将django-allauth作为端点插入django-rest-framework [英] Plug in django-allauth as endpoint in django-rest-framework
问题描述
澄清:Facebook登录部分由本机SDK处理。我需要一个端点,像 POST / user
(即创建一个新用户),但是将Facebook oauth令牌作为输入而不是电子邮件/密码等。
您可以使用此库来进行社交认证 django-rest-framework-social-oauth2 。尝试这个django-allauth相关代码
urls.py
urlpatterns = [
url(
r'^ rest / facebook-login / $',
csrf_exempt(RestFacebookLogin.as_view()),
name ='rest-facebook-login'
),
]
serializers.py
class EverybodyCanAuthentication(SessionAuthentication):
def authenticate(self,request):
return None
views.py
class RestFacebookLogin APIView)
根据来自Facebook的身份验证令牌
登录或注册用户
返回包含会话ID的用户数据
#这是一个公开的api!
permission_classes =(AllowAny,)
authentication_classes =(EverybodyCanAuthentication,)
def dispatch(self,* args,** kwargs):
return super(RestFacebookLogin,自我).dispatch(* args,** kwargs)
def get(self,request,* args,** kwargs):
try:
original_request = request._request
auth_token = request.GET.get('auth_token','')
#查找与传递的Auth令牌相匹配的令牌
app = SocialApp.objects.get(provider = 'facebook')
fb_auth_token = SocialToken(app = app,token = auth_token)
#检查令牌对于Facebook
login = fb_complete_login(original_request,app,fb_auth_token)
login.token = fb_auth_token
login.state = SocialLogin.state_from_request(original_request)
#将用户添加或更新到用户表
complete_so cial_login(original_request,login)
#创建或获取此用户的会话ID
令牌,_ = Token.objects.get_or_create(user = original_request.user)
#如果我们到这里'成功
data = {
'username':original_request.user.username,
'objectId':original_request.user.pk,
'firstName':original_request.user。 first_name,
'lastName':original_request.user.last_name,
'sessionToken':token.key,
'email':original_request.user.email,
}
返回响应(
status = 200,
data = data
)
除了:
返回响应(status = 401,data = {
'detail':'Bad Access Token',
})
I'm using django-allauth on my website for social logins. I also have a REST API powered by django-rest-framework that serves as the backend of a mobile app. Is there a way I can directly plug in allauth's authentication backend to the REST api so that I can validate (and register) users who use Facebook login in the mobile app?
To clarify: The Facebook login part is handled by native SDKs. I need an endpoint that works like POST /user
(that is, creates a new user), but takes Facebook oauth token as input instead of email/password etc.
You can use this libray for social authentication django-rest-framework-social-oauth2. Try this django-allauth related code
urls.py
urlpatterns = [
url(
r'^rest/facebook-login/$',
csrf_exempt(RestFacebookLogin.as_view()),
name='rest-facebook-login'
),
]
serializers.py
class EverybodyCanAuthentication(SessionAuthentication):
def authenticate(self, request):
return None
views.py
class RestFacebookLogin(APIView):
"""
Login or register a user based on an authentication token coming
from Facebook.
Returns user data including session id.
"""
# this is a public api!!!
permission_classes = (AllowAny,)
authentication_classes = (EverybodyCanAuthentication,)
def dispatch(self, *args, **kwargs):
return super(RestFacebookLogin, self).dispatch(*args, **kwargs)
def get(self, request, *args, **kwargs):
try:
original_request = request._request
auth_token = request.GET.get('auth_token', '')
# Find the token matching the passed Auth token
app = SocialApp.objects.get(provider='facebook')
fb_auth_token = SocialToken(app=app, token=auth_token)
# check token against facebook
login = fb_complete_login(original_request, app, fb_auth_token)
login.token = fb_auth_token
login.state = SocialLogin.state_from_request(original_request)
# add or update the user into users table
complete_social_login(original_request, login)
# Create or fetch the session id for this user
token, _ = Token.objects.get_or_create(user=original_request.user)
# if we get here we've succeeded
data = {
'username': original_request.user.username,
'objectId': original_request.user.pk,
'firstName': original_request.user.first_name,
'lastName': original_request.user.last_name,
'sessionToken': token.key,
'email': original_request.user.email,
}
return Response(
status=200,
data=data
)
except:
return Response(status=401, data={
'detail': 'Bad Access Token',
})
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