我想和你分享我的问题。
因此，我有3个需要在查询中访问的实体：
评价：
[
{
_id: 1,
questionary: 1,
subject: 1
},
{
_id: 1,
questionary: 1,
subject: 2
},
]
用户
[
{
_id: 1
name: "John Doe",
photo: "photo1.jpg"
},
{
_id: 2
name: "Paul Smith",
photo: "photo2.jpg"
},
]
疑问的
[
{
_id: 1,
title: "questionary 1",
"date": "2020-02-08T00:00:00.000Z"
},
{
_id: 2,
title: "questionary 2",
"date": "2020-02-09T00:00:00.000Z"
}
]
所以我的目标是获取这样的数据：其中的问题清单，与问题相关的评估清单以及评估中我需要一个用户对象。像这样：
[
{
"_id": "1",
"title": "questionary 1",
"evaluations": [
{
"_id": "1",
"date": "2020-04-05T18:53:46.948Z"
"user": {
_id: 1,
"name": "John Doe",
"photo": "photo1.jpg"
}
},
{
"_id": "2",
"date": "2020-04-06T18:53:46.948Z",
"user": {
_id: 1,
"name": "John Doe",
"photo": "photo1.jpg"
}
}
]
}
]
我的查询是：
return await Questionary.aggregate([{
$lookup: {
from: "evaluation",
localField: "_id",
foreignField: "questionary",
as: "evaluations",
}
},
{
$lookup: {
from: "user",
localField: "evaluations.user",
foreignField: "_id",
as: "user",
}
},
{
$project: {
_id: 1,
title: 1,
status: 1,
evaluations: {
_id: 1,
date: 1,
user: "$user"
}
},
},
]);
我的结果是：
[
{
"_id": "1",
"title": "questionary 1",
"evaluations": [
{
"_id": "1",
"date": "2020-04-05T18:53:46.948Z"
"user": [
{
_id: 1,
"name": "John Doe",
"photo": "photo1.jpg"
},
{
_id: 2,
"name": "Paul Smith",
"photo": "photo2.jpg"
}
]
},
{
"_id": "2",
"date": "2020-04-06T18:53:46.948Z",
"user": [
{
_id: 1,
"name": "John Doe",
"photo": "photo1.jpg"
},
{
_id: 2,
"name": "Paul Smith",
"photo": "photo2.jpg"
}
]
}
]
}
]
我的评估用户正在合并，但这不是我想要的，我只想评估内部用户的内部信息。
有什么建议吗？
您应该能够$lookup通过使用不相关的子查询来使用嵌套，如https://docs.mongodb.com/manual/reference/operator/aggregation/lookup/#join-conditions-and-uncorrelated-sub-queries所述
您的情况是：
return await Questionary.aggregate([{
$lookup: {
from: "evaluation",
let: {questId: "$_id"},
pipeline: [{
$match: {
$expr: {
$eq: ["$$questId", "$questionary"]
}
},
}, {
$lookup: {
from: "user",
localField: "subject",
foreignField: "_id",
as: "user",
}
}],
as: "evaluations"
}
}
]);
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