Pass or refer parent query value into subquery

[Sencer Hamarat]

Hi,

I need to pass the parent query book_id value into the subquery.
Here is the model and the current state of the query I reached.

class Publisher(models.Model):

    book = models.ForeignKey(Book)

    bundle = models.JSONField()

current_sales_count_query = Publisher.objects.filter(
            Q(book_id=F("book_id")) | Q(bundle__contains=F("book_id")),
        ).values('id')

query = Publisher.objects.filter(
             **query_params,
        ).select_related(
             'book',
        ).annotate(
            current_sales_count=Count(Subquery(current_sales_count_query)),
        )

Sql Output:

SELECT "publisher"."id",

       ......

       COUNT((SELECT U0."id"
              FROM "publisher" U0
              WHERE (U0."book_id" = (U0."logbook_id") OR U0."current_tree_level" @> (U0."book_id") "current_rejected_count",
       "books"."id",
       ........

FROM "publisher"
         INNER JOIN "books" ON ("publisher"."book_id" = "books"."id")
WHERE "publisher"."book_id" IN (1, 2, 12)
GROUP BY "publisher"."id", "books"."id"

But subquery insists on using book_id from itself with that query.
What should I do to make things right?

King regards,

Sencer HAMARAT

[Ross Meredith]

https://docs.djangoproject.com/en/4.0/ref/models/expressions/#django.db.models.OuterRef

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[Sencer Hamarat]

Thanks Ross,

Very appreciated.

That's what I need.

I'm also having a problem with this query after I implemented the solution:

I believe, 
Q(bundle__contains=OuterRef("book_id"))
part of the query is expecting an array instead of direct value.

I'm getting this error:

ProgrammingError: operator does not exist: jsonb @> bigint
LINE 1: ...lisher"."book_id") OR U0."bundle" @> ("publisher_...
                                                             ^
HINT:  No operator matches the given name and argument types. You might need to add explicit type casts.

Is there a way to handle this error? How can I cast the value into an array?

Regards,

Sencer HAMARAT

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[Ross Meredith]

Sorry I haven't used the JSONField myself before.

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