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133 lines
4.1 KiB
Plaintext
133 lines
4.1 KiB
Plaintext
########################
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One-to-one relationships
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########################
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.. highlight:: pycon
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To define a one-to-one relationship, use :ref:`ref-onetoone`.
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In this example, a ``Place`` optionally can be a ``Restaurant``:
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.. code-block:: python
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from django.db import models
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class Place(models.Model):
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name = models.CharField(max_length=50)
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address = models.CharField(max_length=80)
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def __str__(self): # __unicode__ on Python 2
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return "%s the place" % self.name
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class Restaurant(models.Model):
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place = models.OneToOneField(Place, primary_key=True)
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serves_hot_dogs = models.BooleanField()
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serves_pizza = models.BooleanField()
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def __str__(self): # __unicode__ on Python 2
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return "%s the restaurant" % self.place.name
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class Waiter(models.Model):
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restaurant = models.ForeignKey(Restaurant)
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name = models.CharField(max_length=50)
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def __str__(self): # __unicode__ on Python 2
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return "%s the waiter at %s" % (self.name, self.restaurant)
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What follows are examples of operations that can be performed using the Python
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API facilities.
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Create a couple of Places::
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>>> p1 = Place(name='Demon Dogs', address='944 W. Fullerton')
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>>> p1.save()
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>>> p2 = Place(name='Ace Hardware', address='1013 N. Ashland')
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>>> p2.save()
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Create a Restaurant. Pass the ID of the "parent" object as this object's ID::
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>>> r = Restaurant(place=p1, serves_hot_dogs=True, serves_pizza=False)
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>>> r.save()
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A Restaurant can access its place::
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>>> r.place
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<Place: Demon Dogs the place>
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A Place can access its restaurant, if available::
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>>> p1.restaurant
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<Restaurant: Demon Dogs the restaurant>
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p2 doesn't have an associated restaurant::
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>>> p2.restaurant
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Traceback (most recent call last):
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...
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DoesNotExist: Restaurant matching query does not exist.
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Set the place using assignment notation. Because place is the primary key on
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Restaurant, the save will create a new restaurant::
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>>> r.place = p2
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>>> r.save()
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>>> p2.restaurant
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<Restaurant: Ace Hardware the restaurant>
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>>> r.place
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<Place: Ace Hardware the place>
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Set the place back again, using assignment in the reverse direction::
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>>> p1.restaurant = r
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>>> p1.restaurant
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<Restaurant: Demon Dogs the restaurant>
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Restaurant.objects.all() just returns the Restaurants, not the Places. Note
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that there are two restaurants - Ace Hardware the Restaurant was created in the
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call to r.place = p2::
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>>> Restaurant.objects.all()
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[<Restaurant: Demon Dogs the restaurant>, <Restaurant: Ace Hardware the restaurant>]
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Place.objects.all() returns all Places, regardless of whether they have
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Restaurants::
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>>> Place.objects.order_by('name')
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[<Place: Ace Hardware the place>, <Place: Demon Dogs the place>]
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You can query the models using :ref:`lookups across relationships <lookups-that-span-relationships>`::
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>>> Restaurant.objects.get(place=p1)
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<Restaurant: Demon Dogs the restaurant>
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>>> Restaurant.objects.get(place__pk=1)
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<Restaurant: Demon Dogs the restaurant>
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>>> Restaurant.objects.filter(place__name__startswith="Demon")
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[<Restaurant: Demon Dogs the restaurant>]
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>>> Restaurant.objects.exclude(place__address__contains="Ashland")
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[<Restaurant: Demon Dogs the restaurant>]
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This of course works in reverse::
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>>> Place.objects.get(pk=1)
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<Place: Demon Dogs the place>
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>>> Place.objects.get(restaurant__place=p1)
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<Place: Demon Dogs the place>
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>>> Place.objects.get(restaurant=r)
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<Place: Demon Dogs the place>
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>>> Place.objects.get(restaurant__place__name__startswith="Demon")
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<Place: Demon Dogs the place>
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Add a Waiter to the Restaurant::
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>>> w = r.waiter_set.create(name='Joe')
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>>> w.save()
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>>> w
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<Waiter: Joe the waiter at Demon Dogs the restaurant>
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Query the waiters::
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>>> Waiter.objects.filter(restaurant__place=p1)
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[<Waiter: Joe the waiter at Demon Dogs the restaurant>]
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>>> Waiter.objects.filter(restaurant__place__name__startswith="Demon")
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[<Waiter: Joe the waiter at Demon Dogs the restaurant>]
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