python - Django查询优化: find a list of objects based on a many-to-one to a many-to-many

Question

我有一个丑陋的循环，它在 Django 中作为创可贴运行，我想将其优化为连接。我已将其分解为最基本的结构，以帮助简化问题。我希望其他人以前遇到过类似的问题 - 如果没有，我会在最终修复它时分享答案。

总结

• 对象 A 与其自身具有非对称的多对多关系。
• 对象 B 与对象 A 具有多对一关系。
• 给定一个 ObjectB，我需要一组其他 ObjectB，它们是与我们的 关联的 ObjectAs 的子级ObjectB 的父对象 ObjectA。

我确信具有更多数据库经验的人可以更好地表达这一点(如果您对此关联有更好的描述，请留下评论)。

这是我在 Django 中使用的结构的简单示例，以及正在运行的循环类型:

class ObjectA(models.Model):
    object_a_rules = models.ManyToManyField("self", symmetrical=False, through='ObjectARule')

class ObjectARule(models.Model):
    object_a_one = models.ForeignKey(ObjectA, related_name="object_a_before")
    object_a_two = models.ForeignKey(ObjectA, related_name="object_a_after")

class ObjectB(models.Model):
    object_a_association = models.ForeignKey(ObjectA)

    def that_annoying_loop_i_mentioned(self):
        object_a_rules_list = ObjectARule.objects.filter(object_a_one = self.object_a_association)
        #A list of all of the ObjectARules that have the ObjectA this ObjectB is associated with
        #as the first half of the many-to-many relationship.

        object_b_list = ObjectB.objects.all()
        #A list of all of the ObjectBs, may also be a filtered list

        for object_a_rule in object_a_rules_list:
            for object_b in object_b_list:
                if (object_a_rule.object_a_two == object_b.object_a_association):
                #if the second half of ObjectARule is the ObjectA of
                #the ObjectB in this list, then do something with that ObjectB.
                pass

Django 如何通过连接获取 ObjectB 列表，从而不必运行这个极其低效的循环？

Answer 1

Given an ObjectB, I need a set of other ObjectBs that are the children of the ObjectAs that are associated with our ObjectB's parent ObjectA.

如果 objb 是给定的 ObjectB，则可以按如下方式执行此操作:

objects = ObjectB.objects.filter(object_a_association__object_a_rules=objb.object_a_association)

或者，

objects = ObjectB.objects.filter(object_a_association__object_a_rules__objectb_set=objb)

另请参阅Lookups that span relationships