python - 如何在 python 中获取 OAuth 访问令牌？

Question

（我在超级用户上问过这个问题，但没有得到回应......）

我正在尝试遵循http://taught-process.blogspot.com/2012/05/asdlasd-asda-sd-asd-asdasd.html上的 Dropbox API 教程

但是当我到达最后一部分时

#Print the token for future reference
print access_token

我得到的是

<dropbox.session.OAuthToken object at 0x1102d4210>

我如何获得实际的令牌？它应该看起来像：

oauth_token_secret=xxxxxxx&oauth_token=yyyyyyy

（我在 Mac 上）

Answer 1

Look around in the properties and methods of the object, to do so apply "dir" on the object. In your case:

dir(access_token)

I'm pretty sure you're gonna find in this object something that will give you the token you need.

Answer 2

你有正确的对象，是的。但是你正在处理一个类的实例。

<dropbox.session.OAuthToken object at 0x1102d4210>

这是 Dropbox SDK 为您创建的 OAuthToken 对象的一个​​实例。这个令牌似乎有两个属性：key和secret。这些将是您的令牌密钥和秘密。这就是你所追求的。

您可以像这样访问它们：

print access_token.key
print access_token.secret

Answer 3

在http://taught-process.blogspot.com/2012/05/asdlasd-asda-sd-asd-asdasd.html使用相同的 Dropbox API 教程

结束了以下对我有用的脚本

# Include the Dropbox SDK libraries
from dropbox import client, rest, session

# Get your app key and secret from the Dropbox developer website
APP_KEY = '3w7xv4d9lrkc7c3'
APP_SECRET = '1v5f80mztbd3m9t'

# ACCESS_TYPE should be 'dropbox' or 'app_folder' as configured for your app
ACCESS_TYPE = 'app_folder'

sess = session.DropboxSession(APP_KEY, APP_SECRET, ACCESS_TYPE)
request_token = sess.obtain_request_token()
url = sess.build_authorize_url(request_token)

# Make the user sign in and authorize this token
print "url:", url
print "Please visit this website and press the 'Allow' button, then hit 'Enter' here."
raw_input()

# This will fail if the user didn't visit the above URL
access_token = sess.obtain_access_token(request_token)

#Print the token for future reference
print access_token.key
print access_token.secret