当模型没有外来对象或外域时，Django过滤器等于

Question

我有两种型号的Community和UserCommunity

社区模型

class Community(models.Model):

    # Community name
    name = models.CharField(max_length=64)
    slug = models.CharField(max_length=40, blank=True)
    admins = models.ManyToManyField(
        settings.AUTH_USER_MODEL, related_name="admins", blank=True
    )
    admins = models.ManyToManyField(
        settings.AUTH_USER_MODEL, related_name="admins", blank=True
    )
    members = models.ManyToManyField(
        settings.AUTH_USER_MODEL,
        related_name="members",
        blank=True,
    )
    ------

UserCommunityModel

class UserCommunity(models.Model):

    user = models.ForeignKey(
        settings.AUTH_USER_MODEL, on_delete=CASCADE, related_name="user"
    )
    group = models.ForeignKey(
        Community, on_delete=CASCADE, related_name="group")
    role = models.CharField(max_length=15, blank=True, null=True)
    # True is in community, false is not in community
    active = models.BooleanField(null=True, blank=True)

我需要获得所有的社区对象，如果用户不是社区的一部分，社区至少应该有1名成员。

我试过用这个

Community.objects.filter(group__user=request.user.id, group__active=False,state=location.state, group_discoverability="public").exclude(
        hofAdmins__isnull=True, admins__isnull=True, members__isnull=True)[:10]

但这将返回具有UserCommunity对象的社区，在用户加入社区时创建UserCommunity对象。任何帮助都将不胜感激，谢谢！

Answer 1

Community.objects.filter(~Q(members__in=[request.user.id]), ~Q(hofAdmins__in=[request.user.id]), ~Q(admins__in=[request.user.id]), state=location.state, group_discoverability="public").exclude(         hofAdmins__isnull=True, admins__isnull=True, members__isnull=True)[:10]