早上好：
我有以下 MongoDB 結構：
{
"_id": 19,
"name": "Fredia Donan",
"faculty": "Zoomzone",
"lectures": [
{
"lecturerID": 25,
"name": "Sigismondo Brecknell",
"email": "[email protected]",
"imparts": [
{
"_id": 76,
"codCourse": 23,
"language": "Malay",
"max_students": 59,
"students": [
{
"studentID": 25
}
],
"classes": [
{
"date": ISODate("2022-02-02T09:23:32.59"),
"cod_classroom": 76
}
]
}
]
}
]
}
有了這個，我想找到給最多學生上課的講師。因此，我必須計算講師教給的學生人數，并從系內所有講師中找出最大值。我怎么能做到？可以在下面聚合的相同代碼中完成嗎？這超出了我對 MongoDB 的了解，如果有人能提供幫助，我將永遠感激不盡。
我已經閱讀了有關 $size 和 $count 的資訊，但是無論我在哪里嘗試它都會給我帶來不同的錯誤。
非常感謝您！
上面的輸出是由兩個收款部門和組的聚合完成的，代碼如下：
[{"$unwind": "$lecturers"}, {"$unwind": "$lecturers.imparts"},
{"$lookup":
{"from": "group",
"localField": "lecturers.imparts.groupID",
"foreignField": "_id",
"as": "lecturers.imparts"}},
{"$group":
{"_id": {"_id": "$_id", "lecturersID": "$lecturers.lecturerID"},
"name": {"$first": "$name"},
"faculty": {"$first": "$faculty"},
"lecturers":
{"$first":
{"lecturerID": "$lecturers.lecturerID",
"name": "$lecturers.name",
"email": "$lecturers.email"}},
"imparts": {"$push": "$lecturers.imparts"}}},
{"$set":
{"lecturers":
{"$mergeObjects":
["$lecturers",
{"imparts":
{"$reduce":
{"input": "$imparts",
"initialValue": [],
"in": {"$concatArrays": ["$$value", "$$this"]}}}}]},
"imparts": "$$REMOVE"}},
{"$group":
{"_id": "$_id._id",
"name": {"$first": "$name"},
"faculty": {"$first": "$faculty"},
"lectures": {"$push": "$lecturers"}}}])
我想要的輸出是讓每個教授最多學生的部門的講師。例如，如果我有這個：
{
"_id": 19,
"name": "Fredia Donan",
"faculty": "Zoomzone",
"lectures": [
{
"lecturerID": 25,
"name": "Sigismondo Brecknell",
"email": "[email protected]",
"imparts": [
{
"_id": 76,
"codCourse": 23,
"language": "Malay",
"max_students": 59,
"students": [
{
"studentID": 25
}
],
"classes": [
{
"date": ISODate("2022-02-09T23:32:59.000Z"),
"cod_classroom": 76
}
]
},
{
"_id": 77,
"codCourse": 24,
"language": "Malayff",
"max_students": 59,
"students": [
{
"studentID": 28
}
],
"classes": [
{
"date": ISODate("2022-02-09T23:32:59.000Z"),
"cod_classroom": 77
}
]
}
]
},
{
"lecturerID": 36,
"name": "Sigismondo Brecknell",
"email": "[email protected]",
"imparts": [
{
"_id": 76,
"codCourse": 23,
"language": "Malay",
"max_students": 59,
"students": [
{
"studentID": 45
}
],
"classes": [
{
"date": ISODate("2022-02-09T23:32:59.000Z"),
"cod_classroom": 76
}
]
},
{
"_id": 77,
"codCourse": 24,
"language": "Malayff",
"max_students": 59,
"students": [
{
"studentID": 54
},
{
"studentID": 435
},
{
"studentID": 45
}
],
"classes": [
{
"date": ISODate("2022-02-09T23:32:59.000Z"),
"cod_classroom": 77
}
]
}
]
}
]
}
]
我想為 _id 19 的部門得到這個，教最多學生的講師是 id 36 的講師，因為他給 4 名學生上課，而 id 為 25 的講師只給 2 名學生上課。
所以我想得到以下輸出：
"_id": 19,
"name": "Fredia Donan",
"lecturerID": 36,
"maxImpartsStudents": 4
}
我想獲得每個現有部門的這些資訊。收集部門的每一份檔案
uj5u.com熱心網友回復：
歡迎克里斯蒂娜·阿蘭達！
編輯：
- 對于每個
lecturer，迭代所有imparts使用$map并創建一個包含所有學生的集合，并將其大小回傳為maxImpartsStudents。$setUnion如果他們在同一講師的多個班級中，我們會避免多次計算同一學生。 $reduce在lecturers每個檔案中僅包含最大值為 的檔案maxImpartsStudents。- 格式
db.collection.aggregate([
{
$project: {
name: 1,
lectures: {
$map: {
input: "$lectures",
as: "item",
in: {
$mergeObjects: [
{
maxImpartsStudents: {
$size: {
$reduce: {
input: "$$item.imparts",
initialValue: [],
in: {$setUnion: ["$$value", "$$this.students"]}
}
}
}
},
{
name: "$$item.name",
lecturerID: "$$item.lecturerID"
}
]
}
}
}
}
},
{
$set: {
lectures: {
$reduce: {
input: "$lectures",
initialValue: {maxImpartsStudents: 0},
in: {
$cond: [
{$gte: ["$$this.maxImpartsStudents", "$$value.maxImpartsStudents"]},
"$$this", "$$value"]
}
}
}
}
},
{
$project: {
lecturerID: "$lectures.lecturerID",
maxImpartsStudents: "$lectures.maxImpartsStudents",
departmentName: "$name"
}
}
])
看看它在操場上的例子是如何作業的
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