How to use 'User' as foreign key in Django 1.5

Question

I have made a custom profile model which looks like this:

from django.db import models
from django.contrib.auth.models import User

class UserProfile(models.Model):
    user = models.ForeignKey('User', unique=True)
    name = models.CharField(max_length=30)
    occupation = models.CharField(max_length=50)
    city = models.CharField(max_length=30)
    province = models.CharField(max_length=50)
    sex = models.CharField(max_length=1)

But when I run manage.py syncdb, I get:

myapp.userprofile: 'user' has a relation with model User, which has either not been installed or is abstract.

I also tried:

from django.contrib.auth.models import BaseUserManager, AbstractUser

But it gives the same error. Where I'm wrong and how to fix this?

Is "django.contrib.auth" in `INSTALLED_APPS` in your settings.py ? — karthikr, Oct 17 '13 at 17:50
This one seem to work: 'user = models.OneToOneField(User, primary_key=True)'. However not sure what's difference. — supermario, Oct 17 '13 at 17:50
The difference is that for a OneToOneField, a user can have only one profile (which is probably what you want). — Mark, Apr 09 '14 at 16:44

score 64 · Answer 1 · edited Mar 01 '17 at 14:26

Exactly in Django 1.5 the AUTH_USER_MODEL setting was introduced, allowing using a custom user model with auth system.

If you're writing an app that's intended to work with projects on Django 1.5 through 1.10 and later, this is the proper way to reference user model (which can now be different from django.contrib.auth.models.User):

class UserProfile(models.Model):
    user = models.ForeignKey(settings.AUTH_USER_MODEL)

See docs for more details.

In case you're writing a reusable app supporting Django 1.4 as well, then you should probably determine what reference to use by checking Django version, perhaps like this:

import django
from django.conf import settings
from django.db import models


def get_user_model_fk_ref():
    if django.VERSION[:2] >= (1, 5):
        return settings.AUTH_USER_MODEL
    else:
        return 'auth.User'


class UserProfile(models.Model):
    user = models.ForeignKey(get_user_model_fk_ref())

I changed the code to note treat the version number as a decimal number. After all, the version 1.10 is considered higher than 1.5, which is not true if we treat them as decimal numbers. — Flimm, Mar 01 '17 at 14:27
@Flimm Looks good to me, didn’t like that dance with the decimals anyway. Tuple comparison beats it (maybe slightly less intuitive but it works!) — Anton Strogonoff, Mar 02 '17 at 16:57

score 57 · Accepted Answer · answered Oct 17 '13 at 17:48

57

Change this:

user = models.ForeignKey('User', unique=True)

to this:

user = models.ForeignKey(User, unique=True)

answered Oct 17 '13 at 17:48

Captain Skyhawk

3,499
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7

Wouldn't it be more clear if we used `OneToOneField` instead of `ForeignKey` with `unique` constraint. – vabhdman Oct 31 '14 at 14:56
5

beware: If you do this and later on try to extend your User model creating a custom user model, you will need to change it to settings.AUTH_USER_MODEL as Anton Strogonoff said in his - more correct IMO - answer – tato Jan 29 '16 at 12:32
2

Use `OneToOneField` if your relation is one to one, and add `primary_key=True` – erajuan Jan 29 '16 at 16:29
1

For newer versions of Django, ForeignKey requires you to specify `on_delete=models.CASCADE` or another [on_delete option](https://docs.djangoproject.com/en/2.0/ref/models/fields/#django.db.models.ForeignKey.on_delete). – krubo Dec 31 '18 at 01:40

How to use 'User' as foreign key in Django 1.5

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