How do we define @list_route that accept arguments

问题:

In my application i have this ModelViewSet with one @list_route() defined function for getting list but with different serializer.

class AnimalViewSet(viewsets.ModelViewSet):     """     This viewset automatically provides `list`, `create`, `retrieve`,     `update` and `destroy` actions.     """     queryset = Animal.objects.all()     serializer_class = AnimalSerializer // Default modelviewset serializer      lookup_field = 'this_id'      @list_route()     def listview(self, request):         query_set = Animal.objects.all()         serializer = AnimalListingSerializer(query_set, many=True) // Serializer with different field included.          return Response(serializer.data) 

The Default AnimalViewSet with this /api/animal/ end point yield this serialized data result as based on AnimalSerializer definition.

{     "this_id": "1001",     "name": "Animal Testing 1",     "species_type": "Cow",     "breed": "Brahman",     ...     "herd": 1 }, {     "this_id": "1004",     "name": "Animal Testing 2",     "species_type": "Cow",     "breed": "Holstien",     ....     "herd": 1 }, {     "this_id": "1020",     "name": "Animal Testing 20",     "species_type": "Cow",     "breed": "Brahman",     ....     "herd": 4 }, 

And the other one which is a @list_route() defined function named listview may have this end point /api/animal/listview/ which yields this result as defined in AnimalListingSerializer structure.

{     "this_id": "1001",     "name": "Animal Testing 1",     "species_type": "Cow",     "breed": "Brahman",     ....     "herd": {         "id": 1,         "name": "High Production",         "description": null     } }, {     "this_id": "1004",     "name": "Animal Testing 2",     "species_type": "Cow",     "breed": "Holstien",     ....     "herd": {         "id": 1,         "name": "High Production",         "description": null     } }, {     "this_id": "1020",     "name": "Animal Testing 20",     "species_type": "Cow",     "breed": "Brahman",     ....     "herd": {         "id": 4,         "name": "Bad Production",         "description": "Bad Production"     } } 

Now what i am trying to do is i want to define another @list_route() function that takes an argument and uses AnimalListingSerializer in order to filter the query_set result of the model object. A work around my help for a beginner like us.

@list_route() def customList(self, request, args1, args2):         query_set = Animal.objects.filter(species_type=args1, breed=args2)         serializer = AnimalListingSerializer(query_set, many=True)          return Response(serializer.data) 

Let us assumed that args1 = "Cow" and args2 = "Brahman". And i am expecting this result.

{     "this_id": "1001",     "name": "Animal Testing 1",     "species_type": "Cow",     "breed": "Brahman",     ....     "herd": {         "id": 1,         "name": "High Production",         "description": null     } }, {     "this_id": "1020",     "name": "Animal Testing 20",     "species_type": "Cow",     "breed": "Brahman",     ....     "herd": {         "id": 4,         "name": "Bad Production",         "description": "Bad Production"     } }, 

But i know my syntax is wrong, but that is what i am talking about. Please help.

回答1:

parameters in the view function are reserved for URL references. ie the route animals/5 would be passed to a view function that has pk as an argument.

def get(self, request, pk):     # get animal with pk     return animal with pk 

You can pass parameters to your url via a query param ie

/animals/listview/?speceis_type=cow&breed=braham

then access it in your view using the request object request.query_params['speceis_type'] and request.query_params['braham'] or you can use the django rest filter middleware that is documented here

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