【发布时间】:2013-12-30 19:54:14
【问题描述】:
我正在编写一个 Twitter 应用程序并尝试使用 tweepy 进行授权。 我收到以下错误,不知道为什么..
谁能帮帮我?我将不胜感激。
Traceback (most recent call last):
File "getconv.py", line 32, in <module>
auth=AppAuthHandler(consumer_token,consumer_secret)
File "getconv.py", line 25, in __init__
response=urllib2.urlopen(req,data)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
response = meth(req, response)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 432, in error
result = self._call_chain(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
result = func(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 619, in http_error_302
return self.parent.open(new, timeout=req.timeout)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 400, in open
response = meth(req, response)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 513, in http_response
'http', request, response, code, msg, hdrs)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 438, in error
return self._call_chain(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 372, in _call_chain
result = func(*args)
File "/System/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/urllib2.py", line 521, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 403: Forbidden
以下是我的代码
import urllib2
import time
import sys
import tweepy
import base64
import urllib
#import twitter
consumer_token='my consumer token'
consumer_secret='my consumer secret'
access_token='my access token'
access_secret='my access secret'
class AppAuthHandler(tweepy.auth.AuthHandler):
TOKEN_URL='http://api.twitter.com/oauth2/token'
def __init__(self,consumer_key,consumer_secret):
token_credential=urllib.quote(consumer_key)+':'+urllib.quote(consumer_secret)
credential=base64.b64encode(token_credential)
value={'grant_type':'client_credentials'}
data=urllib.urlencode(value)
req=urllib2.Request(self.TOKEN_URL)
req.add_header('Authorization','Basic'+credential)
req.add_header('Content_Type','application/x-www-form-urlencoded;charset=UTF-8')
response=urllib2.urlopen(req,data)
json_response=json.loads(response.read())
self._access_token=json_response['access_token']
def apply_auth(self,url,method,headers,parameters):
headers['Authorization']='Bearer'+self._access_token
auth=AppAuthHandler(consumer_token,consumer_secret)
oauth_api=tweepy.API(auth)
【问题讨论】:
-
您的问题标题有点误导,因为(假设您的身份验证信息正确)它似乎是 tweepy 的问题。另一方面,您的 Header 意味着直接使用 urllib2 存在问题
-
@AndreasKlebinger 我现在不能确定错误是来自 urllib2 还是 tweepy。无论如何,我在标题上添加了 tweepy。
标签: python python-2.7 twitter tweepy http-error