Re: Getting "most"

On May 27, 2011, at 9:58 AM, baxter@gretschpages.com wrote:

> I have a "watch" model that lets users keep an eye on various things
> through a generic relation:
>
> class Watch(models.Model):
> subscriber = models.ForeignKey(User, verbose_name="Subscriber")
> content_type = models.ForeignKey(ContentType)
> content_object = generic.GenericForeignKey()
> object_id = models.IntegerField('object ID')
> created = models.DateTimeField(auto_now_add=True)
>
>
> What I'm trying to do is get the most-watched objects.
> I saw James Bennett's snippet from 2007 (http://djangosnippets.org/
> snippets/108/) which looks like it would work (subbing my Watch model
> for comments), but I'm wondering if there's a better way to do it with
> newer versions of django, possibly through annotate or aggregate?
>

If you were to do something like this:

from django.models import count
most = Watch.objects.values('content_type', 'object_id').annotate(Count('object_id')).order_by('-object_id__count')

[
{'object_id__count': 15, 'object_id': 1, 'content_type': 10},
{'object_id__count': 2, 'object_id': 1, 'content_type': 5},
...
]

# this isn't efficient, but you get the idea
# you need to resolve the content types and object id's into models
from django.contrib.contenttypes.models import ContentType
for object in most:
foo = ContentType.objects.get(id= object['content_type']).get_object_for_this_type(object['object_id'])

You can also see
https://github.com/coleifer/django-generic-aggregation

This will get you close but I think it only if you want to annotate the count on a single content type at once.

Jason Culverhouse
http://www.mischievous.org

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