我需要在 View 响应中传递自定义模型更新方法的返回值。
在我的序列化程序中，我想做的是:
class Serializer(ModelSerializer):
def update(self, instance, validated_data):
something_special = validated_data.pop('something_special')
important_info = model.update_something_special(something_special)
for attr, value in validated_data.items():
setattr(instance, attr, value)
instance.save()
return instance
现在在我看来，我想在响应中返回 important_info:
class View(ModelViewSet):
def update(self, request, *args, **kwargs):
partial = kwargs.pop('partial', False)
instance = self.get_object()
serializer = self.get_serializer(instance, data=request.data, partial=partial)
serializer.is_valid(raise_exception=True)
self.perform_update(serializer)
important_info = ???
return Response(serializer.data)
这在 Django REST 中可行还是死路一条？如果是这样，如何以不同的方式做到这一点？
请您参考如下方法:
class Serializer(ModelSerializer):
important_info = None
def update(self, instance, validated_data):
something_special = validated_data.pop('something_special')
self.important_info = model.update_something_special(something_special)
for attr, value in validated_data.items():
setattr(instance, attr, value)
instance.save()
return instance
class View(ModelViewSet):
def update(self, request, *args, **kwargs):
partial = kwargs.pop('partial', False)
instance = self.get_object()
serializer = self.get_serializer(instance, data=request.data, partial=partial)
serializer.is_valid(raise_exception=True)
self.perform_update(serializer)
important_info = serializer.important_info
return Response(serializer.data)
