mirror of https://github.com/django/django.git
Fixed #10666 -- Corrected the handling of inherited fields with aggregate() and annotate(). Thanks to julienb for the report.
git-svn-id: http://code.djangoproject.com/svn/django/trunk@10446 bcc190cf-cafb-0310-a4f2-bffc1f526a37
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@ -1403,8 +1403,17 @@ class BaseQuery(object):
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field_name = field_list[0]
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field_name = field_list[0]
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col = field_name
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col = field_name
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source = self.aggregates[field_name]
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source = self.aggregates[field_name]
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elif (len(field_list) > 1 or
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elif ((len(field_list) > 1) or
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field_list[0] not in [i.name for i in opts.fields]):
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(field_list[0] not in [i.name for i in opts.fields]) or
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self.group_by is None or
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not is_summary):
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# If:
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# - the field descriptor has more than one part (foo__bar), or
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# - the field descriptor is referencing an m2m/m2o field, or
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# - this is a reference to a model field (possibly inherited), or
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# - this is an annotation over a model field
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# then we need to explore the joins that are required.
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field, source, opts, join_list, last, _ = self.setup_joins(
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field, source, opts, join_list, last, _ = self.setup_joins(
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field_list, opts, self.get_initial_alias(), False)
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field_list, opts, self.get_initial_alias(), False)
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@ -1419,15 +1428,11 @@ class BaseQuery(object):
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col = (join_list[-1], col)
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col = (join_list[-1], col)
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else:
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else:
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# Aggregate references a normal field
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# The simplest cases. No joins required -
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# just reference the provided column alias.
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field_name = field_list[0]
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field_name = field_list[0]
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source = opts.get_field(field_name)
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source = opts.get_field(field_name)
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if not (self.group_by is not None and is_summary):
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col = field_name
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# Only use a column alias if this is a
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# standalone aggregate, or an annotation
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col = (opts.db_table, source.column)
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else:
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col = field_name
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# Add the aggregate to the query
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# Add the aggregate to the query
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alias = truncate_name(alias, self.connection.ops.max_name_length())
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alias = truncate_name(alias, self.connection.ops.max_name_length())
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@ -1659,7 +1664,6 @@ class BaseQuery(object):
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last.append(len(joins))
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last.append(len(joins))
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if name == 'pk':
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if name == 'pk':
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name = opts.pk.name
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name = opts.pk.name
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try:
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try:
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field, model, direct, m2m = opts.get_field_by_name(name)
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field, model, direct, m2m = opts.get_field_by_name(name)
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except FieldDoesNotExist:
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except FieldDoesNotExist:
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@ -239,5 +239,19 @@
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"friends": [8],
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"friends": [8],
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"name": "Stuart Russell"
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"name": "Stuart Russell"
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}
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}
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},
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{
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"pk": 5,
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"model": "aggregation_regress.hardbackbook",
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"fields": {
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"weight": 4.5
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}
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},
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{
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"pk": 6,
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"model": "aggregation_regress.hardbackbook",
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"fields": {
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"weight": 3.7
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}
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}
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}
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]
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]
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@ -58,6 +58,12 @@ class Clues(models.Model):
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EntryID = models.ForeignKey(Entries, verbose_name='Entry', db_column = 'Entry ID')
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EntryID = models.ForeignKey(Entries, verbose_name='Entry', db_column = 'Entry ID')
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Clue = models.CharField(max_length=150)
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Clue = models.CharField(max_length=150)
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class HardbackBook(Book):
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weight = models.FloatField()
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def __unicode__(self):
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return "%s (hardback): %s" % (self.name, self.weight)
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__test__ = {'API_TESTS': """
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__test__ = {'API_TESTS': """
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>>> from django.core import management
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>>> from django.core import management
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>>> from django.db.models import get_app, F
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>>> from django.db.models import get_app, F
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@ -137,17 +143,17 @@ __test__ = {'API_TESTS': """
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>>> Book.objects.all().aggregate(num_authors=Count('foo'))
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>>> Book.objects.all().aggregate(num_authors=Count('foo'))
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Traceback (most recent call last):
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Traceback (most recent call last):
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...
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...
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, id, isbn, name, pages, price, pubdate, publisher, rating, store
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, hardbackbook, id, isbn, name, pages, price, pubdate, publisher, rating, store
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>>> Book.objects.all().annotate(num_authors=Count('foo'))
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>>> Book.objects.all().annotate(num_authors=Count('foo'))
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Traceback (most recent call last):
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Traceback (most recent call last):
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...
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...
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, id, isbn, name, pages, price, pubdate, publisher, rating, store
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, hardbackbook, id, isbn, name, pages, price, pubdate, publisher, rating, store
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>>> Book.objects.all().annotate(num_authors=Count('authors__id')).aggregate(Max('foo'))
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>>> Book.objects.all().annotate(num_authors=Count('authors__id')).aggregate(Max('foo'))
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Traceback (most recent call last):
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Traceback (most recent call last):
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...
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...
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, id, isbn, name, pages, price, pubdate, publisher, rating, store, num_authors
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FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, contact, hardbackbook, id, isbn, name, pages, price, pubdate, publisher, rating, store, num_authors
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# Old-style count aggregations can be mixed with new-style
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# Old-style count aggregations can be mixed with new-style
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>>> Book.objects.annotate(num_authors=Count('authors')).count()
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>>> Book.objects.annotate(num_authors=Count('authors')).count()
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@ -276,6 +282,22 @@ FieldError: Cannot resolve keyword 'foo' into field. Choices are: authors, conta
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>>> publishers
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>>> publishers
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[<Publisher: Apress>, <Publisher: Sams>]
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[<Publisher: Apress>, <Publisher: Sams>]
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# Regression for 10666 - inherited fields work with annotations and aggregations
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>>> HardbackBook.objects.aggregate(n_pages=Sum('book_ptr__pages'))
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{'n_pages': 2078}
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>>> HardbackBook.objects.aggregate(n_pages=Sum('pages'))
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{'n_pages': 2078}
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>>> HardbackBook.objects.annotate(n_authors=Count('book_ptr__authors')).values('name','n_authors')
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[{'n_authors': 2, 'name': u'Artificial Intelligence: A Modern Approach'}, {'n_authors': 1, 'name': u'Paradigms of Artificial Intelligence Programming: Case Studies in Common Lisp'}]
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>>> HardbackBook.objects.annotate(n_authors=Count('authors')).values('name','n_authors')
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[{'n_authors': 2, 'name': u'Artificial Intelligence: A Modern Approach'}, {'n_authors': 1, 'name': u'Paradigms of Artificial Intelligence Programming: Case Studies in Common Lisp'}]
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"""
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"""
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}
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}
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