Python 将基于类的视图转换为'；正常'；看法

我编写了以下django rest框架视图：

# urls.py
url(r'user/company', views.UserViewSet.as_view({"get": "companyInfo"}), name="company_info"),

# views.py
class UserViewSet(viewsets.ModelViewSet):
    serializer_class = UserSerializer
    queryset = User.objects.all()
    permission_classes = (permissions.IsAuthenticated,)

    @action(methods=["get"], detail=True)
    def companyInfo(self, request):
        user = request.user
        company = user.get_company()
        detail = {}
        detail['company'] = company.name
        detail['num_users'] = company.num_licenses if company else None        
        return Response(detail) 

如何使用django rest框架编写与“普通”django视图完全相同的视图。例如，类似这样的内容：

# urls.py
path('user/company/', views.company_info, name='company_info'),

# views.py
@require_GET
def company_info(request):
    user = request.user
    company = user.get_company()
    detail = {}
    detail['company'] = company.name
    detail['num_users'] = company.num_licenses if company else None        
    return Response(detail) 

---

我想以上只是一个开始，但我认为我仍然需要正确地进行身份验证（使用jwt），这是

UserViewSet

使用

permission\u classes=（permissions.IsAuthenticated，）

自动完成的。您可能希望添加两个decorator，使其成为一个经过身份验证的API视图，就像在类视图中一样：

# path('user/company/', views.company_info, name='company_info'),

@api_view(['GET'])
@permission_classes((permissions.IsAuthenticated,))
def company_info(request):
    user = request.user
    company = user.get_company()

    detail = {}
    detail['company'] = company.name
    detail['num_users'] = company.num_licenses if company else None
    detail['company_admin'] = company.admin_user.name if company else None

    return Response(detail)         

---

@VaibhavVishal是的，我知道我的函数是不完整的——它在这里没有身份验证，这就是为什么我要问——看看如何使它“正确”。身份验证的意思是你只想让那些已经登录的用户访问视图？@VaibhavVishal correct，已经直接登录到应用程序的人（在django中）或者通过api通过jwt