Question

我有一个由另一个用户录制的喜欢的模型：

class Like(models.Model):
   owner = models.ForeignKey(User, related_name="like_owner")
   likee = models.ForeignKey(User, related_name="like_likee")

我想知道最好的方法来获得至少有1个相似的用户的输出。现在我有一个解决方案，我基本上遍历每个用户的“like_owner”列表，计算它们也被命名为匹配的次数，但这似乎效率低下。

知道如何通过巧妙的聚合来解决这个问题吗？

编辑以添加我的（错误）代码：

users #list of preselected users I wish to find mutual matches for
for u in users:
    #check if user has requested contacts and is also at least 1 requested contact to reduce costly queries
    if u.like_owner.count() > 0 and u.like_likee.count() > 0:
        all = u.like_owner.all()
        for r in all:
            m = Like.objects.filter(r.likee=u, owner=r.likee).count()
            print m

Answer 1

是的，有更好的解决方案。试试gt或gte

User.objects.filter(like__likee__gte=1)

Answer 2

对于Django queryset我在想，但如果你对原始查询没问题。这可能是最好的方式。

SELECT DISTINCT * FROM (select fv1.owner_id as ui1, fv2.owner_id as ui2 from Like fv1 LEFT JOIN Like fv2 ON fv1.likee_id = fv2.likee_id where fv1.owner_id != fv2.owner_id) as a

Django：为所有用户找到共同喜欢的东西

2 个答案: