CODEWITHSUNDEEP
pythondjangodjango-models
Apreche
Apreche

Reputation: 32979

How to Model a Foreign Key in a Reusable Django App?

In my django site I have two apps, blog and links. blog has a model blogpost, and links has a model link. There should be a one to many relationship between these two things. There are many links per blogpost, but each link has one and only one blog post. The simple answer is to put a ForeignKey to blogpost in the link model.

That's all well and good, however there is a problem. I want to make the links app reusable. I don't want it to depend upon the blog app. I want to be able to use it again in other sites and perhaps associate links with other non-blogpost apps and models.

A generic foreign key seems like it might be the answer, but not really. I don't want links to be able to associate with any model in my site. Just the one that I explicitly specify. And I know from prior experience that there can be issues using generic foreign keys in terms of database usage because you can't do a select_related over a generic foreign key the way you can with a regular foreign key.

What is the "correct" way to model this relationship?

Upvotes: 17

Views: 3641

Answers (6)

Evstifeev Roman
Evstifeev Roman

Reputation: 634

Anoher way to solve this is how django-mptt does this: define only an abstract model in a reusable app(MPTTModel), and require to inherit it with defining some fields (parent=ForeignKey to self, or whatever your app usecase will require)

Upvotes: 1

vikingosegundo
vikingosegundo

Reputation: 52237

This question and Van Gale's answer lead me to the question, how it could be possible, to limit contenttypes for GFK without the need of defining it via Q objects in the model, so it could be completly reuseable

the solution is based on

  • django.db.models.get_model
  • and the eval built-in, that evaluates a Q-Object from settings.TAGGING_ALLOWED. This is necessary for usage in the admin-interface

My code is quite rough and not fully tested

settings.py

TAGGING_ALLOWED=('myapp.modela', 'myapp.modelb')

models.py:

from django.db import models
from django.db.models import Q
from django.contrib.contenttypes.models import ContentType
from django.contrib.contenttypes import generic
from django.db.models import get_model
from django.conf import settings as s
from django.db import IntegrityError

TAGABLE = [get_model(i.split('.')[0],i.split('.')[1]) 
        for i in s.TAGGING_ALLOWED if type(i) is type('')]
print TAGABLE

TAGABLE_Q = eval( '|'.join(
    ["Q(name='%s', app_label='%s')"%(
        i.split('.')[1],i.split('.')[0]) for i in s.TAGGING_ALLOWED
    ]
))

class TaggedItem(models.Model):
    content_type = models.ForeignKey(ContentType, 
                    limit_choices_to = TAGABLE_Q)                               
    object_id = models.PositiveIntegerField()
    content_object = generic.GenericForeignKey('content_type', 'object_id')

    def save(self, force_insert=False, force_update=False):
        if self.content_object and not type(
            self.content_object) in TAGABLE:
            raise IntegrityError(
               'ContentType %s not allowed'%(
                type(kwargs['instance'].content_object)))
        super(TaggedItem,self).save(force_insert, force_update)

from django.db.models.signals import post_init
def post_init_action(sender, **kwargs):
    if kwargs['instance'].content_object and not type(
        kwargs['instance'].content_object) in TAGABLE:
        raise IntegrityError(
           'ContentType %s not allowed'%(
            type(kwargs['instance'].content_object)))

post_init.connect(post_init_action, sender= TaggedItem)

Of course the limitations of the contenttype-framework affect this solution

# This will fail
>>> TaggedItem.objects.filter(content_object=a)
# This will also fail
>>> TaggedItem.objects.get(content_object=a)

Upvotes: 0

Mikhail Korobov
Mikhail Korobov

Reputation: 22248

I'd go with generic relations. You can do something like select_related, it just require some extra work. But I think it's worth it.

One possible solution for generic select_related-like functionality:

http://bitbucket.org/kmike/django-generic-images/src/tip/generic_utils/managers.py

(look at GenericInjector manager and it's inject_to method)

Upvotes: 0

Van Gale
Van Gale

Reputation: 43932

If you think the link app will always point to a single app then one approach would be to pass the name of the foreign model as a string containing the application label instead of a class reference (Django docs explanation).

In other words, instead of:

class Link(models.Model):
    blog_post = models.ForeignKey(BlogPost)

do:

from django.conf import setings
class Link(models.Model):
    link_model = models.ForeignKey(settings.LINK_MODEL)

and in your settings.py:

LINK_MODEL = 'someproject.somemodel'

Upvotes: 24

John Paulett
John Paulett

Reputation: 15824

I think TokenMacGuy is on the right track. I would look at how django-tagging handles a similar generic relationship using the content type, generic object_id, and generic.py. From models.py

class TaggedItem(models.Model):
    """
    Holds the relationship between a tag and the item being tagged.
    """
    tag          = models.ForeignKey(Tag, verbose_name=_('tag'), related_name='items')
    content_type = models.ForeignKey(ContentType, verbose_name=_('content type'))
    object_id    = models.PositiveIntegerField(_('object id'), db_index=True)
    object       = generic.GenericForeignKey('content_type', 'object_id')

    objects = TaggedItemManager()

    class Meta:
        # Enforce unique tag association per object
        unique_together = (('tag', 'content_type', 'object_id'),)
        verbose_name = _('tagged item')
        verbose_name_plural = _('tagged items')

Upvotes: 2

SingleNegationElimination
SingleNegationElimination

Reputation: 156258

Probably you need to use the content types app to link to a model. You might then arrange for your app to check the settings to do some additional checking to limit which content types it will accept or suggest.

Upvotes: 0

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