如何从IO异常中检测404响应代码？

Question

您如何从IO异常中检测404.我可以只搜索错误消息"404",但这是正确的方法吗？有什么更直接的吗？

  import com.google.api.services.drive.model.File;
  import com.google.api.services.drive.Drive.Files.Update;
  import com.google.api.services.drive.Drive;


    File result = null;

    try {

        update = drive.files().update(driveId, file , mediaContent);
        update.setNewRevision(true);

        result = update.execute();

    } catch (IOException e) {

        Log.e(TAG, "file update exception, statusCode: " + update.getLastStatusCode());
        Log.e(TAG, "file update exception, e: " + e.getMessage());

    }   

Log.e(TAG, "file update exception, statuscode " + update.getLastStatusCode());

03-03 05:04:31.738: E/System.out(31733): file update exception, statusCode: -1
03-03 05:04:31.738: E/System.out(31733): file update exception, e: 404 Not Found
03-03 05:04:31.738: E/System.out(31733): "message": "File not found: FileIdRemoved",

答:以下Aegan的评论是正确的,事实证明你可以将异常子类化为GoogleJsonResponseException并从那里获取状态代码.在这种情况下,答案最终取决于我使用的是GoogleClient,它生成包含状态代码的IO Exception的子类.

例:

Try{
    ...
    }catch (IOException e) {

      if(e instanceof GoogleJsonResponseException){
          int statusCode = ((GoogleJsonResponseException) e).getStatusCode();
          //do something
   }
 }

Answer 1

处理HttpResponseException:

catch (HttpResponseException hre) {
   if (hre.getStatusCode() == 404) {
      // TODO: Handle Http 404 
   }
}

细节: AbstractGoogleClientRequest创建例外.请参阅源代码

execute方法调用executeUnparsed.executeUnparsed创建异常newExceptionOnError.在那里你会看到,它抛出一个HttpResponseException(这是一个子类IOException)