Hello!
I have two models related one-to many:
class Person(models.Model):
name = models.CharField(max_length=255);
surname = models.CharField(max_length=255);
age = models.IntegerField();
class Dog(models.Model):
name = models.CharField(max_length=255);
owner = models.ForeignKey('Person');
I want to output a list of persons and, below each person a list of dogs he has. Here's how I can do it:
in view:
persons = Person.objects.all()[0:100];
in template:
{% for p in persons %}
{{ p.name }} has dogs:<br />
{% for d in persons.dog_set.all %}
- {{ d.name }}<br />
{% endfor %}
{% endfor %}
But if I do it like that, Django will execute 101 SQL queries which is very inefficient.
I tried to make a custom manager, which will get all the persons, then all the dogs and links them in python, but then I can't use paginator (my another question: http://stackoverflow.com/questions/2532475/django-paginator-raw-sql-query ) and it looks quite ugly.
Is there a more graceful way doing this?
UPDATE
Solution based on blog entry by @Daniel Roseman
all_objects = Person.objects.select_related().all();
paginator = Paginator(all_objects, 100);
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
try:
current_page = paginator.page(page)
except (EmptyPage, InvalidPage):
current_page = paginator.page(paginator.num_pages)
person_list = dict([(obj.id, obj) for obj in current_page.object_list])
id_list = [obj.id for obj in current_page.object_list]
objects = Dog.objects.filter(owner__id__in=id_list)
relation_dict = {}
for obj in objects:
relation_dict.setdefault(obj.owner_id, []).append(obj)
for id, related_items in relation_dict.items():
person_list[id].dogs = related_items