从Twitter API获取推文的Python方法

0 投票
1 回答
722 浏览
提问于 2025-04-17 21:25

我正在尝试运行一个脚本来收集推特上的推文,但它一直无法运行,总是提示语法错误。我想把这些推文收集起来并保存到亚马逊的服务器上,但现在这个脚本就是不工作。我已经创建并插入了所有的推特密钥和访问令牌到下面的脚本中。

这是我使用的Python版本:3.3.4(v3.3.4:7ff62415e426,2014年2月10日,18:13:51)[MSC v.1600 64位(AMD64)],运行在win32系统上。你可以输入“copyright”、“credits”或“license()”来获取更多信息。

>>> import oauth2 as oauth
import urllib2 as urllib



access_token_key = "****"
access_token_secret = "******"

consumer_key = "******"
consumer_secret = "******"

_debug = 0

oauth_token    = oauth.Token(key=access_token_key, secret=access_token_secret)
oauth_consumer = oauth.Consumer(key=consumer_key, secret=consumer_secret)

signature_method_hmac_sha1 = oauth.SignatureMethod_HMAC_SHA1()

http_method = "GET"


http_handler  = urllib.HTTPHandler(debuglevel=_debug)
https_handler = urllib.HTTPSHandler(debuglevel=_debug)

'''
Construct, sign, and open a twitter request
using the hard-coded credentials above.
'''
def twitterreq(url, method, parameters):
  req = oauth.Request.from_consumer_and_token(oauth_consumer,
                                             token=oauth_token,
                                             http_method=http_method,
                                             http_url=url, 
                                             parameters=parameters)

  req.sign_request(signature_method_hmac_sha1, oauth_consumer, oauth_token)

  headers = req.to_header()

  if http_method == "POST":
    encoded_post_data = req.to_postdata()
  else:
    encoded_post_data = None
    url = req.to_url()

  opener = urllib.OpenerDirector()
  opener.add_handler(http_handler)
  opener.add_handler(https_handler)

  response = opener.open(url, encoded_post_data)

  return response

def fetchsamples():
  url = "https://stream.twitter.com/1/statuses/sample.json"
  parameters = []
  response = twitterreq(url, "GET", parameters)
  for line in response:
    print line.strip()

if __name__ == '__main__':

  fetchsamples()

1 个回答

0

你代码的第一行有语法错误,因为出现了三个大于号:

$ cat z.py
>>> import oauth2 as oauth
$ python z.py
  File "z.py", line 1
    >>> import oauth2 as oauth
     ^
SyntaxError: invalid syntax

撰写回答