By output_field Django means to provide field type for the result of the Sum. from django.db.models import FloatField, F total_group=Sum(F(‘total_sold’)*F(‘final_price’), output_field=FloatField()) should do the trick.
What you want is: Visit.objects.filter(stuff).values(“ip_address”).annotate(n=models.Count(“pk”)) What this does is get all ip_addresses and then it gets the count of primary keys (aka number of rows) for each ip address.
You want something similar to “count … group by”. You can do this with the aggregation features of django’s ORM: from django.db.models import Count fieldname=”myCharField” MyModel.objects.values(fieldname) .order_by(fieldname) .annotate(the_count=Count(fieldname)) Previous questions on this subject: How to query as GROUP BY in django? Django equivalent of COUNT with GROUP BY
They are equal: >>> str(Person.objects.filter(age__isnull=True).query) == str(Person.objects.filter(age=None).query) True >>> print(Person.objects.filter(age=None).query) SELECT “person_person”.”id”, “person_person”.”name”, “person_person”.”yes”, “person_person”.”age” FROM “person_person” WHERE “person_person”.”age” IS NULL >>> print(Person.objects.filter(age__isnull=True).query) SELECT “person_person”.”id”, “person_person”.”name”, “person_person”.”yes”, “person_person”.”age” FROM “person_person” WHERE “person_person”.”age” IS NULL Exclusion: the Postgres JSON field (see the answer of @cameron-lee)
The Django 1.7 released a new and simple way to create combined queryset and model manager: class InquiryQuerySet(models.QuerySet): def for_user(self, user): return self.filter( Q(assigned_to_user=user) | Q(assigned_to_group__in=user.groups.all()) ) class Inquiry(models.Model): objects = InqueryQuerySet.as_manager() See Creating Manager with QuerySet methods for more details.
Define a related name: class Price(models.Model): book = models.ForeignKey(Book, related_name=”prices”) and then use it: books = Book.objects.prefetch_related(‘prices’)
You cannot query against model methods or properties. Either use the criteria within it in the query, or filter in Python using a list comprehension or genex.
If your DB is postgres which supports distinct() on field you can try Score.objects.order_by(‘student__username’, ‘-date’).distinct(‘student__username’)
Ok…this post is now old BUT what you could do is get all the ids of the objects in your list, then perform a model.objects.filter(pk__in=list_of_ids)
You should use aggregates. from django.db.models import Count User.objects.annotate(page_count=Count(‘page’)).filter(page_count__gte=2).count()