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Mongodb $ lookup不使用_id

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我们尝试使用此查询，返回查找为空

db.getCollection('tests').aggregate([
    {$match: {typet:'Req'}},
    {$project: {incharge:1}},
    {$lookup:{
            from: "users",
            localField: "incharge", //this is the _id user from tests
            foreignField: "_id", //this is the _id from users
            as: "user"
    }}
])

返回json

[
    {
        "_id": "57565d2e45bd27b012fc4db9",
        "incharge": "549e0bb67371ecc804ad23ef",
        "user": []
    },
    {
        "_id": "57565d2045bd27b012fc4cbb",
        "incharge": "549e0bb67371ecc804ad21ef",
        "user": []
    },
    {
        "_id": "57565d2245bd27b012fc4cc7",
        "incharge": "549e0bb67371ecc804ad24ef",
        "user": []
    }
]

我尝试使用这篇文章，但没有任何事情发生MongoDB aggregation project string to ObjectId并与此MongoDB $lookup with _id as a foreignField in PHP

UPDATE

这是文档“用户”

{
        "_id" : ObjectId("549e0bb67371ecc804ad24ef"),
        "displayname" : "Jhon S."
    },
    {
        "_id" : ObjectId("549e0bb67371ecc804ad21ef"),
        "displayname" : "George F."
    },
    {
        "_id" : ObjectId("549e0bb67371ecc804ad23ef"),
        "displayname" : "Franc D."
    }
node.js mongodb mongoose

3 回答

  • 9

    我最终找到了解决方案，我的Schema在使用ObjectId的mongoose中出现了问题

    我改变了

    var Schema = new Schema({
    name: { type: String, required: true},
    incharge: { type: String, required: true},
});

    有了这个

    var Schema = new Schema({
    name: { type: String, required: true},
    incharge: { type: mongoose.Schema.ObjectId, required: true},
});

    并且正在工作

    回复于
  • 0

    您的查询查询是完美的，但问题是您将 incharge 作为字符串存储到数据库中，而_id：ObjectId（'theID'）是一个对象而不仅仅是字符串，您无法将 string （' '）与 object （{}进行比较） . 因此，最好的方法是将 incharge 键存储为对象（ mongoose.Schema.ObjectId ）而不是架构中的字符串 .

    回复于
  • 0

    您只需使用 "_id.str" 即可完成工作 .

    db.getCollection('tests').aggregate([
{$match: {typet:'Req'}},
{$project: {incharge:1}},
{$lookup:{
        from: "users",
        localField: "incharge", //this is the _id user from tests
        foreignField: "_id.str", //this is the _id from users
        as: "user"
}}

    ]）

    对我来说很好 .

    回复于

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