Question

我有一个休息的网络服务。如果引发任何异常，Web服务将返回http 500错误。但是我不想发送带有异常堆栈跟踪的错误响应。我只想发送错误代码和错误消息。我没有实现。我该怎么办？

我已经尝试过@ControllerAdvice和@ExceptionHandler批注，但是我不能。当我使用@ResponseStatus批注时，请始终发送静态的“ reason”值。如何设置该值？感谢您的帮助。

public class SendMessageController{
    private Logger log = LogManager.getLogger(getClass());
    @Autowired
    private QueueService queueService;

    @RequestMapping(value="/message/check", method = RequestMethod.POST, headers={ "content-type=application/json"})
    public @ResponseBody
    ApiResponse sendMessage(@RequestBody String requestMessage) throws Exception {
        try {
            return new ApiResponse(queueService.processRequestForJSONString(requestMessage);
        } catch (Exception e) {
            throw new GenericException(HttpStatus.INTERNAL_SERVER_ERROR, e.getMessage());
            //throw e;
        }
    }

    @ResponseStatus(value=HttpStatus.INTERNAL_SERVER_ERROR, reason="Exception Message")
    public class GenericException extends Exception {
        public HttpStatus httpCode;
        public String errorMessage;

        public GenericException(HttpStatus httpCode, String errorMessage){
            this.httpCode = httpCode;
            this.errorMessage = errorMessage;
            //I can't set "reason"
        }
    }
}

Answer 1

有许多可能的解决方案，我很确定ErrorHandler是一种更好的选择。

@GetMapping(value="/{empId}", produces=MediaType.APPLICATION_JSON_VALUE)    
public ResponseEntity<EmployeeInfoItem> getEmployeeInfo(@PathVariable("empId") Integer empId) {
    try {
        ...
    } catch (Exception e) {
        logger.error( e.getMessage() );
        return ResponseEntity.status(HttpStatus.FAILED_DEPENDENCY).build();
    }
}

其余-如何发送不带堆栈跟踪的Http错误响应

1 个答案: