MongoDB Aggregation:双重查找，并将查找响应合并到各个对象 [英] MongoDB Aggregation : Double lookup, and merge lookup response to respective object

问题描述

我正在尝试进行汇总，但是找不到合适的管道来完成它.

I'm trying an aggregation but I can't find the right pipeline to do it.

所以，这是我的文档模型的一部分:

So, this is a part of my document model :

//company.js
{
    "_id" : "5dg8aa8c435b1e2868c841f6",
    "name" : "My Corp",
    "externalId" : "d7f348c9-c69b-69c4-923c-91458c53dc22",
    "professionals_customers" : [ 
        {
            "company" : "6f4d01eb3b948150c2aad9c0"
        }, 
        {
            "company" : "5dg7aa8c366b1e2868c841f6",
            "contact" : "5df8ab5c355b1e2999c841f7"
        }
    ],
}

我试图返回专业客户领域，其中充满了数据，就像传统的填充一样.

I try to return the professionnal customers fields hydrated with data, like a classic populate would do.

公司字段来自公司集合，联系方式由用户集合提供

Company field came from the company collection and contact is provided by the user collection

所需的输出必须类似于:

The desired output must look like :

{
"professionals_customers" : [ 
        {
            "company": {
                "_id": "6f4d01eb3b948150c2aad9c0",
                "name": "Transtar",
                "externalId": "d7f386c9-c79b-49c5-905c-90750c42dc22",
            },
        }, 
        {
            "company": {
                "_id": "5dg7aa8c366b1e2868c841f6",
                "name": "Aperture",
                "externalId": "d7f386c9-c69b-49c4-905c-90750c53dc22", 
            },
            "contact" : {
                "_id": "5df8ab5c355b1e2999c841f7",
                "firstname": "Caroline",
                "lastname": "Glados",
                "externalId": "d7f386c9-c69b-49c4-905c-90750c53dc22", //same externalId as above, the user belongs to the company
            },
        }
    ]
}

目前，我已经尝试了多种解决方案，但仍达不到目标.

At this point I've tried multiple solutions but I can't reach my goal.

let query = [{
        $match : { _id : companyId }
    },{
        $lookup : {
            from: 'companies',
            localField : 'professionals_customers.company',
            foreignField : '_id',
            as : 'professionalsCustomers'
        }
    },{
        $lookup : {
            from: 'users',
            localField : 'professionals_customers.contact',
            foreignField : '_id',
            as : 'contacts'
        }
    }]

在这一点上，我已经有了两个包含所有必需信息的新阵列，但是我不知道如何将正确的联系人分组到正确的公司.另外，也许尝试使用$ lookup填充数据以保留初始结构要比尝试通过共享的externalId重新组合professionalCustomers和联系人要容易.

At this, point I' ve got two new arrays with all the needed informations, but I don't know how to get the right contact grouped with the right company. Also, maybe it's easier to try to populate the data (with $lookup) keeping the initial struct than trying to regroup professionalCustomers and contacts through the shared externalId.

其他信息:

-属于公司的用户具有相同的externalId.

-An user that belongs to a company has the same externalId.

-我不想使用经典的填充，在那之后，我需要执行其他一些操作

-I don't want to use a classical populate, after that, I need to do some other operations

推荐答案

尝试以下查询:

db.companies.aggregate([
    { $match: { _id: companyId } },
    { $unwind: "$professionals_customers" },
    {
        $lookup: {
            from: "companies",
            localField: "professionals_customers.company",
            foreignField: "_id",
            as: "professionals_customers.company"
        }
    },
    {
        $lookup: {
            from: "users",
            localField: "professionals_customers.contact",
            foreignField: "_id",
            as: "professionals_customers.contact"
        }
    },
    {
        $addFields: {
            "professionals_customers.company": {
                $arrayElemAt: ["$professionals_customers.company", 0]
            },
            "professionals_customers.contact": {
                $arrayElemAt: ["$professionals_customers.contact", 0]
            }
        }
    },
    {
        $group: { _id: "$_id", professionals_customers: { $push: "$professionals_customers" }, data: { $first: "$$ROOT" } }
    },
    { $addFields: { "data.professionals_customers": "$professionals_customers" } },
    { $replaceRoot: { newRoot: "$data" } }
])

测试: MongoDB-Playground

注意::如果需要，您需要将类型为 string 的字段/输入转换为 ObjectId().基本的事情是您需要检查两个要比较的字段的类型，或者是否要与数据库中的输入字段匹配.

Note : If needed you need to convert fields/input which is of type string to ObjectId(). Basic thing is you need to check types of two fields being compared or input-to-field-in-DB matches or not.

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