Python Django - ForeignKey Class Choice for Model Field
<p>I want to make it so that i can choose a foreignkey for a field model/class when creating and instance of a model. So something like this:</p>

<pre><code>class Posts(models.Model):

    OwnerType = (
        (Accounts, 'User'),
        (Groups, 'Group'),
    )

    PostTypes = (
        ('Text', 'Text'),
        ('Photo', 'Photo'),
        ('Audio', 'Audio'),
        ('Video', 'Video'),
        ('Link', 'Link'),
    )

    owner = models.ForeignKey(OwnerType, default = 0, related_name = "post_owner")
    ownerid = models.IntegerField(blank = False, default = 0)
    owner_type = models.CharField(blank = False, default = '', max_length = 50)

    title = models.CharField(max_length = 500, default = '')
    contents = models.CharField(max_length = 500, default = '')
    attachment = models.CharField(max_length = 500, default = '')
    attachment_type = models.CharField(max_length = 500, default = '')
    link = models.CharField(max_length = 500, default = '')

    post_type = models.CharField(max_length = 20, choices = PostTypes, default = '')
    status = models.CharField(max_length = 20, default = 'public') # either public, private, or deleted

    date_created = models.DateField(auto_now_add=True)
    last_active = models.DateField(auto_now=True)

    @property
    def serialize(self):
        return {
            'p_id': self.id,

            'ownerid': self.ownerid,
            'owner': self.owner.serialize,
            'owner_type': self.owner_type,
            'title': self.title,
            'contents': self.contents,
            'attachment': self.attachment,
            'attachment_type': self.attachment_type,
            'link': self.link,
            'post_type': self.post_type,
            'status': self.status,
            'date_created': self.date_created,
            'last_active': self.last_active
        }

    class Meta:
        db_table = "posts"

# ---
</code></pre>

<p>but then i get this error:</p>

<pre><code>    Traceback (most recent call last):
  File "manage.py", line 9, in &lt;module&gt;
    execute_from_command_line(sys.argv)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 353, in execute_from_command_line
    utility.execute()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/core/management/__init__.py", line 327, in execute
    django.setup()
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/__init__.py", line 18, in setup
    apps.populate(settings.INSTALLED_APPS)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/apps/registry.py", line 108, in populate
    app_config.import_models(all_models)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/apps/config.py", line 202, in import_models
    self.models_module = import_module(models_module_name)
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/importlib/__init__.py", line 37, in import_module
    __import__(name)
  File "/Users/RyanWaite/Desktop/epsity/webapp/models.py", line 320, in &lt;module&gt;
    class Posts(models.Model):
  File "/Users/RyanWaite/Desktop/epsity/webapp/models.py", line 335, in Posts
    owner = models.ForeignKey(OwnerType, default = 0, related_name = "post_owner")
  File "/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/django/db/models/fields/related.py", line 754, in __init__
    RECURSIVE_RELATIONSHIP_CONSTANT,
AssertionError: ForeignKey(((&lt;class 'webapp.models.Accounts'&gt;, u'User'), (&lt;class 'webapp.models.Groups'&gt;, u'Group'))) is invalid. First parameter to ForeignKey must be either a model, a model name, or the string u'self'
</code></pre>

<p>Is that even possible? can i choose the model for a foreignkey when creating a new instance of a model? I have two type of models that can make posts: users and groups. a user can make a post or a group can make a post.</p>