Django

Question

所以我在下面有这段代码用于检查 AWS Cognito 令牌。我显然不想将这 6 行代码添加到每个端点。我也不知道这是否是验证我正在做的所有事情的正确方法是期望令牌的格式为“”，解析它并只解码 JWT 令牌部分。我如何验证每个请求附带的 AWS 放大令牌以确保用户正确登录。我想将此身份验证添加到 APIView 端点和 DRF api_view 装饰端点。

views.py

import django.db.utils
from rest_framework import authentication, permissions, status
from rest_framework.views import APIView
from .serializers import *
from .models import *
from rest_framework.response import Response
from django.http import JsonResponse
from drf_yasg.utils import swagger_auto_schema
from drf_yasg import openapi
from .core.api import jwt
from django.core.exceptions import ObjectDoesNotExist
class LoginView(APIView):
    def post(self, request):
        # 'Bearer z324weroko2iorjqoi=+3r3+3ij.2o2ij4='
        token = request.META['HTTP_AUTHORIZATION'].split(' ')[1]
        print(token)
    
        # TODO this should be separated out to a login module
        try:
            res = jwt.decode_cognito_jwt(token)
            return Response(status=status.Http_200_OK)
        except:
            return Response("Invalid JWT", status=status.HTTP_401_UNAUTHORIZED)

@api_view(['GET'])
@swagger_auto_schema(
    operation_description="Get Goals joined by User"
)
def get_goals_by_user(request, user_id):
    try:
        # Get goal ids of user with id
        goals_query = JoinGoal.objects.filter(
            joiner_id=user_id).values_list('goal_id', flat=True)
        goals_list = list(goals_query)
        # Get Goals using list of goals PK with descriptions and uuid
        data = list(Goal.objects.filter(
            pk__in=goals_list).values('description', 'uuid'))
        response_data = dict(goals=data)
        return JsonResponse(response_data, status=status.HTTP_200_OK)
    except JoinGoal.DoesNotExist:
        return Response(dict(error=does_not_exist_msg(JoinGoal.__name__, 'joiner_id', user_id)), status=status.HTTP_400_BAD_REQUEST)

Answer 1

因为您似乎正在使用 DRF，您可以创建自己的身份验证 class 并在那里应用 JWT 的处理：

from django.contrib.auth.models import AnonymousUser
from rest_framework.authentication import BaseAuthentication
from rest_framework import exceptions


class MyCustomJWTAuthentication(BaseAuthentication):
    def authenticate(self, request):
        token = request.META['HTTP_AUTHORIZATION'].split(' ')[1]
        try:
            jwt.decode_cognito_jwt(token)
        except Exception:
            raise exceptions.AuthenticationFailed('Invalid JWT')

        return AnonymousUser(), None


class MyCustomAPIView(APIView):
    authentication_classes = (MyCustomJWTAuthentication, )

或者如果你想将它应用到所有 APIViews:

REST_FRAMEWORK = {
    'DEFAULT_AUTHENTICATION_CLASSES': (
        'path.to.MyCustomJWTAuthentication',
    ),
}

请注意，一旦 JWT 解码失败，将不会检查其他身份验证 classes。如果您不想这样做，请将 except 子句的处理更改为不引发 AuthenticationFailed.

Answer 2

如果使用 djangorestframework，@bdbd 的回答将是您的最佳选择。否则，您可能需要探索以下选项：

实施您自己的装饰器来执行身份验证。这与 @login_required decorator or the @user_passes_test decorator. When writing such decorator for class-based views, you maybe interested with django.utils.decorators.method_decorator.

from functools import partial, wraps

from django.utils.decorators import method_decorator


def cognito_authenticator(view_func=None):
    if view_func is None:
        return partial(cognito_authenticator)

    @wraps(view_func)
    def wrapped_view(request, *args, **kwargs):
        # Check the cognito token from the request.
        token = request.META['HTTP_AUTHORIZATION'].split(' ')[1]

        try:
            res = jwt.decode_cognito_jwt(token)
            # Authenticate res if valid. Raise exception if not.
        except Exception:
            # Fail if invalid
            return HttpResponseForbidden("You are forbidden here!")
        else:
            # Proceed with the view if valid
            return view_func(request, *args, **kwargs)

    return wrapped_view


# We can decorate it here before the class definition but can also be done before the class method itself. See https://docs.djangoproject.com/en/3.2/topics/class-based-views/intro/#decorating-the-class
@method_decorator(
    name="post",
    decorator=[
        cognito_authenticator,
    ],
)
class SomeView(View):
    @method_decorator(cognito_authenticator)  # As explained above, this is another way of putting the decorator
    def get(self, request):
        return HttpResponse("Allowed entry!")

    def post(self, request):
        return HttpResponse("Allowed entry!")


# Or if using function-based views
@api_view(['POST'])
@cognito_authenticator
def some_view(request):
    return HttpResponse(f"Allowed entry!")

写一个custom middleware. Be aware that the order matters. Same idea as the default AuthenticationMiddleware which populates the request.user field. In your case, implement the __call__ method where you would check the Cognito tokens. Do not proceed to the view when the token is invalid by returning e.g. HttpResponseForbidden as in this reference.

class CognitoAuthenticatorMiddleware:
    def __init__(self, get_response):
        self.get_response = get_response

    def __call__(self, request):
        token = request.META['HTTP_AUTHORIZATION'].split(' ')[1]

        try:
            res = jwt.decode_cognito_jwt(token)
            # Authenticate res if valid. Raise exception if not.
        except Exception:
            # Fail if invalid
            return HttpResponseForbidden("You are forbidden here!")

        # Proceed if valid
        response = self.get_response(request)

        return response

MIDDLEWARE = [
    ...
    'path.to.CognitoAuthenticatorMiddleware',
    ...
]

更新

这是一个使用 Option-1 的示例运行。为简单起见，settings.py 只是默认设置。

views.py

from functools import partial, wraps

from django.http import HttpResponse, HttpResponseForbidden
from django.utils.decorators import method_decorator
from django.views import View  # If using django views
from rest_framework.views import APIView  # If using djangorestframework views


def cognito_authenticator(view_func=None):
    if view_func is None:
        return partial(cognito_authenticator)

    @wraps(view_func)
    def wrapped_view(request, *args, **kwargs):
        # To simplify the authentication, we would check if there is a query parameter "name=me". If none, it is forbidden.
        if request.GET.get('name') == "me":
            return view_func(request, *args, **kwargs)
        return HttpResponseForbidden("You are forbidden here!")

    return wrapped_view


@method_decorator(  # Try this style-1
    name="get",
    decorator=[
        cognito_authenticator,
    ],
)
class SomeView(View):  # If using djangorestframework view, this can also inherit from APIView or others e.g. class SomeView(APIView):
    @method_decorator(cognito_authenticator)  # Or try this style-2
    def get(self, request):
        return HttpResponse(f"Allowed entry!")

urls.py

from django.urls import path

from my_app import views

urlpatterns = [
    path("some-view/", views.SomeView.as_view()),
]

样本运行:

$ curl http://127.0.0.1:8000/my_app/some-view/?name=notme
You are forbidden here!
$ curl http://127.0.0.1:8000/my_app/some-view/?name=me
Allowed entry!

Django - 在处理端点请求之前验证 AWS Cognito 令牌是否有效

Django - Verify AWS Cognito token is valid before processing endpoint request

python

amazon-web-services

amazon-cognito

更新