http.client.BadStatusLine:''在使用python3尝试app-only twitter oauth时

时间:2013-09-29 04:46:06

标签: python twitter oauth

我正在尝试创建一个处理Twitter个流的脚本。不幸的是,OAuth流程阻碍了我。采用我在互联网上找到的一些代码,我收到https://api.twitter.com/oauth/token的空白回复。为了更好地理解这个过程,我试图在没有特殊模块的情况下做到这一点。这是我的代码,缺少什么?任何帮助将不胜感激。

代码:

import http.client
import urllib
import base64

CONSUMER_KEY = 'yadayadayada'
CONSUMER_SECRET = 'I am really tired today'

encoded_CONSUMER_KEY = urllib.parse.quote(CONSUMER_KEY)
encoded_CONSUMER_SECRET = urllib.parse.quote(CONSUMER_SECRET)
concat_consumer_url = encoded_CONSUMER_KEY + ':' + encoded_CONSUMER_SECRET


host = 'api.twitter.com'
url = '/oauth2/token/'
params = urllib.parse.urlencode({'grant_type' : 'client_credentials'})
req = http.client.HTTPSConnection(host, timeout = 100)
req.set_debuglevel(1)
req.putrequest("POST", url)
req.putheader("Host", host)
req.putheader("User-Agent", "My Twitter 1.1")
req.putheader("Authorization", "Basic %s" % base64.b64encode(b'concat_consumer_url'))
req.putheader("Content-Type", "application/x-www-form-urlencoded;charset=UTF-8")
req.putheader("Content-Length", "29")
req.putheader("Accept-Encoding", "identity")
req.endheaders()

req.send(b'params')

resp = req.getresponse()


print ("{} {}".format(resp.status, resp.reason))

错误讯息:

C:\Python33>app_only_test_klug.py
Traceback <most recent call last>:
    File "C:\Python33\app_only_test_klug.py", line 31, in <module>
        resp = req.getresponse()
    File "C:\Python33\lib\http\client.py", line 1131, in getresponse
        response.being()
    File "C:\Python33\lib\http\client.py", line 354, in begin
        version, status, reason = self._read_status()
    File "C:\Python33\lib\http\client.py", line 324, in _read_status
        raise BadStatusLine(line)
http.client.BadStatusLine: ''

非常感谢任何帮助。

更新:

经过一些修补,我相信问题在于我的base64编码:

req.putheader(“授权”,“基本%s”%base64.b64encode(b'concat_consumer_url'))

当我解码上面得到的编码时,我得到“b'concat_consumer_url”,而不是在冒号周围组合的encoded_CONSUMER_KEY和encoded_CONSUMER_SECRET的串联。如何让base64对btadcode表示concat_comsumer_url所代表的值而不是字符串“concat_consumer_url”以便我可以继续前进?提前谢谢。

1 个答案:

答案 0 :(得分:0)

我相信问题也存在 - 您应该将变量传递给编码函数,而不是将变量的名称作为字节传递,如下所示:

req.putheader("Authorization", "Basic %s" % base64.b64encode(concat_consumer_url))

再次尝试更改。