Question

Mongodb版本2.6.9

我试图计算一家公司参与消息传递互动的总数。我可以使用汇总$ group获得交互的一方，但我基本上看了两个字段，并将这些字段聚合在一起，为每个唯一的公司ID提供了空白。

sender_id和receiver_id与同一公司ID相关。

{ "_id" : a, "sender_id" : 1, "receiver_id" : 2, payload: "data" }
{ "_id" : b, "sender_id" : 2, "receiver_id" : 5, payload: "data" }
{ "_id" : c, "sender_id" : 2, "receiver_id" : 4, payload: "data" }
{ "_id" : d, "sender_id" : 3, "receiver_id" : 2, payload: "data" }
{ "_id" : e, "sender_id" : 4, "receiver_id" : 1, payload: "data" }

使用上述数据结构，我尝试生成类似于

的结果集

{ "_id" : 1, count:  2}
{ "_id" : 2, count:  4}
{ "_id" : 3, count:  1}
{ "_id" : 4, count:  2}
{ "_id" : 5, count:  1}

例如，公司2涉及消息a，b，c，d。

Answer 1

您的选项仅限于2.6管道。你可以尝试下面的管道。

<$group $push sender_id为receiver_id和$project创建单值数组。

带有$setUnion的

$unwind将ID合并为单个数组。

$group和db.collection.aggregate({ "$group": { "_id": "$_id", "sender_id": { "$push": "$sender_id" }, "receiver_id": { "$push": "$receiver_id" } } }, { "$project": { "id": { "$setUnion": ["$sender_id", "$receiver_id"] } } }, { "$unwind": "$id" }, { "$group": { "_id": "$id", "count": { "$sum": 1 } } })计算出现次数。

[]

您可以使用以下管道获取更新版本。使用db.collection.aggregate({ $project: { id: ["$sender_id", "$receiver_id"] } }, { $unwind: "$id" }, { $group: { _id: "$id", count: { $sum: 1 } } })创建数组。

import java.util.Scanner;

public class DistanceFinder {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);

        System.out.println("Enter x1 or <crtl + z> to quit");

        while (input.hasNext()) {
            double x1 = input.nextDouble();

            System.out.println("Please input the x2 number:");
            double x2 = input.nextDouble();

            System.out.println("Please input the y1 number:");
            double y1 = input.nextDouble();

            System.out.println("Please input the y2 number:");
            double y2 = input.nextDouble();

            double x3 = 0;
            double y3 = 0;
            double xy1 = 0;

            x3 = Math.pow((x2 - x1), 2);
            y3 = Math.pow((y2 - y1), 2);

            xy1 = x3 + y3;

            double distance = Math.sqrt(xy1);

            System.out.printf("Distance is %.6f %n%n%n", distance);

            System.out.println("Enter x1 or <crtl + z> to quit");
        }
        input.close();
    }
}

发送的所有消息的总和

1 个答案: