Fixed #31568 -- Fixed alias reference when aggregating over multiple subqueries.
691def10a0
made all Subquery() instances
equal to each other which broke aggregation subquery pushdown which
relied on object equality to determine which alias it should select.
Subquery.__eq__() will be fixed in an another commit but
Query.rewrite_cols() should haved used object identity from the start.
Refs #30727, #30188.
Thanks Makina Corpus for the report.
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@ -390,7 +390,7 @@ class Query(BaseExpression):
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else:
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# Reuse aliases of expressions already selected in subquery.
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for col_alias, selected_annotation in self.annotation_select.items():
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if selected_annotation == expr:
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if selected_annotation is expr:
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new_expr = Ref(col_alias, expr)
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break
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else:
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@ -15,3 +15,6 @@ Bugfixes
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* Fixed a regression in Django 3.0 where ``QuerySet.values()`` and
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``values_list()`` crashed if a queryset contained an aggregation and a
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subquery annotation (:ticket:`31566`).
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* Fixed a regression in Django 3.0 where aggregates used wrong annotations when
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a queryset has multiple subqueries annotations (:ticket:`31568`).
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@ -2,7 +2,8 @@ import datetime
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from decimal import Decimal
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from django.db.models import (
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Avg, Case, Count, F, OuterRef, Q, StdDev, Subquery, Sum, Variance, When,
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Avg, Case, Count, Exists, F, Max, OuterRef, Q, StdDev, Subquery, Sum,
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Variance, When,
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)
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from django.test import TestCase
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from django.test.utils import Approximate
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@ -120,3 +121,23 @@ class FilteredAggregateTests(TestCase):
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cnt=Count('pk', filter=Q(earliest_book_year=2008)),
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)
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self.assertEqual(aggs['cnt'], 2)
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def test_filtered_aggregate_ref_multiple_subquery_annotation(self):
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aggregate = Book.objects.values('publisher').annotate(
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has_authors=Exists(
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Book.authors.through.objects.filter(book=OuterRef('pk')),
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),
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authors_have_other_books=Exists(
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Book.objects.filter(
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authors__in=Author.objects.filter(
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book_contact_set=OuterRef(OuterRef('pk')),
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)
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).exclude(pk=OuterRef('pk')),
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),
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).aggregate(
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max_rating=Max(
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'rating',
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filter=Q(has_authors=True, authors_have_other_books=False),
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)
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)
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self.assertEqual(aggregate, {'max_rating': 4.5})
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