Re: Creating model (table) inheritance in Django 1.8 using PostgreSQL

You could dynamically create a ModelForm based on the model class:

from django.forms import ModelForm

MetaClass = type('Meta', (), {'model': ProductItemText})

MetaModelForm = type('ProductModelForm', (ModelForm, ), {'Meta': MetaClass})

form = MetaModelForm()

The only thing you would need is the model class itself ("ProductItemText" in this particular case),  I think you could even add it as a classmethod to the base Product class.

On Thu, Jul 2, 2015 at 9:34 AM, Some Developer <someukdeveloper@gmail.com> wrote:

I have a model which defines an item and many sub-models which inherit from it to define the type of the model. Something like this:

Product(models.Model):
        owner = models.ForeignKey(settings.AUTH_USER_MODEL)
        name = models.CharField(max_length=255)

ProductItemText(Product):
        type = models.TextField()

ProductItemBoolean(Product):
        type = models.BooleanField()

etc etc. There are a lot more different sub-models for different types.

I'm a bit confused about how to generate a form based on the type that the user wants to add to the database. Do you have to have a form each individual sub-type or is there a nice workaround for this specific problem?

Any help is appreciated.

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