Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
356 views
in Technique[技术] by (71.8m points)

python - 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.
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

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"

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...