[英]Django foreign key count filter
我有以下兩種模型:
Class Foo(models.model):
param1 = ...
param2 = ...
...
paramN = ...
Class Bar(models.model):
foo = models.ForeignKey(Foo)
...
...
目標:計算所有Foo實例的QuerySet,以便將多個Bar實例連接到它
我一直在尋找解決方案,這似乎對其他所有人都有效
Foo.objects.annotate(num_bar=Count('bar')).filter(num_bar__gt=1)
這給了我一個FieldError
說'bar'
對於Foo
不是一個可能的字段,然后我嘗試了'bar_set'
並且也得到了同樣的錯誤
我是否有可能錯誤地實施它們,或者因為它們很舊而現在已經貶值了? 任何幫助,將不勝感激!
Traceback (most recent call last):
File "<console>", line 1, in <module>
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/manager.py", line 127, in manager_method
return getattr(self.get_queryset(), name)(*args, **kwargs)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/query.py", line 794, in annotate
obj.query.add_annotation(annotation, alias, is_summary=False)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/sql/query.py", line 982, in add_annotation
summarize=is_summary)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/aggregates.py", line 20, in resolve_expression
c = super(Aggregate, self).resolve_expression(query, allow_joins, reuse, summarize)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/expressions.py", line 491, in resolve_expression
c.source_expressions[pos] = arg.resolve_expression(query, allow_joins, reuse, summarize, for_save)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/expressions.py", line 448, in resolve_expression
return query.resolve_ref(self.name, allow_joins, reuse, summarize)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1532, in resolve_ref
self.get_initial_alias(), reuse)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1471, in setup_joins
names, opts, allow_many, fail_on_missing=True)
File "/home/ryan/.virtualenvs/project/local/lib/python2.7/site-packages/django/db/models/sql/query.py", line 1396, in names_to_path
"Choices are: %s" % (name, ", ".join(available)))
FieldError: Cannot resolve keyword 'bar' into field. Choices are: param1, param2, param3, ..., paramN
我的Django版本是1.8.3
該錯誤可能有多種原因,我將嘗試說明可能的原因:
num_model
。 還檢查是否可以得到它們的數量(骯臟的方式:)):
for foo in Foo.objects.all(): if foo.bar_set.count() < 2: #do sth like : foo.bar_set.get() or temp = temp + 1
從您對模型的簡短描述(不是您的主代碼)開始,找不到其他原因。 您的查詢應該可以。
因此,在嘗試了很多之后,這是一個可行的解決方案:
Bar.objects.values("foo_id").annotate(Count("foo_id")).filter(pk__count__gt=1)
不能完全確定為什么這樣做有效,而另一個則無效,但實際上,它只是獲取具有相同foo_id
的Bar
對象的數量,並確保存在多個對象。
如果有人想解釋一個可行的原因,而另一個卻不可行,那將不勝感激。
由於輸入錯誤,我遇到了同樣的問題。 這是我的用例的解決方案
# from django.db.models.sql.aggregates import Count # wrong import
from django.db.models import Count # correct one
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