我们如何定义接受参数的@list_route [英] How do we define @list_route that accept arguments

问题描述

在我的应用程序中，我有一个 ModelViewSet 与一个 @list_route（）定义的函数来获取列表，但是不同序列化器。

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)

默认 AnimalViewSet 与此 / api / animal / 终点产生此序列化数据结果基于 AnimalSerializer 定义。

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
},

另一个是 @list_route ）定义的函数名为 listview 可能有这个终点 / api / animal / listview / 哪个啊符合 AnimalListingSerializer 结构中定义的结果。

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"
    }
}

现在我想要做的是我想定义另一个 @list_route（）函数，它接受一个参数并使用 AnimalListingSerializer 以过滤模型对象的 query_set 结果。对于像我们这样的初学者，我的帮助工作。

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)

让我们假设 args1 =Cow code>和 args2 =婆罗门。我希望这个结果。

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"
    }
},

但是我知道我的语法错误，但这正是我在说的。
请帮助。

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

推荐答案

视图功能中的参数保留用于URL引用。即路线动物/ 5将传递给具有pk作为参数的视图函数。

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

通过使用@list_route装饰，您可以选择通过您的URL提供任何其他参数。在您的情况下，您可以通过URL请求的唯一资源是listview;所以 / animals / listview 。你有其他的选择，但如果你想继续下去这个路径。您可以通过url将参数传递给您的url，查询参数如 / animals / listview /？speceis_type = cow& breed = braham ，然后通过 request.query_params ['speceis_type'] 和 request.query_params ['braham'] 或者您可以使用django rest过滤器中间件记录在案 here

by decorating with an @list_route you over ride the option to provide any additional params via your URL. In your case the only resource you can request via the url is listview; so /animals/listview. You have other options though if you want to continue down this path. You can pass parameters to your url via a url, query param like /animals/listview/?speceis_type=cow&breed=braham then access it in your view via 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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