Question

情况与Django prefetch_related children of children相同，但问题不同：

我有一个模型Node，看起来像这样：

class Node(models.Model):
    parent = models.ForeignKey('self', related_name='children', on_delete=models.CASCADE, null=True)

一个节点可以有几个孩子，每个孩子可以有自己的孩子。

我想做这样的事情：

def cache_children(node):
    for child in node.children.all():
        cache_children(child)

root_node = Node.objects.prefetch_related('children').get(pk=my_node_id) 

all_nodes = Node.objects.all()  # get all the nodes in a single query

# Currently: hit database for every loop
# Would like: to somehow use the already loaded data from all_nodes
cache_children(root_node)

由于我已经获取了all_nodes查询中的所有节点，因此我想重用该查询中的缓存数据，而不是每次都执行一个新的查询。

有什么办法可以实现？

Answer 1

树状结构中的数据并不是很适合关系数据库，但是有一些策略可以解决此问题-请参阅tree implemenations in the docs of django-treebeard上的章节。

如果树不是太大，则可以将树完全存储在python dict中并缓存结果。

示例（未经测试-使数据结构适合您的喜好...）：

from django.core.cache import cache

# ...

def get_children(nodes, node):
    node['children'] = [n for n in nodes if n['parent']==node['id']]
    for child_node in node['children']:
        child_node = get_children(nodes, child_node)
    return node


def get_tree(timeout_in_seconds=3600)
    tree = cache.get('your_cache_key')
    if not tree:
        # this creates a list of dicts with the instances values - one DB hit!
        all_nodes = list(Node.objects.all().values())
        root_node = [n for n in nodes if n['parent']==None][0]
        tree = get_children(all_nodes, root_node)

        cache.set('your_cache_key', tree, timeout_in_seconds)
    return tree

当然，您必须拥有cache enabled
您可以使Node.save方法中的缓存无效

Answer 2

我设法使其工作，并用2个db调用填充了整个树：

def populate_prefetch_cache(node, all_nodes):
    children = [child for child in all_nodes if child.parent_id==node.id]

    # will not have the attribute if no prefetch has been done
    if not hasattr(node, '_prefetched_objects_cache'):
        node._prefetched_objects_cache = {}

    # Key to using local data to populate a prefetch!
    node._prefetched_objects_cache['children'] = children
    node._prefetch_done = True

    for child in node.children.all():
        populate_prefetch_cache(child , all_nodes )


all_nodes = list(Node.objects.all())  # Hit database once
root_node = Node.objects.get(pk=my_node_id)  # Hit database once

# Does not hit the database and properly populates the children field
populate_prefetch_cache(root_node, all_nodes)

由于以下答案，我发现了_prefetched_objects_cache属性：Django: Adding objects to a related set without saving to DB

Django-从缓存的查询中填充与模型实例相关的字段

2 个答案: