Mongo Db获得单个集合的两个文档作为singel文档

Question

我只有一个集合雇员： 集合中的数据是

{
    "_id" : ObjectId("5b44cb8b44a531b12aa69678"),
    "Empid" : 1,
    "Empname" : "Raghu",
    "Deptid" : "D1",
    "Depname" : "IT",
    "Salary" : 15000,
    "Time" : "8 Hours"
}

{
    "_id" : ObjectId("5b44cbda44a531b12aa69679"),
    "Empid" : 1,
    "Empname" : "Raghu",
    "Deptid" : "D2",
    "Depname" : "Soft",
    "Salary" : 25000,
    "Time" : "9 Hours"
}

我希望输出为

{

    "Empid" : 1,
    "Empname" : "Raghu",
{
    "Deptid" : "D1",
    "Depname" : "IT",
        {
    "Salary" : 15000,
         "Time" : "8 Hours"
    }
}
{

    "Deptid" : "D2",
    "Depname" : "Soft",
    {
         "Salary" : 25000,
         "Time" : "9 Hours"
    }
}

}

Answer 1

根据上述问题的描述，请尝试在MongoDB shell中执行以下聚合操作。

db.employee.aggregate(

    // Pipeline
    [
        // Stage 1
        {
            $group: {
                _id: '$Empid',
                department: {
                    $addToSet: {
                        Deptid: '$Deptid',
                        Depname: '$Depname',
                        Salary: '$Salary',
                        Time: '$Time'
                    }
                },
                doc: {
                    $first: '$$ROOT'
                },
            }
        },

        // Stage 2
        {
            $project: {
                Empid: '$_id',
                EmpName: '$doc.Empname',
                Department: '$department',
                _id: 0
            }
        }

    ]



);