如何修复“ on_exception()缺少1个必需的位置参数:异常”?

时间:2019-05-24 05:43:53

标签: python twitter tweepy

我正在尝试使用tweepy从Streaming API流推文,并打印已流化的每个推文的文本。我收到以下错误:

  

TypeError:on_exception()缺少1个必需的位置参数:   “例外”

我为什么要得到它,如何解决?

我遇到了类似的错误,它说将请求库降级到2.7。不用说,它没有用。

    class Authenticator():
             def authenticate_and_get_API_object(self):
                     auth = tweepy.OAuthHandler(consumerkey,consumersecret)               
                auth.set_access_token(accesstoken,accesstokensecret)

                     api = tweepy.API(auth)
                     return api

    class MyStreamListener(tweepy.StreamListener):
            def on_status(self,status):
                    print(status.text)

            def on_error(self,status_code):
                    if status_code==420:
                            return false

    if __name__ == "__main__":

            my_streamer = MyStreamListener()
            the_api = Authenticator().authenticate_and_get_API_object()
            myStream = tweepy.Stream(auth = the_api.auth, listener = MyStreamListener)
            myStream.filter(track=['python'])

任何帮助将不胜感激

错误回溯:

Traceback (most recent call last):
  File "streamdemtweets.py", line 24, in <module>
    myStream.filter(track=['python'])
  File "/home/ansuman/.local/lib/python3.7/site-packages/tweepy/streaming.py", line 453, in filter
    self._start(is_async)
  File "/home/ansuman/.local/lib/python3.7/site-packages/tweepy/streaming.py", line 368, in _start
    self._run()
  File "/home/ansuman/.local/lib/python3.7/site-packages/tweepy/streaming.py", line 299, in _run
    self.listener.on_exception(exc_info[1])
TypeError: on_exception() missing 1 required positional argument: 'exception'

1 个答案:

答案 0 :(得分:1)

我解决了实例化时在MyStreamListener中添加()的问题:

myStream = tweepy.Stream(auth = the_api.auth, listener = MyStreamListener())