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MongoDB - 根据内部字段分组

来源:stackoverflow

时间:2024-03-05 21:45:26 306浏览 收藏

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问题内容

我用一个简单的例子来解释,看看我的 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。

查询结果应如下所示:

[

    { 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 } }
]

正确答案


  1. $set:通过将 leveldetails 转换为键值对来添加新字段 _leveldetails
  2. $unwind:解构 _leveldetails 数组。
  3. $group:根据级别有条件地按 _leveldetails.v.userid$sum 进行分组 (_leveldetails.k)。
  4. $project:格式化显示的文档。
  5. $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': { // 结果 } }

步骤 1 到 3 与之前的解决方案相同。

  1. $sort(可选):按 _id 升序排序。
  2. $project:显示带有 array 字段的文档(具有属性 kv)。
  3. $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)

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