我试图连接到任何地方的Python上的Twitter流媒体API,但总是得到一个连接拒绝错误。PythonAnywhere拒绝Twitter API的连接
我在我的应用程序中使用Tweepy,并测试了我正在使用可在回购中找到的流式示例的连接。
这里是代码的总和式:
from tweepy.streaming import StreamListener
from tweepy import OAuthHandler
from tweepy import Stream
# Go to http://dev.twitter.com and create an app.
# The consumer key and secret will be generated for you after
consumer_key=""
consumer_secret=""
# After the step above, you will be redirected to your app's page.
# Create an access token under the the "Your access token" section
access_token=""
access_token_secret=""
class StdOutListener(StreamListener):
""" A listener handles tweets 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__':
l = StdOutListener()
auth = OAuthHandler(consumer_key, consumer_secret)
auth.set_access_token(access_token, access_token_secret)
stream = Stream(auth, l)
stream.filter(track=['basketball'])
当运行此线在bash控制台python anywhere(在已填充的过程的令牌)
12:02 ~/tweepy/examples (master)$ python streaming.py
我得到出现以下错误:
Traceback (most recent call last):
File "streaming.py", line 33, in <module>
stream.filter(track=['basketball'])
File "/usr/local/lib/python2.7/site-packages/tweepy/streaming.py", line 228, in filter
self._start(async)
File "/usr/local/lib/python2.7/site-packages/tweepy/streaming.py", line 172, in _start
self._run()
File "/usr/local/lib/python2.7/site-packages/tweepy/streaming.py", line 106, in _run
conn.connect()
File "/usr/local/lib/python2.7/httplib.py", line 1157, in connect
self.timeout, self.source_address)
File "/usr/local/lib/python2.7/socket.py", line 571, in create_connection
raise err
socket.error: [Errno 111] Connection refused
域.twitter.com位于pythonanywhere whithel ist虽然,所以我不明白为什么连接会被拒绝:s。
非常相同的代码就像我的Ubuntu上的魅力。
任何想法都会超过欢迎,谢谢!
嘿,thx为您的答案!我看到一个月前关闭了代理支持。我可以这样看吗? – jlengrand
是的,封闭的拉不合并:s。意味着它已被拒绝。但我不知道为什么! – jlengrand