python - 使用 API key 进行 Urllib2 身份验证

Question

我正在尝试连接到 radian6 api ，它需要 auth_appkey、auth_user 和 auth_pass 作为 md5 加密。

当我尝试使用 telnet 连接时，我可以成功获取响应 xml

telnet sandboxapi.radian6.com 80
Trying 142.166.170.31...
Connected to sandboxapi.radian6.com.
Escape character is '^]'.
GET /socialcloud/v1/auth/authenticate HTTP/1.1
host: sandboxapi.radian6.com
auth_appkey: 123456789
auth_user: xxx@xxx.com
auth_pass: 'md5encryptedpassword'

HTTP/1.1 200 OK
Server: Apache-Coyote/1.1
Date: Thu, 26 Aug 2010 14:17:52 GMT
Content-Type: application/xml
Content-Length: 471

但是当我使用以下代码在 python 中尝试相同的操作时，

import urllib
import urllib2

user = 'xxx@xxx.com'
password = 'md5encryptedpasswrod'
base_url = 'http://sandboxapi.radian6.com/'
api_key = '123456789'

pwman = urllib2.HTTPPasswordMgrWithDefaultRealm()
pwman.add_password(None, base_url, user, password)
auth_handler = urllib2.HTTPBasicAuthHandler(pwman)
opener = urllib2.build_opener(auth_handler)
urllib2.install_opener(opener)

req = urllib2.Request('http://sandboxapi.radian6.com/socialcloud/v1/auth/authenticate')
req.add_header('auth_appkey', api_key)
xml = urllib2.urlopen(req).read()

它抛出以下错误跟踪，

Traceback (most recent call last):
File "connectapi.py", line 17, in <module>
xml = urllib2.urlopen(req).read()
 File"/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 126, in urlopen
return _opener.open(url, data, timeout)
File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 397, in open
response = meth(req, response)
File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 510, in http_response
'http', request, response, code, msg, hdrs)
File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 435, in error
return self._call_chain(*args)
File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 369, in _call_chain
result = func(*args)
File "/opt/local/Library/Frameworks/Python.framework/Versions/2.6/lib/python2.6/urllib2.py", line 518, in http_error_default
raise HTTPError(req.get_full_url(), code, msg, hdrs, fp)
urllib2.HTTPError: HTTP Error 401: Unauthorized

我不知道我错过了什么。是 API key 还是 md5 加密密码导致我未经授权？
非常感谢您的智慧来挽救我的生命。

Answer 1

在您的 telnet session 中，您没有设置 Authorization: header ，但这是 HTTPBasicAuthHandler 使用的 header 。 (您可以使用 wireshark 或类似工具监听此内容。)大概该 API 不使用 HTTP 基本身份验证，而是使用其自制变体。您可能想要删除该行并手动设置 HTTP header 。