如何使用Python库获取Facebook访问令牌？ [英] How to get Facebook access token using Python library?

问题描述

我发现这个图书馆似乎是官方的，那么<一个href =https://stackoverflow.com/questions/10488913/how-to-obtain-a-user-access-token-in-python>发现这个，但每次我找到一个答案它是指向 Facebook API文档的链接，该文档涉及到Javascript或PHP和如何从链接中提取！

I've found this Library it seems it is the official one, then found this, but everytime i find an answer the half of it are links to Facebook API Documentation which talks about Javascript or PHP and how to extract it from links!

我如何使用简单的python脚本？

How do i make it on simple python script?

NB ：我真的不明白，为什么使用一个库并且不能提取令牌如果我们可以使用 urllib 和 regex 提取信息？

NB: what i realy dont understand, why using a library and cant extract token if we can use urllib and regex to extract informations?

推荐答案

Javascript和PHP可以用作Web开发语言。您需要一个Web前端才能授予用户权限，以便您可以获取访问令牌。

Javascript and PHP can be used as web development languages. You need a web front end for the user to grant permission so that you can obtain the access token.

重新启动：您无法以编程方式获取访问令牌，那里必须是手动用户交互

在Python中，它将涉及设置一个Web服务器，例如使用facepy更新feed的脚本

In Python it will involve setting up a web server, for example a script to update feed using facepy

import web
from facepy import GraphAPI
from urlparse import parse_qs

url = ('/', 'index')

app_id = "YOUR_APP_ID"
app_secret = "APP_SECRET"
post_login_url = "http://0.0.0.0:8080/"

user_data = web.input(code=None)

if not user_data.code:
    dialog_url = ( "http://www.facebook.com/dialog/oauth?" +
                               "client_id=" + app_id +
                               "&redirect_uri=" + post_login_url +
                               "&scope=publish_stream" )

    return "<script>top.location.href='" + dialog_url + "'</script>"
else:
    graph = GraphAPI()
    response = graph.get(
        path='oauth/access_token',
        client_id=app_id,
        client_secret=app_secret,
        redirect_uri=post_login_url,
        code=code
    )
    data = parse_qs(response)
    graph = GraphAPI(data['access_token'][0])
    graph.post(path = 'me/feed', message = 'Your message here')

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