Tweepy Stream error：__init __（）需要刚好3个参数（给出4个参数）

Question

我试图使用Tweepy python api创建一个twitter搜索流，但我面临一个错误。这是我想执行的代码，我得到的错误 -

from tweepy.streaming import StreamListener 
from tweepy import OAuthHandler 
from tweepy import Stream 

# Twitter Credentials 
access_token_key = "*****" 
access_token_secret = "*****" 
consumer_key = "*****" 
consumer_secret = "*****" 

class StdOutListener(StreamListener): 
    """ A listener handles tweets that are the received from the stream. 
    This is a basic listener that just prints received tweets to stdout. 
    """ 

    def on_data(self, data): 
     print data 
     return True 

    def on_error(self, status): 
     print status 

if __name__ == '__main__': 
    x = StdOutListener() 
    auth = OAuthHandler(consumer_key, consumer_secret) 
    auth.set_access_token(access_token_key, access_token_secret) 

    stream = Stream(auth, x, "microsoft") 
    response = stream.filter(track=['microsoft']) 

print response 

Answer 1

问题出在流intitialization>

File "code.py", line 28, in <module> 
    stream = Stream(auth, x, "microsoft") 
__init__() takes exactly 3 arguments (4 given) 

（感谢您的帮助对不起，我是初学者。）行：

stream = Stream(auth, x, "microsoft") 

，应该是

stream = Stream(auth, x) 

跟踪的单词不通过构造函数传递，而是通过filter方法传递。

Answer 2

Stream()只需要2个参数（加self的绑定方法）：

stream = Stream(auth, x) 
stream.filter(track=['microsoft']) 

见的Tweepy项目给出的streaming example。

应该将307重定向处理为非错误;你可以告诉Tweepy从on_error处理程序返回True继续：

def on_error(self, status): 
    print status 
    if status == 307: 
     return True