Django查询优化：根据多对一到多对多查找对象列表

I have an ugly loop that is operating as a band-aid in Django that I'd like to optimize into a join. I've broken it down to the most basic structure to help simplify the problem. I'm hoping someone else has run into something like this before -- if not, I'll share the answer when I finally fix it.

我有一个丑陋的循环,作为Django的创可贴,我想优化成一个连接。我把它分解为最基本的结构,以帮助简化问题。我希望其他人之前碰到过这样的事情 - 如果没有,我会在最后修复时分享答案。

To summarize

Object A has a non-symmetrical, many-to-many relationship with itself.

对象A与其自身具有非对称的多对多关系。

Object B has a many-to-one relationship with Object A.

对象B与对象A具有多对一关系。

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.

给定一个ObjectB,我需要一组其他ObjectB,它们是与ObjectB的父ObjectA相关联的ObjectAs的子代。

I'm sure someone with more database experience could phrase that better (leave a comment if you have a better description of this association).

我确信拥有更多数据库经验的人可能会更好地说出来(如果您对此关联有更好的描述,请留言)。

Here is a bare bones example of the structure I'm working with in Django, and the sort of loop that's running:

这是我在Django中使用的结构的一个简单的例子,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

How could Django get a list of ObjectBs through a join, so this painfully inefficient loop wouldn't have to run?

Django如何通过连接获得ObjectB的列表,所以这个非常低效的循环不必运行?

1 个解决方案

#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.

给定一个ObjectB,我需要一组其他ObjectB,它们是与ObjectB的父ObjectA相关联的ObjectAs的子代。

If objb is the ObjectB you are given, you could do this as follows:

如果objb是您给出的ObjectB,您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)

or alternatively,

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

See also Lookups that span relationships

另请参阅跨越关系的查找

1 个解决方案

#1

更多相关文章

随机推荐