Django Rest Swagger APIView

Question

我制作了一个 API 并想制作 swagger 文档。我不为此开发任何序列化。

Views.py

class DeliveryView(APIView):

    renderer_classes = (XMLRenderer,)

    def get_campaign_vast(self, request, *args):
        return response

    def get(self, request):
        return self.get_campaign_vast(request, data)

    def post(self, request):
        """ 
        This text is the description for this API
        ---
        param1 -- A first parameter
        param2 -- A second parameter
        """
        data = request.data
        return self.get_campaign_vast(request, data) 

urls.py

from django.conf.urls import url,include
from django.contrib import admin
from rest_framework_swagger.views import get_swagger_view

schema_view = get_swagger_view(title='Add Delivery')

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^$',schema_view),
    url(r'^', include('deliverymanagment.urls')),
]

我想在 Swagger 中获取我没有得到的所有参数。

i am not able to get parameters

我在用:

django-rest-swagger==2.1.1

djangorestframework==3.5.3

Answer 1

考虑使用 GenericAPIView相反，因为这将生成文档。当端点与模型无关时，使用它有点麻烦，但它确实有效。
例如，以下代码将创建一个仅接受发布请求的端点，并使用序列化程序在 swagger 中进行记录。

from rest_framework.generics import GenericAPIView

class SomeThing(GenericAPIView):
    serializer_class = MySerializer

    def post(self, request, *args, **kwargs):
        serializer = MySerializer(data=request.data)
        if serializer.is_valid() is False:
            return Response(serializer.errors, status=status.HTTP_400_BAD_REQUEST)

        res = do_magic(**serializer.data)
        return Response(res)