python - How do we define @list_route that accept arguments -

in application have modelviewset 1 @list_route() defined function getting list different serializer.

class animalviewset(viewsets.modelviewset):     """     viewset automatically provides `list`, `create`, `retrieve`,     `update` , `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 different field included.          return response(serializer.data) 

the default animalviewset /api/animal/ end point yield serialized data result 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 other 1 @list_route() defined function named listview may have end point /api/animal/listview/ yields result 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 trying want define @list_route() function takes argument , uses animallistingserializer in order filter query_set result of model object. work around beginner 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 assumed args1 = "cow" , args2 = "brahman". , expecting 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 know syntax wrong, talking about. please help.

parameters in view function reserved url references. ie route animals/5 passed view function has pk argument.

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

by decorating @list_route on ride option provide additional params via url. in case resource can request via url listview; /animals/listview. have other options though if want continue down path. can pass parameters url via url, query param /animals/listview/?speceis_type=cow&breed=braham access in view via request.query_params['speceis_type'] , request.query_params['braham'] or can use django rest filter middleware documented here