Mongodb 聚合 -unwind 然后分组（查找唯一计数）

Question

我有以这种格式存储的数据。我想在日期找到唯一的 msisdn。即，依靠日期

上的唯一 msisdn

{
    "_id" : ObjectId("5e551b2e63a0f1fc55f026dd"),
    "dateHour" : "2020-02-25 13",
    "detail" : [
        {
            "campaignId" : "5e4a937b459418737cc1dfa5",
            "timestamp" : "2020-02-25 13:03:626",
            "msisdn" : "050920305373677038659253736080036024-1582133696781",
            
        },
        {
            "campaignId" : "5e4a937b459418737cc1dfa5",
            "timestamp" : "2020-02-2025 13:04:626",
            "msisdn" : "050920305373677038659253736080036024-1582133696781",
            
        },
        {
            "campaignId" : "5e4a937b459418737cc1dfa5",
            "timestamp" : "2020-02-25 13:06:626",
            "msisdn" : "050920305373677038659253-1582133696781",
            
        }
        
    ]
}

Answer 1

您可以执行以下操作

[
  {
    $unwind: "$detail"
  },
  {
    $addFields: {
      "dHr": {
        $substr: [
          "$dateHour",
          0,
          10
        ]
      }
    }
  },
  {
    $group: {
      _id: "$dHr",
      msisdn: {
        $addToSet: "$detail.msisdn"
      }
    }
  },
  {
    $project: {
      _id: 0,
      date: "$_id",
      count: {
        $size: "$msisdn"
      }
    }
  }
]

工作Mongo playground