如何在mongo db聚集中将数据作为数组获取

如何解决如何在mongo db聚集中将数据作为数组获取

假设这是我的数据模式

{
    "_id" : ObjectId("5f90ed2954d61feed1720603"),"date" : ISODate("2020-10-22T00:00:00Z"),"worker_full_name" : "JOHN DOE","worker_department" : "Worker Department","type" : "Worker","site_name" : "DL LIMITED","in_time" : ISODate("2020-10-22T07:53:35Z"),"attendance_points" : 2,"out_time" : ISODate("2020-10-22T22:03:41Z"),"duration" : 14,"attendance_count" : 2,"wage" : 580,"in_location" : "Countryside Avenue","worker_aadhar_card_number" : "xxxxxxxxxxxx","out_location" : "Golf Drive","worker_id" : ObjectId("5f12ea794fdb64e82ce68fac")
}
{
    "_id" : ObjectId("5f90ed2754d61feed17205fe"),"worker_full_name" : "JOHN DOE 2","in_time" : ISODate("2020-10-22T07:53:34Z"),"out_time" : ISODate("2020-10-22T22:24:02Z"),"wage" : 0,"out_location" : "Countryside Avenue","worker_id" : ObjectId("5f688cf2df29927bfb8531eb")
}

我正在执行以下汇总操作：

my_collection.aggregate([{'$match': {'date': {"$gte": start_date,"$lte": end_date},'worker_full_name': {"$exists": 'true'},"site_name": site_name
                                                   }},{"$group": {'_id': {
                                           'worker_id': '$worker_id','worker_full_name': '$worker_full_name'
                                       },'present_days': {'$sum': 1},'total_shift_points': {'$sum': '$attendance_points'}
                                       }}
])

这样做，我可以实现以下输出：

{'_id': {'worker_id': ObjectId('5f688cf2df29927bfb8531d6'),'worker_full_name': 'JOHN DOE'},'present_days': 22,'total_shift_points': 38.25}
{'_id': {'worker_id': ObjectId('5f66130f94c75522f314dbc0'),'worker_full_name': 'JOHN DOE 2'},'present_days': 19,'total_shift_points': 35.25}
{'_id': {'worker_id': ObjectId('5f66130e94c75522f314db99'),'worker_full_name': 'JOHN DOE 3'},'present_days': 23,'total_shift_points': 42.75}
{'_id': {'worker_id': ObjectId('5f27b678749921225e5df98c'),'worker_full_name': 'JOHN DOE 4 '},'total_shift_points': 38.25}
{'_id': {'worker_id': ObjectId('5f6f2ac0b112533f081c3bae'),'worker_full_name': 'JOHN DOE 5'},'present_days': 21,'total_shift_points': 36.75}

但是有什么办法可以使数组中的所有点像下面的期望输出一样，其中我通过单个查询返回每日Attenance_points数组及其对应的日期：

 {'_id': {'worker_id': ObjectId('5f688cf2df29927bfb8531d6'),'total_shift_points': 38.25,'daily_points_stats':[
                {ISODate("2020-10-01T00:00:00Z"):2},{ISODate("2020-10-02T00:00:00Z"):1.5}
                {ISODate("2020-10-03T00:00:00Z"):1}
                {ISODate("2020-10-04T00:00:00Z"):1.25}
                ..
                ..for all the days

 ]}

或者类似这样的东西：

{'_id': {'worker_id': ObjectId('5f688cf2df29927bfb8531d6'),'daily_points_stats':[
                    {"date":ISODate("2020-10-01T00:00:00Z"),"points":2},{"date":ISODate("2020-10-02T00:00:00Z"),"points":1.5},{"date":ISODate("2020-10-03T00:00:00Z"),"points":1},{"date":ISODate("2020-10-04T00:00:00Z"),"points":1.25},..
                    ..for all the days

     ]}

解决方法

您可以使用$group在$push内创建一个数组，

  {
    "$group": {
      "_id": {
        "worker_id": "$worker_id","worker_full_name": "$worker_full_name"
      },// like this
      daily_points_stats: {
        $push: {
          date: "$date",points: "$attendance_count"
        }
      },"present_days": { "$sum": 1 },"total_shift_points": { "$sum": "$attendance_points" }
    }
  }

Playground

---

如果您想要一个对象，则可以尝试

• $ push k（key）和v（value），将键转换为字符串类型，因为其日期类型使用$toString
• $addFields使用$arrayToObject
将数组转换为对象

  {
    "$group": {
      "_id": {
        "worker_id": "$worker_id",daily_points_stats: {
        $push: {
          k: { $toString: "$date" },v: "$attendance_count"
        }
      },"total_shift_points": { "$sum": "$attendance_points" }
    }
  },{
    $addFields: {
      daily_points_stats: {
        $arrayToObject: "$daily_points_stats"
      }
    }
  }

Playground