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python - Google 混合(openid + oauth)协议(protocol)不适用于 127.0.0.1

转载 作者:太空宇宙 更新时间:2023-11-04 06:36:27 24 4
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我正在尝试实现谷歌混合协议(protocol)(oauth over openid)。谷歌不要求 oauth 权限的问题(尝试使用 gmail),只有 openid。我在 google api 控制台注册:

Client ID for web applications
Client ID: 248141267047.apps.googleusercontent.com
Email address:248141267047@developer.gserviceaccount.com
Client secret:
Redirect URIs: http://127.0.0.1:8000/oauth2callback
JavaScript origins: http://127.0.0.1:8000

这是我生成 openid url 的 python 代码:

class OpenIDOAuthRequest(Extension):

ns_alias = 'oauth'

def __init__(self, consumer, scope, ns_uri=None):
Extension.__init__(self)
self.consumer = consumer
self.scope = scope
self.ns_uri = ns_uri or oauth_ns_uri

def getExtensionArgs(self):
return {
'consumer': self.consumer,
'scope': ' '.join(self.scope),
}


def google():
#define google openid url
openid_session = {}
openid_store = filestore.FileOpenIDStore('.')
consumer = Consumer(openid_session, openid_store)
openid = u"https://www.google.com/accounts/o8/id"
URLS = {
'ax_last': "http://axschema.org/namePerson/last",
'ax_first': "http://axschema.org/namePerson/first",
'ax_email': "http://axschema.org/contact/email",
"country":"http://axschema.org/contact/country/home",
"timezone":"http://axschema.org/pref/timezone",
"language":"http://axschema.org/pref/language",
"person":"http://axschema.org/namePerson",
}
#defining what fields we're going to get
ax_request = ax.FetchRequest()
for k,v in URLS.iteritems():
ax_request.add(ax.AttrInfo(v, required = True))
oa = OpenIDOAuthRequest("248141267047.apps.googleusercontent.com",["https://mail.google.com/",])
try:
authrequest = consumer.begin(openid)
except DiscoveryFailure, e:
print e
print "some errror happened"
else:
authrequest.addExtension(ax_request)
authrequest.addExtension(oa)



redirecturl = authrequest.redirectURL("http://127.0.0.1:8000",
return_to = "http://127.0.0.1:8000/oauth2callback",
immediate=False)
print redirecturl

它生成以下 url:

https://accounts.google.com/o/openid2/auth?openid.assoc_handle=AMlYA9Vr6Biwp-rCAr4TLbf8CtItR-zr3bs0LT7oYQ3Pakg93ivCS_6C&openid.ax.mode=fetch_request&openid.ax.required=ext0,ext1,ext2,ext3,ext4,ext5,ext6&openid.ax.type.ext0=http://axschema.org/contact/email&openid.ax.type.ext1=http://axschema.org/namePerson&openid.ax.type.ext2=http://axschema.org/namePerson/first&openid.ax.type.ext3=http://axschema.org/pref/timezone&openid.ax.type.ext4=http://axschema.org/pref/language&openid.ax.type.ext5=http://axschema.org/contact/country/home&openid.ax.type.ext6=http://axschema.org/namePerson/last&openid.claimed_id=http://specs.openid.net/auth/2.0/identifier_select&openid.identity=http://specs.openid.net/auth/2.0/identifier_select&openid.mode=checkid_setup&openid.ns=http://specs.openid.net/auth/2.0&openid.ns.ax=http://openid.net/srv/ax/1.0&openid.ns.oauth=http://specs.openid.net/extensions/oauth/1.0&openid.oauth.consumer=248141267047.apps.googleusercontent.com&openid.oauth.scope=https://mail.google.com/&openid.realm=http://127.0.0.1:8000&openid.return_to=http://127.0.0.1:8000/oauth2callback?janrain_nonce%3D2012-05-01T22%253A50%253A33ZUW7vcj

并且它具有所有必要的扩展。但是,如果我转到此网址,它不会询问我对 Gmail 的许可。我还比较了来自 sanebox.com 的类似网址。它按预期工作,请求 gmail 的许可。但是我没有看到为什么他们的网址有效而我的网址无效的任何本质区别。此外,我将我的网址中的 127.0.0.1 替换为 sanebox 网址,并将其他部分保持不变。而且...现在它正在请求 gmail 的许可。切换回 127.0.0.1 - 停止询问。这是 sanebox 网址:

https://accounts.google.com/o/openid2/auth?openid.assoc_handle=AMlYA9UV4Ud714HHaFJ0fpItabA8v-zw0QuReEPcn61ilJzyFrFia5PO&openid.ax.mode=fetch_request&openid.ax.required=ext0,ext1,ext2,ext3,ext4,ext5,ext6&openid.ax.type.ext0=http://axschema.org/pref/timezone&openid.ax.type.ext1=http://axschema.org/contact/country/home&openid.ax.type.ext2=http://axschema.org/pref/language&openid.ax.type.ext3=http://axschema.org/namePerson/last&openid.ax.type.ext4=http://axschema.org/namePerson/first&openid.ax.type.ext5=http://axschema.org/namePerson&openid.ax.type.ext6=http://axschema.org/contact/email&openid.claimed_id=http://specs.openid.net/auth/2.0/identifier_select&openid.identity=http://specs.openid.net/auth/2.0/identifier_select&openid.mode=checkid_setup&openid.ns=http://specs.openid.net/auth/2.0&openid.ns.ax=http://openid.net/srv/ax/1.0&openid.ns.oauth=http://specs.openid.net/extensions/oauth/1.0&openid.ns.sreg=http://openid.net/extensions/sreg/1.1&openid.oauth.consumer=www.sanebox.com&openid.oauth.scope=https://mail.google.com/+http://www.google.com/m8/feeds&openid.realm=https://www.sanebox.com/&openid.return_to=https://www.sanebox.com/users?_method%3Dpost%26open_id_complete%3D1

所以我缺少什么?如果我在 api 控制台中注册了这个 url,为什么它对 127.0.0.1 不起作用。它与 openid 一起工作正常。并且在没有 openid 的情况下与 oauth 本身一起工作正常。但是现在,当我尝试在 openid 上使用 oauth 时,它不会询问我对 gmail 的许可。

最佳答案

我已经运行了 PHP、ASP、DOT NET 代码,它们运行得非常好,因此可以确保 python 应用程序也能正确运行。您必须更改默认应用程序。首先运行他们提供的默认应用程序并尝试更改。如果默认应用程序无法运行,那么也可以向谷歌打开错误报告。

关于python - Google 混合(openid + oauth)协议(protocol)不适用于 127.0.0.1,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10413367/

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