我想使用$ lookup在mongoDB中加入3个不同的集合,并用ref的核心响应数据(即客户端集合)填充ref数组。
对话:
$Date = "TEXT"
Write-Host "The script was started $Date"
$fileOut = Join-Path -Path $tsvEingang -ChildPath ("$Date.csv")
$MosaicSummary | Export-Csv -Path $fileOut -Delimiter "`t" -NoTypeInformation
#Result: The writtens csv-file-name is TEXT.csv
消息:
{
_id: mongoose.Schema.Types.ObjectId,
participants: [{ type: mongoose.Schema.Types.ObjectId, ref: "client" }],
created_at: { type: Date, default: Date.now },
}
客户
:{
_id: mongoose.Schema.Types.ObjectId,
conversation_id: { type: mongoose.Schema.Types.ObjectId, ref: "Conversation",
message: { type: String, default: "" },
sender: {type: mongoose.Schema.Types.ObjectId,ref: "Client",default: null},
reciever: {type: mongoose.Schema.Types.ObjectId,ref: "Client",default: null,},
created_at: { type: Date, default: Date.now }
}
这是我现在没有人口的地方:
{
_id: mongoose.Schema.Types.ObjectId,
email: {type: String, required: true, unique: true},
name: { type: String, default: "" },
job: { type: String, default: "" },
company: { type: String, default: "" },
school: { type: String, default: "" },
}
这是我得到的结果:
Conversation.aggregate([
{ $match: { participants: { $all: [mongoose.Types.ObjectId(myId)] } } },
{
$lookup: {
from: "messages",
localField: "_id",
foreignField: "conversation_id",
as: "messages",
},
},
{
$unwind: { path: "$messages", preserveNullAndEmptyArrays: true },
},
{ $sort: { "messages.created_at": -1 } },
{
$group: {
_id: "$_id",
messages: { $first: "$messages" },
doc: { $first: "$$ROOT" },
},
},
{ $replaceRoot: { newRoot: "$doc" } },
])
当我将对话与消息进行汇总时,我想populate个参与者数组。
谢谢。
答案 0 :(得分:1)
使用aggregate(),
在这里您可以使用lookup with pipeline,它可以添加所有管道查找级别,
不需要$ unwind和$ group
Conversation.aggregate([
{
$match: {
participants: {
$all: [
mongoose.Types.ObjectId(myId)
]
}
}
},
{
$lookup: {
from: "Message",
let: {
conversation_id: "$_id"
},
pipeline: [
{
$match: {
$expr: {
$eq: [
"$$conversation_id",
"$conversation_id"
]
}
}
},
{
$sort: {
"created_at": -1
}
}
],
as: "messages"
}
},
{
$lookup: {
from: "Client",
let: {
participants: "$participants"
},
pipeline: [
{
$match: {
$expr: {
$in: [
"$_id",
"$$participants"
]
}
}
},
{
$sort: {
"created_at": -1
}
}
],
as: "participants"
}
}
])
游乐场:https://mongoplayground.net/p/Pz5ojJ_dGGY
注意:我已经假定了集合的名称,以便您可以根据需要进行更正和更改。
答案 1 :(得分:0)
这是我得到的代码,我将id更改为仅字符串:
Conversation.aggregate([
{ $match: { participants: { $all: [mongoose.Types.ObjectId(myId)] } } },
{
$lookup: {
from: "messages",
let: { conversationId: { $toObjectId: "$_id" }},
pipeline: [
{
$match: { $expr: {$eq: ["$$conversationId", "$conversation_id"]}},
},
{
$sort: { create_at: -1 },
},
{ $limit: 1 },
],
as: "messages",
},
},
{
$lookup: {
from: "clients",
let: {participants: "$participants"},
pipeline: [
{
$match: {
$expr: {
$and: [
{ $in: ["$_id", "$$participants"] },
{ $ne: ["$_id", mongoose.Types.ObjectId(myId)] },
]},
},
},
{
$sort: { created_at: -1 },
},
],
as: "participants",
},
},
{
$project: {
_id: 1,
messages: 1,
"participants.name": 1,
"participants.company": 1,
"participants.school": 1,
},
},
])