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。
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