-response-status-exception