我要数据库MongoDB中的每个第n个文档

这是我收集的大约25万份文件：

{ 
    "_id" : ObjectId("5ccfbae47dd9094e30998954"), 
    "date" : "05/02/2019"
}

{ 
    "_id" : ObjectId("5ccfbaff7dd9094e30998956"), 
    "date" : "05/04/2019"
}

{ 
    "_id" : ObjectId("5ccfd7dc6432426a9b4a4785"), 
    "date" : "05/06/2019"
}

{ 
    "_id" : ObjectId("5ccfd8d96432426a9b4a4788"), 
    "date" : "05/08/2019"
}

{ 
    "_id" : ObjectId("5cd01ab6c3bb270aa5c12ade"), 
    "date" : "05/04/2019"
}

{ 
    "_id" : ObjectId("5cd0227abf056b4b82d46ba9"), 
    "date" : "05/06/2019"
}

我收集了超过25万份文档。我必须从我的收集中获取每N个数据，这样就不会花费更多的时间。 作为经验：-我需要收集的每2份文件，预期的文件如下：

{ 
    "_id" : ObjectId("5ccfbaff7dd9094e30998956"), 
    "date" : "05/04/2019"
}

{ 
    "_id" : ObjectId("5ccfd8d96432426a9b4a4788"), 
    "date" : "05/08/2019"
}

{ 
    "_id" : ObjectId("5cd0227abf056b4b82d46ba9"), 
    "date" : "05/06/2019"
}

这个问题有什么解决办法吗？

您可以尝试以下方法：

db.collection.aggregate([
    {
        "$group": { 
            "_id": null,
            data: { $push: "$$ROOT"}
        }     
    },
    {
        $project: {
            requiredResult: {
                $map: {
                    input: { $range: [ 0, { $size: "$data" }, 2 ] },    // here you can replace 2 with any number (N)                        
                    as: "index",
                    in: { $arrayElemAt: [ "$data", "$$index" ] }
                }
            }
        }
    }
])