MongoDB 聚合以获取统计信息答案

【问题标题】：MongoDB aggregate to get statsMongoDB 聚合以获取统计信息
【发布时间】：2023-01-12 19:42:21
【问题描述】：

我去年某个时候做过这个，但现在我真的不记得也找不到任何有用的资源。我想根据类型获取我的收藏的统计信息。

这是我的数据对象

{
  "_id": {
    "$oid": "63bfc374378c59a5328f229e"
  },
  "amountEarned": 11500,
  "amountPaid": 10350,
  "relianceCommission": 1150,
  "receiverType": "RESTAURANT",
  "__v": 0
}

我只需要每个receiverType的amountPaid之和，它可以是STORE、RESTAURANT或SHOPPER。然后我还需要所有的relianceCommission 的总和。导致形状像

{
 storeEarnings: 500,
 restaurantEarnings: 30,
 shopperEarnings: 40,
 totalRelianceCommission: 45
}

我试过了

aggregate([
  {
    $group: {_id: "$receiverType", total: {$sum: "amountPaid"}} 
}
])

然后加入另一个管道来计算totalRelianceCommission，但我觉得应该有一种更简洁的方法来做到这一点。我也不确定如何进行投影以产生所需的形状。请帮忙。

【问题讨论】：

标签： mongodb statistics aggregation-framework group

【解决方案1】：

query:
 {
    $group: {
      _id: "$receiverType",
      total: {
        $sum: "$amountPaid"
      },
      commissions: {
        $sum: "$relianceCommission"
      }
    }
  }
result:[
  {
    "_id": "STORE",
    "commissions": 1150,
    "total": 10350
  },
  {
    "_id": "RESTAURANT",
    "commissions": 2300,
    "total": 20700
  }
]

loop through the array to get a sum of commissions

【讨论】：

【解决方案2】：

你需要有条件的总和。

db.collection.aggregate([
  {
    $group: {
      _id: null,
      storeEarnings: {
        $sum: {
          $cond: [{$eq: ["$receiverType","STORE"]},"$amountPaid",0]
        }
      },
      restaurantEarnings: {
        $sum: {
          $cond: [{$eq: ["$receiverType","RESTAURANT"]},"$amountPaid",0]
        }
      },
      shopperEarnings: {
        $sum: {
          $cond: [{$eq: ["$receiverType","SHOPPER"]},"$amountPaid",0]
        }
      },
      totalRelianceCommission: {
        $sum: "$relianceCommission"
      }
    }
  }
])

Demo

【讨论】：