1

I am using Django Rest Framework to create my API. I am using @link to return information about a particular object.

class SomeClassView(viewsets.ReadOnlyModelViewSet):
   @link
   def info(self, request, pk=None):
       obj = models.SomeClass.objects.get(pk=pk)
       return Response({"info": object.info()})

 GET: /someclass/1/info

What I would like to do is extend the method so I can access it at the "class level" so that my api could accept a list of objects

class SomeClassView(viewsets.ReadOnlyModelViewSet):
   @link
   def info(self, request, pk=None):
       if isinstance(s, str):
           obj = models.SomeClass.objects.get(pk=pk)
           return Response({"info": obj.info()})
       else:
           objs = models.SomeClass.objects.filter(pk__in=pk).all()
           return Response({"infos": [obj.info() for obj in objs]})

GET: /someclass/1/info
GET: /someclass/info?pk=1&pk=2&pk=3

Is there a way I can add a class level method to my api? Or will I need to create a new class to handle the one api call?

PS: I don't mind if I need to have a separate method to make this work

4

1 回答 1

3

The dynamically generated routes using the @link or @action decorators are hard-coded to look like /someclass/<pk>/<methodname>/. You can expose a /someclass/info endpoint by adding a custom route, e.g.:

class MyRouter(DefaultRouter):
    routes = [
        Route(
            url=r'^{prefix}/((?P<pk>\d+)/)?info$',
            mapping={'get': 'info'},
            name='{basename}-info',
            initkwargs={}
        )
    ] + DefaultRouter.routes

Then your info method might look like this:

def info(self, request, pk=None):
    if pk:
        obj = SomeClass.objects.get(pk=pk)
        return Response({'info': obj.info()})
    else:
        objs = SomeClass.objects.filter(pk__in=request.GET.getlist('pk'))
        return Response({'infos': [obj.info() for obj in objs]})

(Note the absence of the @link decorator.)

于 2013-10-01T09:12:19.320 回答