CODEWITHSUNDEEP
djangodjango-modelsdjango-querysetdjango-aggregationdjango-annotate
Eric
Eric

Reputation: 601

django annotate models with an aggregate value based on query

Let's say I have the following model structure:

Parent():

Child():
parent = ForeignKey(Parent)

GrandChild():
child = ForeignKey(Child)
state = BooleanField()
num = FloatField()

I'm trying to from the Parent ViewSet, recover the following:

  1. The number of children.
  2. The SUM of the 'num' fields when 'state' is True.

I can do the following:

queryset = Parent.objects\
    .annotate(child_count=Count('child'))\
    .annotate(sum_total=Sum('child__grandchild__num'))

This gives me (1) but instead of (2) it gives me the SUM for ALL grandchildren. How can I filter the grandchildren appropriately while ensuring I have all of the Parent objects still in the QuerySet?

Upvotes: 6

Views: 7107

Answers (3)

Arpit Solanki
Arpit Solanki

Reputation: 9931

Try using filter before the annotate

queryset = Parent.objects.filter(child__grandchild__state='True')\
    .annotate(child_count=Count('child'))\
    .annotate(sum_total=Sum('child__grandchild__num'))

Upvotes: 1

ZaherSarieddine
ZaherSarieddine

Reputation: 126

You can do the following:

qs = Parents.objects.all()
child_count = Count('children')
num_sum = Sum('children__grandchildren__num', filter=Q(children__grandchildren__state=True))
qs = qs.annotate(child_count=child_count).annotate(num_sum=num_sum)

chidlren and grandchildren are the related names that you can define in your models

Upvotes: 0

Pati Ram Yadav
Pati Ram Yadav

Reputation: 935

Which version of django are you using? You can use subquery as well if version is supported.

from django.db.models import OuterRef, Subquery

Parent.objects
.annotate(child_count=Count('child'))
.annotate(
    grandchild_count_for_state_true=Subquery(
        GrandChild.objects.filter(
            state=True,
            child=OuterRef('pk')
        ).values('parent')
        .annotate(cnt=Sum('child__grandchild__num'))
        .values('cnt'),
        num=models.IntegerField()
    )
)

You can optimise this through aggregation query.

Upvotes: 5

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