Empresas
Empregos
  • Sobre nós
  • Soluções
    • Publicação de vagas
      Publique sua vaga e receba candidatos qualificados em 48h.
    • Avaliações de candidatos
      Mais de 500 testes técnicos e psicológicos, mais anti-fraude.
    • Headhunting
      Busca executiva personalizada do início ao fim.
    • Folha de Pagamento + EOR
      Dispersão de folha e EOR em mais de 15 países da LATAM.
  • Preços
  • Empregos

0

539
Visualizações
Simple Subquery with OuterRef

I am trying to make a very simple Subquery that uses OuterRef (not for practical purposes, but just to get it working), but I keep running into the same error.

posts/models.py code

from django.db import models

class Tag(models.Model):
    name = models.CharField(max_length=120)
    def __str__(self):
        return self.name

class Post(models.Model):
    title = models.CharField(max_length=120)
    tags = models.ManyToManyField(Tag)
    def __str__(self):
        return self.title

manage.py shell code

>>> from django.db.models import OuterRef, Subquery
>>> from posts.models import Tag, Post
>>> tag1 = Tag.objects.create(name='tag1')
>>> post1 = Post.objects.create(title='post1')
>>> post1.tags.add(tag1)
>>> Tag.objects.filter(post=post1.pk)
<QuerySet [<Tag: tag1>]>
>>> tags_list = Tag.objects.filter(post=OuterRef('pk'))
>>> Post.objects.annotate(count=Subquery(tags_list.count()))

The last two lines should give me number of tags for each Post object. And here I keep getting the same error:

ValueError: This queryset contains a reference to an outer query and may only be used in a subquery.
over 4 years ago · Santiago Trujillo
2 Respostas
Responde à pergunta

0

One of the problems with your example is that you cannot use queryset.count() as a subquery, because .count() tries to evaluate the queryset and return the count.

So one may think that the right approach would be to use Count() instead. Maybe something like this:

Post.objects.annotate(
    count=Count(Tag.objects.filter(post=OuterRef('pk')))
)

This won't work for two reasons:

  1. The Tag queryset selects all Tag fields, while Count can only count on one field. Thus: Tag.objects.filter(post=OuterRef('pk')).only('pk') is needed (to select counting on tag.pk).

  2. Count itself is not a Subquery class, Count is an Aggregate. So the expression generated by Count is not recognized as a Subquery (OuterRef requires subquery), we can fix that by using Subquery.

Applying fixes for 1) and 2) would produce:

Post.objects.annotate(
    count=Count(Subquery(Tag.objects.filter(post=OuterRef('pk')).only('pk')))
)

However if you inspect the query being produced:

SELECT 
    "tests_post"."id",
    "tests_post"."title",
    COUNT((SELECT U0."id" 
            FROM "tests_tag" U0 
            INNER JOIN "tests_post_tags" U1 ON (U0."id" = U1."tag_id") 
            WHERE U1."post_id" = ("tests_post"."id"))
    ) AS "count" 
FROM "tests_post" 
GROUP BY 
    "tests_post"."id",
    "tests_post"."title"

you will notice a GROUP BY clause. This is because COUNT is an aggregate function. Right now it does not affect the result, but in some other cases it may. That's why the docs suggest a different approach, where the aggregation is moved into the subquery via a specific combination of values + annotate + values :

Post.objects.annotate(
    count=Subquery(
        Tag.objects
            .filter(post=OuterRef('pk'))
            # The first .values call defines our GROUP BY clause
            # Its important to have a filtration on every field defined here
            # Otherwise you will have more than one group per row!!!
            # This will lead to subqueries to return more than one row!
            # But they are not allowed to do that!
            # In our example we group only by post
            # and we filter by post via OuterRef
            .values('post')
            # Here we say: count how many rows we have per group 
            .annotate(count=Count('pk'))
            # Here we say: return only the count
            .values('count')
    )
)

Finally this will produce:

SELECT 
    "tests_post"."id",
    "tests_post"."title",
    (SELECT COUNT(U0."id") AS "count" 
            FROM "tests_tag" U0 
            INNER JOIN "tests_post_tags" U1 ON (U0."id" = U1."tag_id") 
            WHERE U1."post_id" = ("tests_post"."id") 
            GROUP BY U1."post_id"
    ) AS "count" 
FROM "tests_post"
over 4 years ago · Santiago Trujillo Relatório

0

The django-sql-utils package makes this kind of subquery aggregation simple. Just pip install django-sql-utils and then:

from sql_util.utils import SubqueryCount
posts = Post.objects.annotate(
            tag_count=SubqueryCount('tag'))

The API for SubqueryCount is the same as Count, but it generates a subselect in the SQL instead of joining to the related table.

over 4 years ago · Santiago Trujillo Relatório
Responde à pergunta
Encontrar trabalhos remotos

Descubra a nova forma de encontrar um emprego!

melhores empregos
Principais categorias de trabalho
Empresas
Postar vaga Preços Comercial
Jurídico
Termos e Condições Política de privacidade
© 2026 PeakU Inc. All Rights Reserved.
Andres GPT
Recomende algumas ofertas para mim
Preciso de ajuda