permission based on the action/url in Django rest framewrok-CodePudding

I have a Modelviewset class that has several functions inside it like list(), destroy, partial_update() and also other custom functions like def get_examples(self,request). The thing is different functions have different permissions based on the user types. I have a separate permission class inside a permission file. Now I need to access the action inside the has_permission function. I know it can be done inside the get_permission function inside modelviewset class. But how can we do it inside has_permission function??

My View:

class ExampleView(viewsets.ModelViewSet):
    queryset = Example.objects.all().order_by("-created_at")
    serializer_class = ExampleSerializer
    
    pagination_class = ExampleNumberPagination
    
    def list(self, request, *args, **kwargs):        

    def partial_update(self, request, *args, **kwargs):        

    def custom_fuctin_1(self,request,*args,**kwargs):
    def custom_fuctin_2(self,request,*args,**kwargs):   

    def destroy(self, request, *args, **kwargs):

Now permission file:

class ExampleClassPermission(BasePermission):

    def has_permission(self, request, view):

        user =request.user
        if request.method == 'GET' or request.method == 'POST':
            return True
        elif request.method == 'DELETE' and (user.role == 'manager' or user.role ==  'admin'):
            return True
        elif request.method == 'PUT':
            return True
        else:
            return False

Here in this permission class, I want to set permission for custom function_1 and 2 in the view class. I cant do self.action == ''? just like in the get_permission.Can I??

CodePudding user response：

You can access action by using view argument:

class ExampleClassPermission(BasePermission):

    def has_permission(self, request, view):
       if view.action == 'list':
           # your code here