如何流式传输特定用户的推文并将它们作为变量返回答案

【问题标题】：how can I stream a specific user's tweets and return them as a variable如何流式传输特定用户的推文并将它们作为变量返回
【发布时间】：2021-03-16 12:42:38
【问题描述】：

我正在尝试流式传输特定用户的推文，然后将它们作为基于事件的触发器的变量返回。我正在使用以下代码，但它似乎没有为我返回任何推文。我只对他们的推文感兴趣而不是转推，并且想跳过不是由指定用户发布的推文。

知道我哪里出错了：

import tweepy

#API login for Twitter
consumer_key= CONSUMER_KEY
consumer_secret= CONSUMER_SECRET
access_key= ACCESS_TOKEN
access_secret= ACCESS_TOKEN_SECRET

class StreamListener(tweepy.StreamListener):
    def on_status(self, status):
        if status.user.id_str != '111111111':
            return
        tweet = status.text
        return tweet

auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_key, access_secret)
api = tweepy.API(auth)

# initialize stream
streamListener = StreamListener()
stream = tweepy.Stream(auth=api.auth, listener=streamListener)
stream.filter(follow=['111111111'])

print(tweet)

【问题讨论】：

标签： python twitter streaming tweepy

【解决方案1】：

以下是流式传输特定用户推文的代码：

import tweepy
import sys


#API login for Twitter
consumer_key= CONSUMER_KEY
consumer_secret= CONSUMER_SECRET
access_key= ACCESS_TOKEN
access_secret= ACCESS_TOKEN_SECRET

# StreamListener class inherits from tweepy.StreamListener and overrides on_status/on_error methods.
class StreamListener(tweepy.StreamListener):
    def on_status(self, status):
        if status.user.id_str != '11111111': 
            return
        text = status.text
        print(text)


    def on_error(self, status_code):
        print("Encountered streaming error (", status_code, ")")
        sys.exit()

if __name__ == "__main__":
    # complete authorization and initialize API endpoint
    auth = tweepy.OAuthHandler(consumer_key, consumer_secret)
    auth.set_access_token(access_key, access_secret)
    api = tweepy.API(auth)

    # initialize stream
    streamListener = StreamListener()
    stream = tweepy.Stream(auth=api.auth, listener=streamListener,tweet_mode='extended')
    stream.filter(follow=['11111111'])

【讨论】：