Django查询会获得精确的多重查询 [英] Django queryset get exact manytomany lookup

问题描述

我有一个标签模型的实例列表，例如

I have a pk list of instances of Tag model, say

pk_list = [10, 6, 3]

我有另一个模型，包含m2m字段的标签和一个实际，只包含3个标签（以上的pks）。

I have another model with m2m field of tags and an instance that contains exactly 3 tags (of above pks).

class Node(models.Model):
    ...
    tags = models.ManyToManyField(Tag, related_name='nodes')

我想检索包含精确标签集的节点在我的pk_list中指定。当我做

I'd like to retrieve a Node that contains exact set of tags as specified in my pk_list. When I do

Node.objects.filter(tags__in=pk_list)

它返回三个相同实例的列表

it returns a list of three same instances

[<Node: My node title>, <Node: My node title>, <Node: My node title>]

调用.get（）不起作用，因为它必须返回单个实例，显然。

Calling .get() doesn't work cause it have to return a single instance, obviously.

那么如何检索单个实例？
我必须注意，如果我的pk_list是不同的例如。 [10,6]或[10，6，3，7]然后我什么也不收。我需要一个精确的匹配。

So, how do I retrieve a single instance? I must note that if my pk_list was different eg. [10, 6] or [10, 6, 3, 7] then I must receive nothing. I need an exact matching.

谢谢

推荐答案

使用过滤器链：

node_query = Node.objects.all()
pk_list = [10, 6, 3]

for pk in pk_list:
    node_query = node_query.filter(tags=pk)

现在，node_query将匹配节点，它具有至少三个具有pk 10,6,3的标签。要精确匹配三个标签：

Now node_query will match node, that has at least three tags with pk 10, 6, 3. To exact matching of three tags:

更新：
感谢 @janos 和<一个href =https://stackoverflow.com/users/1959899/adrian-lopez> @AdriánLópez，正确的答案是：

UPDATE: Thanks to @janos and @Adrián López, the correct answer is:

from django.db.models import Count

pk_list = [10, 6, 3]
node_query = Node.objects.annotate(count=Count('tags')).filter(count=len(pk_list))

for pk in pk_list:
    node_query = node_query.filter(tags__pk=pk)

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