在Spring 3中是否可以动态设置@ResponseStatus的原因？

Question

我有一个自定义的异常类注释返回给定HttpStatus:

@ResponseStatus(value=HttpStatus.BAD_REQUEST, reason="Invalid parameter")
public class BadRequestException extends RuntimeException
{
  public BadRequestException(String msg)
  {
    super(msg);
  }
}

当我BadRequestException从我的控制器中抛出一个但是原因总是"无效的参数"当然有效.有没有办法在这个类中设置返回的原因？我想传递一个字符串作为原因.

谢谢!

Answer 1

如果在自定义异常的@ResponseStatus注释中省略'reason'属性,

@ResponseStatus(value = HttpStatus.CONFLICT)  // 409
public class ChildDataExists extends RuntimeException {
...

然后抛出异常

throw new ChildDataExists("Can't delete parent if child row exists.");

异常消息作为JSON输出中"数据"的"消息"传递.注释中的"原因"似乎会覆盖自定义行为.

Answer 2

您可以使用 response.sendError(HttpServletResponse.SC_BAD_REQUEST, "Invalid foo");

Answer 3

正确的方法是在控制器中引入异常处理程序,然后可以设置任何状态代码的响应主体:

@Controller
@RequestMapping(produces = MediaType.APPLICATION_JSON_VALUE)
public class SomeController {
...
  @ExceptionHandler(BadRequestException.class)
  @ResponseStatus(HttpStatus.BAD_REQUEST)
  public @ResponseBody
   Map<String,Object> handleIndexNotFoundException(BadRequestException bre,
                                           HttpServletRequest request, HttpServletResponse resp) {
     HashMap<String, Object> result = new HashMap<>();
     result.put("error", true);
     result.put("error_message", bre.getMessage());
     return result;
  }
}

移过您不必使用任何Spring Web MVC注释和依赖项污染您的模型/异常类.

如果你想与所有控制器共享处理程序,请查看@ControllerAdvice.

Answer 4

从 spring 5.0 开始，您可以使用ResponseStatusException可用的

// From https://www.baeldung.com/spring-response-status-exception
@GetMapping("/actor/{id}")
public String getActorName(@PathVariable("id") int id) {
    try {
        return actorService.getActor(id);
    } catch (ActorNotFoundException ex) {
        throw new ResponseStatusException(
          HttpStatus.NOT_FOUND, "Actor Not Found", ex);
    }
}