Django model inheritance: Create a subclass using existing super class

Question

I'm using multi-table-inheritance, and want to know how to create an inherited type from an instance of the superclass.

Using the example given in the documentation:

class Place(models.Model): name = models.CharField(max_length=50) address = models.CharField(max_length=80) class Restaurant(Place): serves_hot_dogs = models.BooleanField() serves_pizza = models.BooleanField()

Now when you create a Restaurant, you automatically make a Place, which is fine, and the behaviour I expect and want.

But what if I make a Place, and later decide I want to convert to a specific type (like Restaurant). How do you create a Restaurant, using an existing Place?

possible duplicate of Django model inheritance: create sub-instance of existing instance (downcast)? — jpic
– jpic, Commented Mar 22, 2012 at 12:47
I saw that question and answer, but was confused because I thought that problem had to do with authentication (because a User is a bit more of a special object) — michel.iamit
– michel.iamit, Commented Mar 22, 2012 at 13:00
I upvoted your duplication remark, however... the question refering to is about Users, which might be confusing. I find this example of places and restaurants (stolen from the Django Documentations) more clear and resembling my problem.... but in base it's the same problem. — michel.iamit
– michel.iamit, Commented Mar 22, 2012 at 13:08

luckydonald · Accepted Answer · 2024-08-12 07:39:31Z

32

Multi-table inheritance is just OneToOneField relation between Place and Restaurant.

place = Place.objects.get(id=1) # Create a restaurant using existing Place restaurant = Resturant(place_ptr=place) # this might be needed to not delete the data: restaurant.__dict__.update(place.__dict__) restaurant.save()

edited Aug 12, 2024 at 7:39

luckydonald

7,0665 gold badges43 silver badges65 bronze badges

answered Mar 22, 2012 at 12:26

Mariusz Jamro

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3 Comments

jpic Over a year ago

This will cause restaurant.name to be emptied. Because restaurant.save() will update the Place with name='' (because restaurant = Resturant())... there must be a better way...

jpic Over a year ago

You should add restaurant.__dict__.update(place.__dict__), see Daniel Roseman's answer.

michel.iamit Over a year ago

Ok, the combination of the answer of secator and addition of jpic does the trick...

Davide Callegari · Accepted Answer · 2012-07-12 14:54:16Z

place = Place.objects.get(id=1) # Create a restaurant using existing Place place.__class__ = Restaurant place.save() restaurant = place

velis · Accepted Answer · 2021-10-12 09:06:09Z

While undocumented, this seems to do the trick:

restaurant(place_ptr=place).save_base(raw=True)

This solves the problem without using any hacks and is the shortest solution, also in terms of processing, using Django APIs.

While searching for this solution, I also found a slightly longer one, but using documented APIs. It is basically the same as Mariusz answer, also see this answer for more details:

from django.forms.models import model_to_dict restaurant(place_ptr=place, **model_to_dict(place)).save()

However, this second one is more risky due to limited field set returned by the model_to_dict (see again the answer explaining the differences among various methods presented). Naturally, it also generates more DB calls because it writes to both tables.

restaurant(place_ptr=place).save_base(raw=True) Worked like a charm, thank you!

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