Question

我有一个类似于收藏集1

{group_name: "group_1", key_name: "key_name_1"},
{group_name: "group_1", key_name: "key_name_2"},
{group_name: "group_2", key_name: "key_name_3"},
{group_name: "group_2", key_name: "key_name_4"},
{group_name: "group_2", key_name: "key_name_5"}

{subtype: "A", key_name: "key_name_1"},
{subtype: "B", key_name: "key_name_1"},
{subtype: "A", key_name: "key_name_2"},
{subtype: "B", key_name: "key_name_2"},
{subtype: "C", key_name: "key_name_2"},
{subtype: "A", key_name: "key_name_3"},
{subtype: "A", key_name: "key_name_4"}

我希望每个组中有collection2个对象。我正在尝试使用聚合，如下所示：

db.collection1.aggregate([
{
    $group : {
        _id: "$group_name",
        total : { $sum : db.collection2.find({key_name : "$key_name"}).count()}
   }
}
])

预期输出：

{_id:"group_1",total:5},
{_id:"group_2",total:2}

实际输出：

{_id:"group_1",total:0},
{_id:"group_2",total:0}

注意：

db.collection1.aggregate([ {
    $group : {
        _id: "$group_name",
        total : { $sum : db.collection2.find({key_name : "key_name_1"}).count()}
        } 
    } 
])

以上代码产生了以下预期的输出。

{_id:"group_1",total:4}, 
{_id:"group_2",total:6}

我需要pymongo或mongoshell中的解决方案。.

Answer 1

这就是$lookup聚合的目的

db.collection1.aggregate([ {   
     $lookup:
       {
         from: "collection2",
         localField: "key_name",
         foreignField: "key_name",
         as: "collection2_docs"
       }
  },{
    $group : {
        _id: "$group_name",
        total : { $sum : { $size: "$collection2_docs" } }
        } 
    } 
])

如何在mongo集合中查询另一个集合

1 个答案: