Foreign keys in django admin list display

I was looking for a solution to the same problem and ran across this question... ended up solving it myself. The OP might not be interested anymore but this could still be useful to someone.

from functools import partialfrom django.forms import MediaDefiningClassclass ModelAdminWithForeignKeyLinksMetaclass(MediaDefiningClass):    def __getattr__(cls, name):        def foreign_key_link(instance, field):            target = getattr(instance, field)            return u'<a href="../../%s/%s/%d">%s</a>' % (                target._meta.app_label, target._meta.module_name, target.id, unicode(target))        if name[:8] == 'link_to_':            method = partial(foreign_key_link, field=name[8:])            method.__name__ = name[8:]            method.allow_tags = True            setattr(cls, name, method)            return getattr(cls, name)        raise AttributeErrorclass Book(models.Model):    title = models.CharField()    author = models.ForeignKey(Author)class BookAdmin(admin.ModelAdmin):    __metaclass__ = ModelAdminWithForeignKeyLinksMetaclass    list_display = ('title', 'link_to_author')

Replace 'partial' with Django's 'curry' if not using python >= 2.5.

I don't think there is a mechanism to do what you want automatically out of the box.

But as far as determining the path to an admin edit page based on the id of an object, all you need are two pieces of information:

a) self.model._meta.app_label

b) self.model._meta.module_name

Then, for instance, to go to the edit page for that model you would do:

'../%s_%s_change/%d' % (self.model._meta.app_label, self.model._meta.module_name, item.id)

Take a look at django.contrib.admin.options.ModelAdmin.get_urls to see how they do it.

I suppose you could have a callable that takes a model name and an id, creates a model of the specified type just to get the label and name (no need to hit the database) and generates the URL a above.

But are you sure you can't get by using inlines? It would make for a better user interface to have all the related components in one page...

Edit:

Inlines (linked to docs) allow an admin interface to display a parent-child relationship in one page instead of breaking it into two.

In the Post/Author example you provided, using inlines would mean that the page for editing Posts would also display an inline form for adding/editing/removing Authors. Much more natural to the end user.

What you can do in your admin list view is create a callable in the Post model that will render a comma separated list of Authors. So you will have your Post list view showing the proper Authors, and you edit the Authors associated to a Post directly in the Post admin interface.

See https://docs.djangoproject.com/en/stable/ref/contrib/admin/#admin-reverse-urls

Example:

from django.utils.html import format_htmldef get_admin_change_link(app_label, model_name, obj_id, name):    url = reverse('admin:%s_%s_change' % (app_label, model_name),              args=(obj_id,))    return format_html('<a href="%s">%s</a>' % (        url, unicode(name)    ))