我做了一个API,并且想要让swagger doc。我不为此开发任何序列化。Django Rest Swagger APIView
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')),
]
我想所有扬鞭对此我没有得到的参数。
i am not able to get parameters
我使用:
Django的休息,招摇== 2.1.1
djangorestframework == 3.5.3
所以你使用yaml文档,这个功能在django-swagger中不再被支持。我有同样的问题,如果你找到解决方案,你可以ping我吗? –
@OleksandrDashkov好心地通过答案,并问是否仍然不回答您的疑问 –
@RagulParani真的,我无法测试它,因为我已经切换到DRF的本机文档 –