Apart from one example in the docs, I can’t find any documentation on how exactly django chooses the name with which one can access the child object from the parent object. In their example, they do the following:
class Place(models.Model):
name = models.CharField(max_length=50)
address = models.CharField(max_length=80)
def __unicode__(self):
return u"%s the place" % self.name
class Restaurant(models.Model):
place = models.OneToOneField(Place, primary_key=True)
serves_hot_dogs = models.BooleanField()
serves_pizza = models.BooleanField()
def __unicode__(self):
return u"%s the restaurant" % self.place.name
# Create a couple of Places.
>>> p1 = Place(name='Demon Dogs', address='944 W. Fullerton')
>>> p1.save()
>>> p2 = Place(name='Ace Hardware', address='1013 N. Ashland')
>>> p2.save()
# Create a Restaurant. Pass the ID of the "parent" object as this object's ID.
>>> r = Restaurant(place=p1, serves_hot_dogs=True, serves_pizza=False)
>>> r.save()
# A Restaurant can access its place.
>>> r.place
<Place: Demon Dogs the place>
# A Place can access its restaurant, if available.
>>> p1.restaurant
So in their example, they simply call p1.restaurant without explicitly defining that name. Django assumes the name starts with lowercase. What happens if the object name has more than one word, like FancyRestaurant?
Side note: I’m trying to extend the User object in this way. Might that be the problem?
If you define a custom
related_namethen it will use that, otherwise it will lowercase the entire model name (in your example.fancyrestaurant). See the else block in django.db.models.related code:And here’s how the
OneToOneFieldcalls it:The opts.object_name (referenced in the django.db.models.related.get_accessor_name) defaults to
cls.__name__.As for
No it won’t, the
Usermodel is just a regular django model. Just watch out forrelated_namecollisions.