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

虽然标记的答案是正确的,但有一种方法可以做到这一点,没有例外。服务返回搜索对象的Optional<T>,这被映射到HttpStatus。如果找到OK,如果为空则返回404。

@Controller
public class SomeController {

    @RequestMapping.....
    public ResponseEntity<Object> handleCall(@PathVariable String param) {
        return  service.find(param)
                .map(result -> new ResponseEntity<>(result, HttpStatus.OK))
                .orElse(new ResponseEntity<>(HttpStatus.NOT_FOUND));
    }
}

@Service
public class Service{
  
    public Optional<Object> find(String param){
        if(!found()){
            return Optional.empty();
        }
        ...
        return Optional.of(data); 
    }
    
}

其他回答

我建议抛出HttpClientErrorException,就像这样

@RequestMapping(value = "/sample/")
public void sample() {
    if (somethingIsWrong()) {
        throw new HttpClientErrorException(HttpStatus.NOT_FOUND);
    }
}

您必须记住,这只能在将任何内容写入servlet输出流之前进行。

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;
        }
}

虽然标记的答案是正确的,但有一种方法可以做到这一点,没有例外。服务返回搜索对象的Optional<T>,这被映射到HttpStatus。如果找到OK,如果为空则返回404。

@Controller
public class SomeController {

    @RequestMapping.....
    public ResponseEntity<Object> handleCall(@PathVariable String param) {
        return  service.find(param)
                .map(result -> new ResponseEntity<>(result, HttpStatus.OK))
                .orElse(new ResponseEntity<>(HttpStatus.NOT_FOUND));
    }
}

@Service
public class Service{
  
    public Optional<Object> find(String param){
        if(!found()){
            return Optional.empty();
        }
        ...
        return Optional.of(data); 
    }
    
}

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。

重写你的方法签名,使它接受HttpServletResponse作为参数,这样你就可以对它调用setStatus(int)。

http://static.springsource.org/spring/docs/3.0.x/spring-framework-reference/html/mvc.html#mvc-ann-requestmapping-arguments