Mongodb按内部元素分组答案

【问题标题】：Mongodb group by inner elementMongodb按内部元素分组
【发布时间】：2021-12-20 00:24:08
【问题描述】：

我用一个简单的例子来解释我的 mongodb 集合看起来像这样， [

    {
        pid: erwer,
        qty: 3,
        LevelDetails: {
            level1: { userId: 1, amount: 10 },
            level2: { userId: 2, amount: 20 },
            level3: { userId: 3, amount: 13 },
        }
    },
    {
        pid: qwsdfg,
        qty: 1,
        LevelDetails: {
            level1: { userId: 1, amount: 10 },
            level2: { userId: 4, amount: 20 },
            level3: { userId: 3, amount: 13 },
        }
    },

]

从集合中，我需要每个用户的级别 1、级别 2 和级别 3 的总和。查询结果应该是这样的 [

        { userId1: { TotalLevel1Amount: 20, TotalLevel2Amount: 0, TotalLevel3Amount: 0 } },

        { userId2: { TotalLevel1Amount: 0, TotalLevel2Amount: 20, TotalLevel3Amount: 0 } },

        { userId3: { TotalLevel1Amount: 0, TotalLevel2Amount: 0, TotalLevel3Amount: 26 } },

        { userId4: { TotalLevel1Amount: 0, TotalLevel2Amount: 20, TotalLevel3Amount: 0 } }
    ]

【问题讨论】：

标签： database mongodb go nosql aggregation-framework

【解决方案1】：

$set：通过将 LevelDetails 转换为键值对来添加新字段 _levelDetails。
$unwind：解构_levelDetails数组。
$group：根据级别 (_levelDetails.k) 有条件地按 _levelDetails.v.userId 和 $sum 分组。
$project: 格式化显示的文档。
$sort（可选）：按userID升序排序。

db.collection.aggregate([
  {
    $set: {
      _levelDetails: {
        $objectToArray: "$LevelDetails"
      }
    }
  },
  {
    $unwind: "$_levelDetails"
  },
  {
    $group: {
      _id: "$_levelDetails.v.userId",
      "TotalLevel1Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level1"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      },
      "TotalLevel2Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level2"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      },
      "TotalLevel3Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level3"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      }
    }
  },
  {
    $project: {
      _id: 0,
      userId: "$_id",
      TotalLevel1Amount: 1,
      TotalLevel2Amount: 1,
      TotalLevel3Amount: 1
    }
  },
  {
    $sort: {
      userId: 1
    }
  }
])

Sample Mongo Playground

至键值对：{ 'userId': { // Result } }

步骤 1 到 3 与之前的解决方案相同。

$sort（可选）：按_id升序排序。
$project：显示带有array 字段的文档（带有属性k 和v）。
$replaceRoot：将整个文档替换为键（userId）和值（结果）。

db.collection.aggregate([
  {
    $set: {
      _levelDetails: {
        $objectToArray: "$LevelDetails"
      }
    }
  },
  {
    $unwind: "$_levelDetails"
  },
  {
    $group: {
      _id: "$_levelDetails.v.userId",
      "TotalLevel1Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level1"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      },
      "TotalLevel2Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level2"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      },
      "TotalLevel3Amount": {
        $sum: {
          $cond: [
            {
              "$eq": [
                "$_levelDetails.k",
                "level3"
              ]
            },
            "$_levelDetails.v.amount",
            0
          ]
        }
      }
    }
  },
  {
    $sort: {
      _id: 1
    }
  },
  {
    $project: {
      array: [
        {
          k: {
            $toString: "$_id"
          },
          v: {
            TotalLevel1Amount: "$TotalLevel1Amount",
            TotalLevel2Amount: "$TotalLevel2Amount",
            TotalLevel3Amount: "$TotalLevel3Amount"
          }
        }
      ]
    }
  },
  {
    "$replaceRoot": {
      newRoot: {
        $arrayToObject: "$array"
      }
    }
  }
])

Sample Mongo Playground (To Key-Value Pair)

【讨论】：