python - 我们如何定义接受参数的@list_route

Question

在我的应用程序中，我有这个 ModelViewSet 和一个 @list_route() 定义的函数来获取列表，但有不同的序列化程序。

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)

具有此 /api/animal/ 端点的默认 AnimalViewSet 会根据 AnimalSerializer 定义产生此序列化数据结果。

{
    "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
},

另一个是 @list_route() 定义的名为 listview 的函数可能有这个终点 /api/animal/listview/ 这会产生在 AnimalListingSerializer 结构中定义的结果。

{
    "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"
    }
}

现在我想做的是定义另一个 @list_route() 函数，它接受一个参数并使用 AnimalListingSerializer 来过滤query_set 模型对象的结果。解决我对像我们这样的初学者的帮助。

@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)

让我们假设 args1 = "Cow" 和 args2 = "Brahman"。我期待着这个结果。

{
    "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.

Answer 1

View 函数中的参数保留用于 URL 引用。即路线 animals/5 将被传递给以 pk 作为参数的 View 函数。

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

您可以通过查询参数将参数传递给您的 url

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

然后使用请求对象在您的 View 中访问它request.query_params['speceis_type'] 和 request.query_params['braham'] 或者您可以使用记录在案的 django rest 过滤器中间件 here