50
votes

I want to edit a django-rest-framwork serializer object before it is saved. This is how I currently do it -

def upload(request):
    if request.method == 'POST':
        form = ImageForm(request.POST, request.FILES)
        if form.is_valid(): # All validation rules pass
             obj = form.save(commit=False)
             obj.user_id = 15
             obj.save()

How can I do it with a django-rest-framework serializer object?

@api_view(['POST','GET'])
def upload_serializers(request):
    if request.method == 'POST':
         serializer = FilesSerializer(data=request.DATA, files=request.FILES)
         if serializer.is_valid():
              serializer.save()
2
Your first paragraph seems incomplete. “This how I do it with” what?Donal Fellows

2 Answers

31
votes

You can edit the serializer's object before save the serializer:

if serializer.is_valid():
    serializer.object.user_id = 15 # <----- this line
    serializer.save()
70
votes

Now edited for REST framework 3

With REST framework 3 the pattern is now:

if serializer.is_valid():
    serializer.save(user_id=15)

Note that the serializers do not now ever expose an unsaved object instance as serializer.object, however you can inspect the raw validated data as serializer.validated_data.

If you're using the generic views and you want to modify the save behavior you can use the perform_create and/or perform_update hooks...

def perform_create(self, serializer):
    serializer.save(user_id=15)