blocks|key|2200357|text|您可以考虑添加一个可读的自定义查找字段，如group__at_least_one=True+by+自定义查找|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2200358|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/2.1/howto/custom-lookups/^0|L|O|1D|5|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

You can consider adding a readable custom lookup field like <code>group__at_least_one=True</code> by <a href="https://docs.djangoproject.com/en/2.1/howto/custom-lookups/" rel="nofollow noreferrer">Custom Lookups</a>

blocks|key|11609|text|我不知道你对groups__isnull=False有什么意见，这是绝对好的。但是请注意，它完全等同于groups=None，您可能会发现它更好一些。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|11610|entityMap^0|6|K|1F|B|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

I'm not sure what you have against <code>groups__isnull=False</code>; it's absolutely fine. Note though that it's exactly equivalent to <code>groups=None</code>, which you might find a bit nicer.

blocks|key|911822|text|你也可以：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|911823|Group.objects.values_list('user').distinct('user')|code-block|syntax|javascript|911824|并且您将获得唯一的用户只有1和更多的Group。|offset|length|style|CODE|911825|entityMap^0|0|0|I|5|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

You can also do:

<pre><code>Group.objects.values_list('user').distinct('user')
</code></pre>

And you will get unique users only with 1 and more <code>Group</code>.

blocks|key|913558|text|User.objects.filter(groups__isnull=False).values_list('username')生成简单的SQL：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|913559|SELECT+"users_user"."username"+FROM+"users_user"+
INNER+JOIN+"users_user_groups"+ON+("users_user"."id"+=+"users_user_groups"."user_id")+
WHERE+"users_user_groups"."group_id"+IS+NOT+NULL|code-block|syntax|javascript|913560|fk上的过滤null在文档中显示完全相同。|913561|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.djangoproject.com/en/2.1/ref/models/querysets/#std:fieldlookup-isnull^0|0|1T|0|0|B|2|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|M|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]]]

<code>User.objects.filter(groups__isnull=False).values_list('username')</code> produces simple SQL: 

<pre><code>SELECT "users_user"."username" FROM "users_user" 
INNER JOIN "users_user_groups" ON ("users_user"."id" = "users_user_groups"."user_id") 
WHERE "users_user_groups"."group_id" IS NOT NULL
</code></pre>

Filtering null on a fk shown exactly the same in <a href="https://docs.djangoproject.com/en/2.1/ref/models/querysets/#std:fieldlookup-isnull" rel="nofollow noreferrer">docs</a>.

blocks|key|2200329|text|较少使用SQL实现的解决方案是使用Exists子查询或Count。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2200330|User.objects.annotate(
++++has_groups=Exists(
++++++++User.groups.through.objects.filter(
++++++++++++user=OuterRef('pk'),
++++++++)
++++)
).filter(has_groups=True)

User.objects.annotate(
++++groups_count=Count('groups'),
).filter(groups__gte=1)|code-block|syntax|javascript|2200331|不过，我怀疑这些解决方案的执行速度是否会快于groups__isnull=True的内部连接。如果m2m字段公开了__exists查找，这样您就可以执行filter(groups__exists=True)并避免使用大多数模板，这将是很好的。|2200332|entityMap^0|H|6|R|5|0|0|M|J|1L|8|24|R|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

A less SQL implementation specific solution would be to use either an <code>Exists</code> subquery or a <code>Count</code>

<pre><code>User.objects.annotate(
 has_groups=Exists(
 User.groups.through.objects.filter(
 user=OuterRef('pk'),
 )
 )
).filter(has_groups=True)

User.objects.annotate(
 groups_count=Count('groups'),
).filter(groups__gte=1)
</code></pre>

I doubt these solution would perform any faster than <code>groups__isnull=True</code>'s inner join though. It'd be great if m2m fields exposed an <code>__exists</code> lookup so you could do <code>filter(groups__exists=True)</code> and avoid most of this boilerplate.

I want to find all users which are at least in one group.

I found a solution

<pre><code>from django import setup

setup()

from django.contrib.auth.models import User
user_without_group = User.objects.update_or_create(username='user-without-group')[0]
user_without_group.groups.clear()
query = User.objects.filter(groups__isnull=False).values_list('username', flat=True)
print (query.query)
print list(query)
assert user_without_group.username not in query
</code></pre>

... but the solution is not nice. Is there a more obvious solution than <code>groups__isnull=False</code>?

update

Why I think this solution is not nice: I think this is not obvious. It is not easy to read and understand. If you show this to someone who knows Python, but has never used the django ORM, I don't think he will immediately understand what this means.

<pre><code>User.objects.filter(groups__isnull=False)
</code></pre>

Django ORM: All users which have at least one group

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想找到至少在一个组中的所有用户。我找到了一个解决办法from django import setupsetup()from django.contrib.auth.models import Useruser_without_group = User.objects.update_or_create(username='user-without-group')[0]user_without_gr

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