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Mongodb groupby on Dictionary里面的字典

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我有一个集合，其中包含以下文档：文档1：

{
 "company": "ABC"
 "application": {
    "app-1": {"earning_from_src_A": 50,
              "earning_from_src_B": 43},
    "app-2": {"earning_from_src_A": 10,
              "earning_from_src_B": 13}
 }
}

文件2：

{
 "company": "ABCD"
 "application": {
    "app-1": {"earning_from_src_A": 150,
              "earning_from_src_B": 413},
    "app-2": {"earning_from_src_A": 110,
              "earning_from_src_B": 113},
    "app-3": {"earning_from_src_A": 1,
              "earning_from_src_B": 34},
 }
}

我想在整个收集文件中获得应用程序的总和 . 喜欢：结果：

{
 "app-1": {"earning_from_src_A": 200,
           "earning_from_src_B": 456},
 "app-2": {"earning_from_src_A": 120,
           "earning_from_src_B": 126},
 "app-3": {"earning_from_src_A": 1,
           "earning_from_src_B": 34},
}

有人请帮帮我 .

我尝试使用聚合函数，但我无法在整个集合中进行分组 .

mongodb mongodb-query aggregation-framework pymongo

1 回答

  • 2

    您可以使用以下聚合

    所以基本上你的钥匙在这里是未知的因此，您必须使用$objectToArray聚合将它们转换为键值对 . 然后你可以轻松$group使用应用程序名称并从 A 和 B 获得收入 .

    db.collection.aggregate([
  { "$addFields": {
    "application": { "$objectToArray": "$application" }
  }},
  { "$unwind": "$application" },
  { "$group": {
    "_id": "$application.k",
    "earning_from_src_A": {
      "$sum": "$application.v.earning_from_src_A"
    },
    "earning_from_src_B": {
      "$sum": "$application.v.earning_from_src_B"
    }
  }},
  { "$group": {
    "_id": null,
    "data": {
      "$push": {
        "k": "$_id",
        "v": {
          "earning_from_src_A": "$earning_from_src_A",
          "earning_from_src_B": "$earning_from_src_B"
        }
      }
    }
  }},
  { "$replaceRoot": { "newRoot": { "$arrayToObject": "$data" }}}
])

    哪个outputs

    [
  {
    "app-1": {
      "earning_from_src_A": 200,
      "earning_from_src_B": 456
    },
    "app-2": {
      "earning_from_src_A": 120,
      "earning_from_src_B": 126
    },
    "app-3": {
      "earning_from_src_A": 1,
      "earning_from_src_B": 34
    }
  }
]
    回复于

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