報告表示例資料
{
"_id" : ObjectId("614415f4a6566a001623b622"),
"record" : [
{
"dateTime" : ISODate("2021-09-17T04:13:39.465Z"),
"status" : "time-in",
"month" : 9,
"day" : 17,
"year" : 2021,
"time" : 1631852019465.0,
"date" : ISODate("2021-09-17T00:00:00.000Z"),
},
{
"dateTime" : ISODate("2021-09-17T04:14:01.182Z"),
"status" : "time-out",
"month" : 9,
"day" : 17,
"year" : 2021,
"time" : 1631852041182.0,
"date" : ISODate("2021-09-17T00:00:00.000Z"),
}
],
"uid" : ObjectId("614415b0a6566a001623b80b"),
"date" : ISODate("2021-09-17T00:00:00.000Z"),
"status" : "time-out",
"createdAt" : ISODate("2021-09-17T04:13:40.102Z"),
"updatedAt" : ISODate("2021-09-17T04:14:01.831Z"),
"__v" : 0
}
用戶表樣本資料
{
"_id" : ObjectId("615c0f6db30aff375cd05ac1"),
"displayName" : "test test",
"firstName" : "test",
"lastName" : "test",
"email" : "[email protected]",
"brand" : "Jollibee",
"phone" : " 632312312312312",
"role" : 1,
"isVerified" : true,
"isArchived" : false,
"createdAt" : ISODate("2021-10-05T08:40:13.208Z"),
"updatedAt" : ISODate("2021-10-05T08:40:13.208Z"),
"__v" : 0
}
我有這樣的資料
db.getCollection('users').aggregate([
{
"$match": { brand: "Jollibee" }
},
{
$lookup: {
from: "orders",
let: { id: 'id' },
pipeline: [
{
$match: {
date: { $gte: ISODate("2020-11-01"), $lt: ISODate("2021-11-31") },
}
}
],
as: "orders",
},
},
{
$project: {
"_id": 1,
"name": 1,
"orders": 1
}
}
])
當我使用這個聚合時，我會為每個用戶插入所有資料。
我想要發生的事情就是這樣。我只會得到屬于用戶的資料，而不是所有用戶的所有資料。
為每個集合添加了示例檔案
uj5u.com熱心網友回復：
您不是在比較兩個集合的 userId。你應該在你的 $match 上添加它。操場
db.users.aggregate([
{
"$match": {
brand: "Jollibee"
}
},
{
$lookup: {
from: "orders",
let: {
id: "$_id"
},
pipeline: [
{
$match: {
date: {
$gte: ISODate("2020-11-01"),
$lt: ISODate("2021-11-30")
},
$expr: {
$eq: [
"$uid",
"$$id"
]
}
}
}
],
as: "orders",
},
},
{
$project: {
"_id": 1,
"name": 1,
"orders": 1
}
}
])
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