Question

鉴于这是来自List的直通模型（我试图使用一些通用模型/字段，因为我的实际模型是行业特定的），

class List( models.Model ):
    members = models.ManyToManyField( User, through='ListMembership')
    ...

class ListMembership( models.Model ):
    member = models.ForeinKey( User )
    list = models.ForeignKey( List )
    member_accepted = models.BooleanField( default=False )
    list_accepted = models.BooleanField( default=False )
    ...

我试图在List的查询集中使用through模型。我想获取当前登录用户所属的所有列表，并且被用户（member_accepted）和列表（list_accepted）接受。

如果我只关心成员并且不关心额外的模型字段，这将有效。

qs = List.objects.filter(members = self.request.user)

我想进入通过领域。

#this is wrong, but this gives an idea of what I am attempting to do
qs = List.objects.filter(members__through__member_accepted=True, members__through__list_accepted=True)

编辑以举例说明我不想做的事情

#this works, but I want to avoid the extra query
through_lists = ListMembership.objects.filter(member = request.user, member_accepted=True, list_accepted=True).values_list('list__pk', flat=True)
qs = List.objects.filter(pk__in=through_lists)

Answer 1

这个应该真的有效：

qs = List.objects.filter(
        members__member=request.user,
        members__member_accepted=True,
        members__list_accepted=True
     )

请注意双下划线。

Answer 2

through_lists = ListMembership.objects.filter(member = request.user, member_accepted=True, list_accepted=True)

然后在through_lists中有ListMembership个实例，只需获取每个实例的列表，无需额外查询。

lists = map(lambda x: x.list,through_lists)

通过queryset中的模型字段使用

2 个答案: