mongodb `$lookup` 或 `join` 与对象数组内的属性

Posted 2023-02-16

【中文标题】mongodb `$lookup` 或 `join` 与对象数组内的属性

【英文标题】：mongodb `$lookup` or `join` with attributes inside array of objects

【发布时间】：2020-08-18 08:19:34

【问题描述】：

我有这个来自 mongodb 的对象

[
    
        "_id": "5eaf2fc88fcee1a21ea0d94d",
        "migration_customer_union_id": 517,
        "__v": 0,
        "account": 1,
        "createdAt": "2020-05-03T20:55:36.335Z",
        "customerUnion": "5eaf2fc7698de8321ccd841d",
        "shaufel_customers": [
            
                "percent": 50,
                "_id": "5eaf2fc8698de8321ccd881f",
                "customer": "5eaf2fb9698de8321ccd68c0"
            ,
            
                "percent": 50,
                "_id": "5eaf2fc9698de8321ccd8a9d",
                "customer": "5eaf2fb9698de8321ccd68c0"
            
        ],
    
]

您可以注意到在 shaufel_customers 数组中有一个名为 customer 的属性，我想用它来加入客户文档，所以这就是我正在做的事情（在 *** 的帮助下编写了这段代码:)）

const aggregate = await CustomerUnionCustomer.aggregate(
        [
            
                $match: migration_customer_union_id: 517
            ,
            
                $lookup: 
                    from: 'customers',
                    localField: 'shaufel_customers.customer',
                    foreignField: '_id',
                    as: 'customers',
                
            ,
            
                $project: 
                    shaufel_customer_union_id: 1,
                    customerUnion: '$customerUnions',
                    shaufel_customers: 
                        $map: 
                            input: "$customers",
                            as: "c",
                            in: 
                                $mergeObjects: [
                                    "$$c",
                                    
                                        $arrayElemAt: [
                                            $filter: 
                                                input: "$shaufel_customers",
                                                cond: $eq: ["$$this.customer", "$$c._id"]
                                            
                                        , 0]
                                    ,

                                ]
                            
                        ,

                    
                
            ,
            
                "$project":  // this project just to get some specific values inside shaufel_customers
                    '_id': 0,

                    "shaufel_customers": 
                        "$map": 
                            "input": "$shaufel_customers",
                            "as": "customer",
                            "in": 
                                "customer_id": "$$customer.shaufel_customer_id",
                                "percent": "$$customer.percent"
                            
                        
                    
                
            

        ]
    )

执行此代码时，我收到以下响应

[
    
        "shaufel_customers": [
            
                "customer_id": "869",
                "percent": 50
            
        ]
    
]

你可以注意到我得到了一个对象，虽然上面的原始数组中有两个对象，这是因为上面的客户属性具有相同的 ObjectId 值5eaf2fb9698de8321ccd68c0，这就是我想问的。即使 id 相同，我也想获得相同的两个对象，所以我在这里期待的结果是

[
    
        "shaufel_customers": [
            
                "customer_id": "869",
                "percent": 50
            ,
            
                "customer_id": "869",
                "percent": 50
            ,
        ]
    
]

我该怎么做:(

【参考方案1】：

您需要恢复您的 $map 并迭代 shaufel_customers 而不是 customer - 这将返回两个结果：

    $project: 
        shaufel_customer_union_id: 1,
        customerUnion: '$customerUnions',
        shaufel_customers: 
            $map: 
                input: "$shaufel_customers",
                as: "sc",
                in: 
                    $mergeObjects: [
                        "$$c",
                        
                            $arrayElemAt: [
                                $filter: 
                                    input: "$customers",
                                    cond: $eq: ["$$this._id", "$$sc.customer"]
                                
                            , 0]
                        ,

                    ]
                
            ,

        
    
,

【讨论】：

成功了！谢谢。你又救了我，感谢你的帮助。