Django 中的左外反向 select_lated？

发布于 2024-09-03 20:48:26 字数 450 浏览 8 评论 0原文

想象一下以下模型：

class Parent(Model):
    ...

class Child(Model)
    father = ForeignKey(Parent)
    ...

有些父母有孩子，有些则没有（他们不是真正意义上的父母，只是一个虚构的名字）。

我想做以下查询：我想列出所有父母，如果他们有孩子，也把孩子带给我。这相当于子表的左外连接，即：

select * from app_parent left join app_child on child_father_id=parent_id

这样，当我在模板中调用 Parent.child_set 时，我不会无数次访问数据库。有办法做到这一点吗？谢谢

原文

Imagine the following model:

class Parent(Model):
    ...

class Child(Model)
    father = ForeignKey(Parent)
    ...

Some parents have children, others do not (they're not parents in the real meaning, it's just a fictional name).

I would like to make the following query: I want to list all the Parents, and if they have children, bring me the children too. That would be the equivalent of a left outer join to Child table, that is:

select * from app_parent left join app_child on child_father_id=parent_id

This way, when I invoke Parent.child_set in my template, I won't hit the database a gazillion times. Is there a way to do that?
Thanks

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地狱即天堂 2024-09-10 20:48:26

从 Django 1.4 prefetch_lated 开始，可以满足您的需求。

Parent.objects.prefetch_related('child_set')

相关（！）django 文档： https://docs.djangoproject .com/en/dev/ref/models/querysets/#prefetch-lated。

Starting from Django 1.4 prefetch_related does what you want.

Parent.objects.prefetch_related('child_set')

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兮子 2024-09-10 20:48:26

在这种情况下，我认为最好的办法是列出子级，然后从它们中获取父级，如下所示：

children = Child.objects.filter(...).select_related('parent').order_by('parent')

然后在模板中，可能使用regroup（注意order_by< /代码> 上面）：

{% regroup children by parent as parents %}
<ul>
{% for parent in parents %}
    <li>{{ parent.grouper }}
    <ul>
    {% for child in parents.list %}
    ...
    {% endfor %}
    </ul>
    </li>
{% endfor %}
</ul>

In this case, I think the best thing to do is list the children, then get the parent from them, like this:

children = Child.objects.filter(...).select_related('parent').order_by('parent')

Then in the template, possibly use a regroup (note the order_by above):

{% regroup children by parent as parents %}
<ul>
{% for parent in parents %}
    <li>{{ parent.grouper }}
    <ul>
    {% for child in parents.list %}
    ...
    {% endfor %}
    </ul>
    </li>
{% endfor %}
</ul>

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