Django - 从缓存查询填充模型实例相关字段

Django - Populate model instance related field from cached query

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)

一个节点可以有多个子节点,每个子节点可以有自己的子节点。

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)  

因为我已经抓取了 all_nodes 查询中的所有节点,所以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数据。

有什么方法可以实现吗?

树状结构中的数据不太适合关系数据库,但是有一些策略可以解决这个问题 - 请参阅有关 tree implemenations in the docs of django-treebeard 的章节。

如果您的树不是太大,您完全可以将树存储在 python 字典中并缓存结果。

示例(未经测试 - 根据您的喜好调整数据结构...):

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
  • 当然你必须cache enabled
  • 您可以在 Node.save 方法中使缓存无效

我设法让它以这种方式工作并用 2 个数据库调用填充整个树:

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)

由于这个答案,我发现了 _prefetched_objects_cache 属性:Django: Adding objects to a related set without saving to DB