Question

我的申请是使用Tweepy转发标签。转发标签工作，我无法让错误2和3工作。错误 1.你自己的id（完成） 2.如果推文已经被RTd 3.如果RT的推文来自受保护的来源

在StdOutListener中访问api.retweet（doTweet）不允许它落入on_error（）。我怎么能这样做？我是Python noob。

class StdOutListener(tweepy.StreamListener):
    def on_data(self, data):      
        all_data = json.loads(data)
        username = all_data["user"]["screen_name"]
        doTweet = all_data["id"]
        if username != our_own_id:
            #make sure you haven't already retweeted
            #make sure tweets aren't protected
            print(username) # just so we know it's working
            api.retweet(doTweet)
        return True

    def on_error(self, status_code):
        print('error')
        read_error = json.loads(status_code)
        print('Got an error with status code: ' + str(read_error))          
        return True # To continue listening

    def on_timeout(self):
        print('Timeout...')
        return True # To continue listening

try:
    if __name__ == '__main__':  
      listener = StdOutListener() 
      stream = tweepy.Stream(auth, listener)
      stream.filter(track=['#love'])

except KeyboardInterrupt:
   sys.exit()

Answer 1

您可以在api.retweet方法中修改对on_data的通话，如下所示：

...
if username != our_own_id:
    print(username) # just so we know it's working
    try:
        api.retweet(doTweet)
    except tweepy.TweepError as e:
        # add here a more complex error handling
        print(e)

希望它有所帮助。

使用api.retweet（）时出错处理错误Tweepy 3.5 / Python3

1 个答案: