Django models.py Circular Foreign Key

Question

I have a django app which basically is just a photo album. Right now I have two models: Image and Album. Among other things, each Album has a foreign key to an Image to be its thumbnail and each Image has a foreign key to the Album it belongs in. However, when I try to use manage.py syncdb or manage.py sqlall I get errors saying the class not defined first in models.py isn't defined when it is used in the first class defined.

models.py (abridged):

from django.db import models
import os

class Album(models.Model):
    thumb = models.ForeignKey(Image, null=True, blank=True)

class Image(models.Model):
    image = models.ImageField(upload_to='t_pics/images')
    thumb = models.ImageField(upload_to='t_pics/images/thumbs')
    album = models.ForeignKey(Album)

Error I get when I do manage.py sqlall appname:

[...]
 File "/path/to/file/appname/models.py", line 4, in ?
    class Album(models.Model):
  File "/path/to/file/appname/models.py", line 5, in Album
    thumb = models.ForeignKey(Image, null=True, blank=True)
NameError: name 'Image' is not defined

I get the same error when I switch the order of the classes in models.py except it says 'Album' undefined instead of 'Image' undefined I also tried commenting the dependancy in the first class then uncommenting after everything else was successfully imported but that didn't help. How should I go about making this work? I'm reluctant to make an entire third class Thumb because it will have a lot of the same code as Image I'm also pretty sure I could manually add the foreign key to the database but I want this to be clean and not hackish.

Answer 1

You don't actually have a circular reference; the issue is that, at the time you define Album, you haven't defined Image yet. You can fix that by using a string instead:

class Album(models.model):
  thumb = models.ForeignKey('Image', null=True, blank=True)

However, in this case, you might want to use a OneToOneField instead of a foreign key. (Note that you'll still have to use the trick with the string, though).

Answer 2

Use '' to force a lazy reference:

models.ForeignKey('Image', null=True, blank=True)

Also, ForeignKey.related_name is your friend (avoids back-reference name clashes).