jle*_*and 3 python twitter streaming twitter-oauth pythonanywhere
我试图在任何地方连接到Python上的Twitter流API,但总是得到连接拒绝错误.
我在我的应用程序中使用Tweepy,并测试连接我正在使用可以在repo中找到的流示例.
HEre是代码的总结:
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'])
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当我在任何地方的python中的bash控制台中运行此行时(当然在填充了令牌之后)
12:02 ~/tweepy/examples (master)$ python streaming.py
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我收到以下错误:
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
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域.twitter.com虽然在pythonanywhere whithelist,所以我不明白为什么连接会被拒绝:s.
完全相同的代码就像我的Ubuntu上的魅力一样.
任何想法都会受到欢迎,谢谢!!
如果您使用的是免费帐户,则tweepy将无效.它不使用环境中的代理设置.
在主线正确使用代理设置之前,您可以使用(http://github.com/ducu/tweepy).
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