Question

我有一个mongo查询

db.memberships.aggregate([{
        '$match': {
            unit: new ObjectId('566fbaa1e63225b10bacac44')
        }
    },

    {
        '$lookup': {
            from: 'seasons',
            localField: 'season',
            foreignField: '_id',
            as: 'season'
        }
    },

    {
        '$unwind': '$season'
    },
    {
        '$project': {
            season: {
                'name': '$season.name',
                'id': '$season._id'
            }
        }
    },
    {
        '$group': {
            _id: '$_id',
            'season': {
                '$addToSet': '$season'
            }
        }
    },

]);

结果是

[
    {
        "_id": ObjectId("56a0e5a860d8a6e41eda2ea3"),
        "season": [{
            "name": "Summer",
            "id": ObjectId("56a0d6c692c26b8c019a2758")
        }]
    },
    {
        "_id": ObjectId("56a0e56d60d8a6e41eda2e9e"),
        "season": [{
            "name": "Summer",
            "id": ObjectId("56a0d6c692c26b8c019a2758")
        }]
    },
    {
        "_id": ObjectId("56a0e53860d8a6e41eda2e9a"),
        "season": [{
            "name": "Summer",
            "id": ObjectId("56a0d6c692c26b8c019a2758")
        }]
    },

    {
        "_id": ObjectId("56a0e4d660d8a6e41eda2e94"),
        "season": [{
            "name": "Winter",
            "id": ObjectId("5680dc01ba1e41f5066526fa")
        }]
    }
]

我需要以这种格式获得结果

{
    _id: [
        // ids here
    ],
    seasons: [{
            id: '56a0d6c692c26b8c019a2758',
            'name': 'Summer'
        },
        {
            id: '5680dc01ba1e41f5066526fa',
            'name': 'Winter'
        }
    ]
}

查询必须返回包含2个属性id和season的单个文档。季节必须是一系列独特的季节。每个季节都应该包含一个id和一个名字。谢谢你的帮助！

Answer 1

使用 $group 步骤重构您的管道，使_id值为null，以计算所有输入文档的累计值，从而将id字段创建为所有唯一ID的数组：

var pipeline = [ 
    { 
        '$match': { 
            unit: new ObjectId('566fbaa1e63225b10bacac44')
        } 
    },

    { 
        '$lookup': { 
            from: 'seasons', localField: 'season', foreignField: '_id', as: 'season' 
        } 
    },
    {
        '$unwind' : '$season'
    },
    {
        '$project': {
            season:  {'name': '$season.name', 'id' : '$season._id'}
        }
    },
        { 
        '$group': {
            "_id": null,  
            "id": { "$addToSet": "$_id" },
            'season': { 
                '$addToSet': '$season' 
            }
        } 
    }       
];
db.memberships.aggregate(pipeline);

选择字段的值作为数组mongoose / mongodb aggregation

1 个答案: