Django model which must have a one to one relationship to one of four possible tables

Question

I'm working on a Django app. I have a table, and one of its columns can have one of four values, and depending on that value, it must have a one-to-one relationship with another table which has columns specific to that value.

Here is my models.py:

from django.db import models

class Item(models.Model):
    id = models.AutoField(primary_key=True)
    type_choices = (
     ('A', 'Type A'),
     ('B', 'Type B'),
     ('C', 'Type C'),
     ('D', 'Type D'),
    )
    type = models.CharField(max_length=1, choices=type_choices)


class TypeA(models.Model):
    item = models.OneToOneField(
        Item,
        primary_key=True,
    )
    # Type A specific fields


class TypeB(models.Model):
    item = models.OneToOneField(
        Item,
        primary_key=True,
    )
    # Type B specific fields


class TypeC(models.Model):
    item = models.OneToOneField(
        Item,
        primary_key=True,
    )
    # Type C specific fields


class TypeD(models.Model):
    item = models.OneToOneField(
        Item,
        primary_key=True,
    )
    # Type D specific fields

Since nothing is stopping me from creating, say, a row in TypeA and a row in TypeB, with links back to the same row in Item, (Which is not something I want) I am sure to screw something up at some point.

Is their anything I can do, in Django or something else, that will let me protect myself against this problem?

Or perhaps should my models be structured in a completely different way?

Cheers

maybe check this out: http://stackoverflow.com/q/929029/2112269 — Aquiles, Apr 05 '16 at 04:17
I'm not sure how this helps me. Are you suggesting that I have a ForeignKey field in my Item model? — 99lives, Apr 05 '16 at 04:27
This doesn't help me avoid creating multiple ForeignKeys to the one Item row, which I don't want. — 99lives, Apr 05 '16 at 04:51

score 1 · Answer 1 · answered Apr 05 '16 at 04:55

Disclaimer: this may not be a complete answer, but will provide some pointers to get there.

This might get more complicated than you think.

What you can do is

Create a base class - TypeBase for the types you have. Add foreign key of your item class
Create required sub-classes per type - class TypeA(TypeBase).
Use django's ContenType and add its foreingn key into the base. This is used to determine which type of instance that particular object is. ContentType creates a loose coupling between the classes.
Override save() method of TypeBase to set the type appropriately.
You may need to add few more methods like
- get_type - to get type from ContentType.
- get_type_object - to get object of particular type from the TypeBase object.

Django model which must have a one to one relationship to one of four possible tables

1 Answers1