MongoDB Mongoose 聚合查询深度嵌套数组删除空结果并填充引用

Question

这个问题是上一个问题的后续，我已经接受了答案。我有一个聚合查询，它根据日期范围返回深度嵌套的子文档数组的结果。查询返回指定日期范围内的正确结果，但对于与查询不匹配的结果，查询也会返回一个空数组。

技术： MongoDB 3.6， Mongoose 5.5， NodeJS 12

问题 1：有没有办法删除与查询不匹配的结果？

问题 2：有没有办法在结果中"填充"人员数据库引用？例如，要获取人员显示名称，我通常使用"填充"，例如find().populate({ path: 'Person', select: 'DisplayName'})

记录架构

let RecordsSchema = new Schema({
RecordID: {
type: Number,
index: true
},
RecordType: {
type: String
},
Status: {
type: String
},
// ItemReport array of subdocuments
ItemReport: [ItemReportSchema],
}, {
collection: 'records',
selectPopulatedPaths: false
});
let ItemReportSchema = new Schema({
// ObjectId reference
ReportBy: {
type: Schema.Types.ObjectId,
ref: 'people'
},
ReportDate: {
type: Date,
required: true
},
WorkDoneBy: [{
Person: {
type: Schema.Types.ObjectId,
ref: 'people'
},
CompletedHours: {
type: Number,
required: true
},
DateCompleted: {
type: Date
}
}],
});

查询

有效，但也返回空结果，还需要填充 Person db 引用的"显示名称"属性

db.records.aggregate([
{
"$project": {
"ItemReport": {
$map: {
input: "$ItemReport",
as: "ir",
in: {
WorkDoneBy: {
$filter: {
input: "$$ir.WorkDoneBy",
as: "value",
cond: {
"$and": [
{ "$ne": [ "$$value.DateCompleted", null ] },
{ "$gt": [ "$$value.DateCompleted", new Date("2017-01-01T12:00:00.000Z") ] },
{ "$lt": [ "$$value.DateCompleted", new Date("2018-12-31T12:00:00.000Z") ] }
]
}
}
}
}
}
}
}
}
])

实际结果

{
"_id": "5dcb6406e63830b7aa5427ca",
"ItemReport": [
{
"WorkDoneBy": [
{
"_id": "5dcb6406e63830b7aa53d8ea",
"PersonID": 111,
"ReportID": 8855,
"CompletedHours": 3,
"DateCompleted": "2017-01-20T05:00:00.000Z",
"Person": "5dcb6409e63830b7aa54fdba"
}
]
}
]
},
{
"_id": "5dcb6406e63830b7aa5427f1",
"ItemReport": [
{
"WorkDoneBy": [
{
"_id": "5dcb6406e63830b7aa53dcdc",
"PersonID": 4,
"ReportID": 9673,
"CompletedHours": 17,
"DateCompleted": "2017-05-18T04:00:00.000Z",
"Person": "5dcb6409e63830b7aa54fd69"
},
{
"_id": "5dcb6406e63830b7aa53dcdd",
"PersonID": 320,
"ReportID": 9673,
"CompletedHours": 3,
"DateCompleted": "2017-05-18T04:00:00.000Z",
"Person": "5dcb6409e63830b7aa54fe88"
}
]
}
]
},
{
"_id": "5dcb6406e63830b7aa5427f2",
"ItemReport": [
{
"WorkDoneBy": []
}
]
},
{
"_id": "5dcb6406e63830b7aa5427f3",
"ItemReport": [
{
"WorkDoneBy": []
}
]
},
{
"_id": "5dcb6406e63830b7aa5427f4",
"ItemReport": [
{
"WorkDoneBy": []
}
]
},
{
"_id": "5dcb6406e63830b7aa5427f5",
"ItemReport": [
{
"WorkDoneBy": []
}
]
},

期望的结果

请注意，将删除具有空"WorkDoneBy"数组的结果(问题 1(，并填充"人员"显示名称(问题 2(。

{
"_id": "5dcb6406e63830b7aa5427f1",
"ItemReport": [
{
"WorkDoneBy": [
{
"_id": "5dcb6406e63830b7aa53dcdc",
"CompletedHours": 17,
"DateCompleted": "2017-05-18T04:00:00.000Z",
"Person": {
_id: "5dcb6409e63830b7aa54fe88",
DisplayName: "Joe Jones"
}
},
{
"_id": "5dcb6406e63830b7aa53dcdd",
"CompletedHours": 3,
"DateCompleted": "2017-05-18T04:00:00.000Z",
"Person": {
_id: "5dcb6409e63830b7aa54fe88",
DisplayName: "Alice Smith"
}
}
]
}
]
},

Answer 1

第一个问题相对容易回答，有多种方法可以做到这一点。我更喜欢将$anyElementTrue与$map一起使用，因为这些运算符是不言自明的。

{
"$match": {
$expr: { $anyElementTrue: { $map: { input: "$ItemReport", in: { $gt: [ { $size: "$$this.WorkDoneBy" }, 0 ] } } } }
}
}

蒙戈游乐场

第二部分有点复杂，但仍然有可能。您需要运行$lookup而不是填充以从其他集合中提取数据。问题是您的Person值是深度嵌套的，因此您需要在使用$reduce和$setUnion之前准备id值的列表。获取数据后，需要使用$map和$mergeObjects将嵌套对象与人员实体合并。

{
$addFields: {
people: {
$reduce: {
input: "$ItemReport",
initialValue: [],
in: { $setUnion: [ "$$value", "$$this.WorkDoneBy.Person" ] }
}
}
}
},
{
$lookup: {
from: "people",
localField: "peopleIds",
foreignField: "_id",
as: "people"
}
},
{
$project: {
_id: 1,
ItemReport: {
$map: {
input: "$ItemReport",
as: "ir",
in: {
WorkDoneBy: {
$map: {
input: "$$ir.WorkDoneBy",
as: "wdb",
in: {
$mergeObjects: [
"$$wdb",
{
Person: { $arrayElemAt: [{ $filter: { input: "$people", cond: { $eq: [ "$$this._id", "$$wdb.Person" ] } } } , 0] }
}
]
}
}
}
}
}
}
}
}

完整解决方案