Django rest framework urls

Question

I'm structuring a Django API with rest framework, I read the docs and DRF only makes a crud (get, post, patch, delete) from a model. Now the deal is how I can make custom actions with DRF.

Example:

api/v1/model/custom_action

Code:

class DistrictViewSet(viewsets.ModelViewSet):
      queryset = District.objects.all()
      serializer_class = DistrictSerializer

      def custom_action(request, param):
          # do many actions and return as Json Object

urls.py

url(r'api/v1/', include(router.urls))

Where router

router.register(r'model',api.ModelViewSet)

I'm correct with this or I need to create another modelview, customize the code and add it to router list?

RHSmith159 · Accepted Answer · 2018-12-19 15:58:34Z

3

You can add custom actions as you have done but you may need the @action decorator to configure the url to apply to a single object, or many. @action(detail=True) adds pk to the url, as it applies to one object.

The url is generated from the action name, so for example

@action(detail=True)
def custom_action(self):
    pass

Would yield the url ^<app_name>/{pk}/custom_action/$

You may find this useful: https://www.django-rest-framework.org/api-guide/viewsets/#marking-extra-actions-for-routing

answered Dec 19, 2018 at 15:58

RHSmith159

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1 Comment

drkpkg Over a year ago

Thanks!! It worked. Reading the docs again to check if I miss some :D

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