我正在尝试在linkedin api中进行身份验证，以便从个人和公司资料中获取一些数据。我对developer.linkedin.com提供的OAuth2.0文档非常困惑。我不确定我是否正在使用正确的api调用访问数据点。例如，我想要了解许多用户的工作历史等等。但在我点击它之前，如何验证用户进行api调用？我使用以下芽不起作用 -
client_id ='.....'
client_secret='.....'
authorization_base_url = 'https://www.linkedin.com/uas/oauth2/authorization'
token_url = 'https://www.linkedin.com/uas/oauth2/accessToken'
linkedin = OAuth2Session(client_id, redirect_uri='https://127.0.0.1')
linkedin = linkedin_compliance_fix(linkedin)
authorization_url, state = linkedin.authorization_url(authorization_base_url)
print 'Please go here and authorize,', authorization_url
redirect_response = raw_input('Paste the full redirect URL here:')
#here I add https://127.0.0.1
linkedin.fetch_token(token_url, client_secret=client_secret,
authorization_response=redirect_response)
r = linkedin.get('https://api.linkedin.com/v1/people/~')
print r.content
我从这里按照教程：https://github.com/requests/requests-oauthlib/blob/master/docs/examples/linkedin.rst
这是我输入的消息 - https://127.0.0.1作为终端请求它。
python ~/linkedin.py
Please go here and authorize, https://www.linkedin.com/uas/oauth2/authorization?response_type=code&client_id=777m76reoggue0&redirect_uri=https%3A%2F%2F127.0.0.1&state=bo91J8MNuMTx1WQWn9nLd5kgGbZYFE
**Paste the full redirect URL here:https://127.0.0.1**
Traceback (most recent call last):
File "~/linkedin.py", line 22, in <module>
authorization_response=redirect_response)
File "/anaconda/lib/python2.7/site-packages/requests_oauthlib/oauth2_session.py", line 187, in fetch_token
state=self._state)
File "/anaconda/lib/python2.7/site-packages/oauthlib/oauth2/rfc6749/clients/web_application.py", line 174, in parse_request_uri_response
response = parse_authorization_code_response(uri, state=state)
File "/anaconda/lib/python2.7/site-packages/oauthlib/oauth2/rfc6749/parameters.py", line 227, in parse_authorization_code_response
raise MissingCodeError("Missing code parameter in response.")
oauthlib.oauth2.rfc6749.errors.MissingCodeError: (missing_code) Missing code parameter in response.
确保：
?=code参数粘贴你在浏览器中获得的完整重定向URL，在我看来如下：
在[19]中：redirect_response
出[19]：'https://127.0.0.1/?code=AQQyB2NCpL5iE_wyOqmqTm5z......iKhx#！'