我正在使用REST

API。接收到带有错误JSON的POST消息(例如{sdfasdfasdf})会使Spring返回默认服务器页面，以显示400错误请求错误。我不想返回页面，我想返回自定义JSON错误对象。

当使用@ExceptionHandler引发异常时，可以执行此操作。因此，如果它是一个空白请求或一个空白JSON对象(例如{})，它将抛出NullPointerException，我可以使用ExceptionHandler捕获它并做我想做的任何事情。

那么问题是，当Spring只是无效的语法时，它实际上并没有引发异常……至少我看不到。它只是从服务器返回默认错误页面，无论是Tomcat，Glassfish等。

所以我的问题是如何“拦截” Spring并使其使用我的异常处理程序，否则将阻止错误页面的显示并返回JSON错误对象？

这是我的代码：

@RequestMapping(value = "/trackingNumbers", method = RequestMethod.POST, consumes = "application/json")

@ResponseBody

public ResponseEntity setTrackingNumber(@RequestBody TrackingNumber trackingNumber) {

HttpStatus status = null;

ResponseStatus responseStatus = null;

String result = null;

ObjectMapper mapper = new ObjectMapper();

trackingNumbersService.setTrackingNumber(trackingNumber);

status = HttpStatus.CREATED;

result = trackingNumber.getCompany();

ResponseEntity response = new ResponseEntity(result, status);

return response;

}

@ExceptionHandler({NullPointerException.class, EOFException.class})

@ResponseBody

public ResponseEntity resolveException()

{

HttpStatus status = null;

ResponseStatus responseStatus = null;

String result = null;

ObjectMapper mapper = new ObjectMapper();

responseStatus = new ResponseStatus("400", "That is not a valid form for a TrackingNumber object " +

"({\"company\":\"EXAMPLE\",\"pro_bill_id\":\"EXAMPLE123\",\"tracking_num\":\"EXAMPLE123\"})");

status = HttpStatus.BAD_REQUEST;

try {

result = mapper.writeValueAsString(responseStatus);

} catch (IOException e1) {

e1.printStackTrace();

}

ResponseEntity response = new ResponseEntity(result, status);

return response;

}