GWT处理StatusCodeException

Question

如果我的服务方法抛出新的MyException（“某些原因”），如何从服务器端获取异常; 我想在onFailure方法中获取MyException，但实际上我得到了StatusCodeException。我是否可以获取MyException以在UI上显示错误消息： Window.alert（exception.getMessage（））; - ＆GT;打印：“某种原因”

@RemoteServiceRelativePath("reportService.rpc")
public interface ReportService extends RemoteService {
    void saveReport(ReportDTO reportDTO) throws MyException;
}

 @Override
    public void saveReport(ReportDTO reportDTO) throws MyException {
        //Report report = ReportFunc.INST.apply(reportDTO);
        //reportRepository.save(report);
        throw new MyException("Some reason");
    }

 reportServiceAsync.saveReport(reportDTO, new AsyncCallback<Void>() {
                    @Override
                    public void onSuccess(Void result) {
                        Window.alert("Successfully saved");
                    }

                    @Override
                    public void onFailure(Throwable e) {
                        Window.alert(e.getMessage());
                    }
                });

class MyException extends RuntimeException implements Serializable{
  ....
}

Answer 1

您的异常应该扩展Exception，而不是RuntimeException。