Question

如何抛出具有特定http状态的异常？我想用403 Forbidden状态生成异常。我使用了jax-rs库，但它只是更改了消息，状态代码仍然有500个。

if (countRequest > 10) {
    throw new WebApplicationException(Response.Status.FORBIDDEN);
}

邮递员的输出：

{
    "timestamp": 1502582674498,
    "status": 500,
    "error": "Internal Server Error",
    "exception": "javax.ws.rs.WebApplicationException",
    "message": "HTTP 403 Forbidden",
    "path": "/timesheets/late/61/cancel"
}

Answer 1

您可以使用ExceptionMapper执行此操作，如下所示：

  @Provider 
  public class CustomExceptionMapper implements ExceptionMapper<WebApplicationException> { 
      @Override public Response toResponse(WebApplicationException weException) { 
           // grab the response from exception
           Response response = weException.getResponse(); 
           // create error object/message as you like
           CustomError error = ...; 
           // return the custom error 
           return Response.status(response.getStatus()).entity(error).build(); 
       }
  }

您可以参考this教程，了解ExceptionMapper的更多详细信息。

如何在JAVA中编写具有特定http状态代码的异常类

1 个答案: