Question

我正在尝试实施谷歌混合协议（oauth over openid）。谷歌没有要求oauth许可（尝试使用gmail）的问题，只有openid。我在google api console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

它会生成以下网址：

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

它有所有必要的扩展。但是，如果我去这个网址，它不会让我获得gmail的许可。此外，我正在与sanebox.com的类似网址进行比较。它按预期工作，要求获得gmail的许可。但我没有看到为什么他们的网址正在工作而我的网站没有的任何本质区别。此外，我在我的网址中将127.0.0.1替换为sanebox网址，并将其他部分保留为相同。并且...现在它要求获得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

那么我错过了什么？如果我在api控制台中重新设置此URL，为什么它不适用于127.0.0.1。它在openid下运行良好。并且在没有openid的情况下与oauth本身一起工作正常。但是现在当我尝试使用oauth而不是openid时，它并没有让我获得gmail的许可。

Answer 1

我运行的PHP，ASP，DOT NET代码运行得非常好，因此确保python应用程序也能正常运行。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。如果默认应用程序不起作用，那么也可以打开谷歌的错误报告。

Google混合（openid + oauth）协议不适用于127.0.0.1

1 个答案: