Question

我已经定义了一个全局异常处理程序，如下所示。它毫无问题地阻止了例外。问题是我使用spring作为我的角度后端，所以我不想返回一个ModelAndView对象。我的问题是;

我不希望在全局处理程序之后由任何其他类处理异常。
我想直接从全局错误处理程序返回响应。

GlobalExceptionHandler。

public class GlobalExceptionHandler implements HandlerExceptionResolver, Ordered {
    public int getOrder() {
        return Integer.MIN_VALUE; 
    }

    public ModelAndView resolveException(
        HttpServletRequest aReq, HttpServletResponse aRes,
        Object aHandler, Exception ex
    ) {
         // i want to add something like below code here
        return null; // if i return null here, the next handler is triggered. I don`t want to do that. I want program to stop here and return a response.

    }
}

我希望能够直接从全局处理程序

           aRes.reset();
           aRes.setHeader("Access-Control-Allow-Origin", "http://localhost:4200");
           aRes.setHeader("Access-Control-Allow-Credentials", "true");
           aRes.setHeader("Access-Control-Allow-Methods","POST, GET, OPTIONS, DELETE");
           aRes.setHeader("Access-Control-Allow-Headers", "Origin, X-Requested-With, Authorization, Content-Type, Accept, enctype");
           aRes.sendError(ErrorCodes.UNAUTHORIZED_USER, "The service is booting.");

Answer 1

您可以扩展类ResponseEntityExceptionHandler并使用@ControllerAdvice注释对其进行注释。在这里，您可以通过覆盖现有的受保护方法来自定义如何处理特定的Spring Web异常，例如：

 @Override
 protected ResponseEntity<Object> handleHttpMessageNotReadable(HttpMessageNotReadableException ex, HttpHeaders headers, HttpStatus status,
  WebRequest request) {
    // Do anything you want here with the httpheaders and just return a ResponseEntity with an null body
 }

如果需要，您可以向此类添加其他方法来处理您自己的异常类型，例如：

@ResponseBody
@ExceptionHandler(ApplicationException.class)
public ResponseEntity<ErrorInfo> handleApplicationException(ApplicationException applicationException) {
   // Called for our own application exception
}

如何在Spring中直接从HandlerExceptionResolver返回错误响应

1 个答案: