如何在mongodb聚合中获取数据作为数组
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【中文标题】如何在mongodb聚合中获取数据作为数组【英文标题】：How to get data as a array in mongo db aggregration 【发布时间】：2021-02-08 10:49:57 【问题描述】：假设这是我的数据架构
"_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"),
"date" : ISODate("2020-10-22T00:00:00Z"),
"worker_full_name" : "JOHN DOE 2",
"worker_department" : "Worker Department",
"type" : "Worker",
"site_name" : "DL LIMITED",
"in_time" : ISODate("2020-10-22T07:53:34Z"),
"attendance_points" : 2,
"out_time" : ISODate("2020-10-22T22:24:02Z"),
"duration" : 14,
"attendance_count" : 2,
"wage" : 0,
"in_location" : "Countryside Avenue",
"worker_aadhar_card_number" : "xxxxxxxxxxxx",
"out_location" : "Countryside Avenue",
"worker_id" : ObjectId("5f688cf2df29927bfb8531eb")
我正在做以下聚合：
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'
])
通过这样做，我能够实现以下输出：
'_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 ', 'present_days': 22, 'total_shift_points': 38.25
'_id': 'worker_id': ObjectId('5f6f2ac0b112533f081c3bae'), 'worker_full_name': 'JOHN DOE 5', 'present_days': 21, 'total_shift_points': 36.75
但是有什么方法可以让我得到一个数组中的所有点，就像下面这个期望的输出一样，我在其中单个查询返回一个每日出席点数及其相应日期的数组：
'_id': 'worker_id': ObjectId('5f688cf2df29927bfb8531d6'),
'worker_full_name': 'JOHN DOE',
'present_days': 22,
'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
]
或者是这样的：
'_id': 'worker_id': ObjectId('5f688cf2df29927bfb8531d6'),
'worker_full_name': 'JOHN DOE',
'present_days': 22,
'total_shift_points': 38.25 ,
'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
]
【问题讨论】：
【参考方案1】：您可以使用$push 在$group 中创建一个数组，
"$group":
"_id":
"worker_id": "$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
如果你想要一个对象，那么你可以尝试，
$push k(key) 和 v(value)，将 key 转换为字符串类型，因为它的日期类型使用$toString
$addFields 使用 $arrayToObject 将数组转换为对象
"$group":
"_id":
"worker_id": "$worker_id",
"worker_full_name": "$worker_full_name"
,
daily_points_stats:
$push:
k: $toString: "$date" ,
v: "$attendance_count"
,
"present_days": "$sum": 1 ,
"total_shift_points": "$sum": "$attendance_points"
,
$addFields:
daily_points_stats:
$arrayToObject: "$daily_points_stats"
Playground
【讨论】：
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