我正在使用REST

API。接收到帶有錯誤JSON的POST消息(例如{sdfasdfasdf})會使Spring返回默認服務器頁面，以顯示400錯誤請求錯誤。我不想返回頁面，我想返回自定義JSON錯誤對象。

當使用@ExceptionHandler引發異常時，可以執行此操作。因此，如果它是一個空白請求或一個空白JSON對象(例如{})，它將拋出NullPointerException，我可以使用ExceptionHandler捕獲它并做我想做的任何事情。

那么問題是，當Spring只是無效的語法時，它實際上并沒有引發異常……至少我看不到。它只是從服務器返回默認錯誤頁面，無論是Tomcat，Glassfish等。

所以我的問題是如何“攔截” Spring并使其使用我的異常處理程序，否則將阻止錯誤頁面的顯示并返回JSON錯誤對象？

這是我的代碼：

@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;

}