DRF Post to ViewSet without writing into Model

Question

I've always written data into database when posting via Django Rest Framework endpoints. This time I would like to process received data and send it somewhere else without writing into DB. I switched from ModelViewSet to ViewSet, I can issue GET request OK but receiving Bad Request 400 when I curl or POST via DRF URL. Here's a working minimal code (removed need for authentication etc):

urls.py

from django.urls import path, include
from .views import ContactView
from rest_framework import routers

router = routers.DefaultRouter()
router.register('message', ContactView, basename='message')

urlpatterns = [
    path('', include(router.urls)),
]

serializers.py

from rest_framework import serializers

class ContactSerializer(serializers.Serializer):
    text = serializers.CharField(max_length=250)

views.py

from rest_framework.response import Response
from .serializers import ContactSerializer
from rest_framework import viewsets

class ContactView(viewsets.ViewSet):
    def list(self, request):
        return Response('Got it')

    def create(self, request):
        serializer = ContactSerializer(data=request.data)
        if serializer.is_valid():
            return Response(serializer.data)
        else:
            return Response('Invalid')

Would greatly appreciate your suggestions.

Mykyta · Accepted Answer · 2020-07-20T20:22:15.277

You can use GenericAPIView for get or post request and do some logic in validate method, for example do something with signals or edit something. Also u can use @detailt_route or @list_route for any ModelViewSet for write special url for instance, example for edit extra data.

how i did rarely:

in urls.py

urlpatterns = [
    url('v1/message', ContactAPIView.as_view(), name='message'),
]

in view.py

class ContactAPIView(GenericAPIView):
    serializer_class = ContactSerializer
    permission_classes = ()

    def post(self, request, *args, **kwargs):
        serializer_class = self.get_serializer_class()
        serializer = serializer_class(data=request.data, context={'request': request})
        serializer.is_valid(raise_exception=True)
        data = {"status": True}

        return Response(data)

in serializers.py

class ContactSerializer(serializers.Serializer):
    text = serializers.TextField()

    def validate(self, attrs):
        write some logic

This solves it, thanks a ton, applied minor amendments following console output. The use case is simple so GenericAPIView is sufficient. — nekton, Jul 21 '20 at 11:35
GenericAPIView has queryset, queryset lookup fields, pagination. It is specifically for use with Django models. — Chris Sattinger, Jul 01 '23 at 06:24

score 1 · Answer 2 · answered Jul 20 '20 at 21:10

1

you are getting this error because you are using Viewsets which uses DefaultRouter to register routers for you. What it does is that it creates 2 urls for your viewset

message
message/id

so in your current case; i.e. viewset you need to send some dummy number in your url to access this post function (which is not a good approach).

So, you should use any class which parent doesn't include ViewSetMixin (which gives functionality of Router Registration) like in your case inherit your view from these classes

ListModelMixin
CreateModelMixin
GenericAPIView

answered Jul 20 '20 at 21:10

K14

181
7

Thank you, how does it affect the set up of `router` in __urls.py__? – nekton Jul 21 '20 at 11:29
you need to register routers manually like ```urlpatterns = [path('message', APIView.as_view(), name='message'), ]``` – K14 Jul 21 '20 at 19:38

DRF Post to ViewSet without writing into Model

2 Answers2