Django聚合：在关系查询集上聚合

Question

Suppose I have following models, which store questions and options for those questions.

PS: I am just writing the basic code just to give you idea.

class Question(models.Model):
    text = models.CharField()

class Option(models.Model):
    text = models.CharField()

class QuestionOption(models.Model):
    question = models.ForeignKey(Question)
    option = models.ForeignKey(Option)

And then I have models which store user Feedback and Options selected for each Question in the Feedback survey.

class Feedback(models.Model):
    full_name = models.CharField()
    cell_phone = models.CharField()
    created_at = models.DateTime()

class FeedbackOption(models.Model):
    feedback = models.ForeignKey(Feedback, related_name='feedback_options')
    option = models.ForeignKey(QuestionOption)

Every feedback will have lots of feedback option objects. Now I want to filter all the feedback whose feedback options contain specific QuestionOption object and then perform aggregate query where I am checking if that feedback's FeedbackOptions option text is 'boo' add count. Can i do this in one step, Something like this

# lets say i want to filter all feedback with QuestionOption id 1
stats = Feedback.objects.filter(feedback_options__option=1).aggregate(
    boo=Count(Case(When(feedback_options__option__option__text='boo', then=1))),
    hoo=Count(Case(When(feedback_options__option__option__text='hoo', then=1))))

It looks like it's applying aggregate on the only feedback option where option id is 1 not the rest of the feedback options for each feedback.

Answer 1

.filter() works as a where clause. When you add another method like .aggregate() to the resulting queryset, it is bounded by the same where clause.

In Django 1.11 you can use Subquery

from django.db.models import OuterRef, Subquery, Count

Feedback.objects.filter(feedback_options__option=o1).annotate(
    boo=Subquery(
        FeedbackOption.objects.filter(option__option__text="boo", feedback=OuterRef('pk')).values('feedback').annotate(count=Count('pk')).values('count'), 
        output_field=models.IntegerField())
    )

Similarly you can add another annotation for hoo .

Doc reference