Mongodb连接多组值

Question

你好，我开始学习mongoDB查询

我对聚合有一些理解上的问题

例如，我有以下文档：

           [
            {
              totalTaxInclusive: 15,
              totalTaxExclusive: 12.5,
              payments:[{
                 method: "CB",
                 amount: 10
              },
              {
                 method: "CASH",
                 amount: 5
              }
            ],
           },
           {
              totalTaxInclusive: 40,
              totalTaxExclusive: 33.33,
              payments:[{
                 method: "CB",
                 amount: 40
              }
            ],
           },
          ]

等等。

我如何提出请求，我将拥有谁：

        $group: {
          _id: "$payments.method",
          amount: { $sum: "$payments.amount"},
        }

和a

$group: {
       _id: null,
       totalCount: { $sum: 1 },
       totalTaxInclusive: { $sum: "$totalTaxInclusive"},
       totalTaxExclusive: { $sum:"$totalTaxExclusive" },
     }

产生如下结果：

{
totalCount: 2,
totalTaxInclusive: 55,
totalTaxExclusive: 45.83,
payments: [{
     method: "CASH",
     amount: 5,
    },
    {
     method: "CB",
     amount: 50,
    }
  ]
}

非常感谢你的帮助。

Answer 1

• 第一阶段对所有文档进行分组并将其数据存储在相关数组中。还计算此阶段中的所有文档。
• 第二阶段，对总额中的所有项目进行求和。独占和总额包含数组
• 展开支付的第三阶段和第四阶段，因为它是数组的数组
• 第四阶段按payment_method分组，并按每个payment_method
获得总和
• 以请求格式输出结果的第五阶段

db.collection.aggregate([
  {
    "$group": {
      "_id": null,
      "totalTaxInclusive": {
        "$addToSet": "$totalTaxInclusive"
      },
      "totalTaxExclusive": {
        "$addToSet": "$totalTaxExclusive"
      },
      "payments": {
        "$addToSet": "$payments"
      },
      "count": {
        "$sum": 1
      }
    }
  },
  {
    "$set": {
      "totalTaxExclusive": {
        "$sum": "$totalTaxExclusive"
      },
      "totalTaxInclusive": {
        "$sum": "$totalTaxInclusive"
      }
    }
  },
  {
    "$unwind": "$payments"
  },
  {
    "$unwind": "$payments"
  },
  {
    "$group": {
      "_id": "$payments.method",
      "amount": {
        "$sum": "$payments.amount"
      },
      "totalTaxInclusive": {
        "$first": "$totalTaxInclusive"
      },
      "totalTaxExclusive": {
        "$first": "$totalTaxExclusive"
      },
      "count": {
        "$first": "$count"
      }
    }
  },
  {
    "$group": {
      "_id": null,
      "payments": {
        "$addToSet": {
          "method": "$_id",
          "amount": "$amount"
        }
      },
      "totalTaxInclusive": {
        "$first": "$totalTaxInclusive"
      },
      "totalTaxExclusive": {
        "$first": "$totalTaxExclusive"
      },
      "count": {
        "$first": "$count"
      }
    }
  }
])

工作示例

Answer 2

• $groupby null并获取所需字段的总和，构造支付数组
• $concatArray连接数组
• $减少以将支付的嵌套数组包含到数组
• $展开解构支付数组
• $groupbymethod并获取金额的总和并获取所需字段的第一个值
• $groupby null并构造支付数组并获取计数字段第一个值

db.collection.aggregate([
  {
    $group: {
      _id: null,
      totalTaxInclusive: { $sum: "$totalTaxInclusive" },
      totalTaxExclusive: { $sum: "$totalTaxExclusive" },
      totalCount: { $sum: 1 },
      payments: { $push: "$payments" }
    }
  },
  {
    $addFields: {
      payments: {
        $reduce: {
          input: "$payments",
          initialValue: [],
          in: { $concatArrays: ["$$this", "$$value"] }
        }
      }
    }
  },
  { $unwind: "$payments" },
  {
    $group: {
      _id: "$payments.method",
      amount: { $sum: "$payments.amount" },
      totalTaxInclusive: { $first: "$totalTaxInclusive" },
      totalTaxExclusive: { $first: "$totalTaxExclusive" },
      totalCount: { $first: "$totalCount" }
    }
  },
  {
    $group: {
      _id: null,
      totalTaxInclusive: { $first: "$totalTaxInclusive" },
      totalTaxExclusive: { $first: "$totalTaxExclusive" },
      totalCount: { $first: "$totalCount" },
      payments: {
        $push: {
          method: "$_id",
          amount: "$amount"
        }
      }
    }
  }
])

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