NodeJs MongoDB 嵌套查询 - 同步运行回调

Question

我有一个如下所示的代码块。我找不到如何实现这一点。
- 我在 mongodb 中有房间集合，该集合中的文档包括用户 ID 数组。每个房间可以有多个用户。
- 所以我想找到房间中包含我的用户 ID 的用户名。

数据库集合：
房间 {roomId:1, users:[99, 100]}, {roomId:2, users:[99, 101]}
用户 {_id:99, name:"Alex"},{_id:100, name:"Harry"},{_id:101, name:"Tom"}

var userId = 99;
var arrUserIds = [];
var arrRooms = [];
var strUserNames = "";
db.collection("rooms").find({"users.userId":userId}).toArray(function(err, rooms) {

    for (var i=0; i<rooms.length; i++) {
        arrUserIds = [];

        for(var j=0; j<rooms[i].users.length; j++){
            arrUserIds.push(new BSON.ObjectID(rooms[i].users[j].userId));   
        }

        db.collection('users').find({"_id": {$in: arrUserIds}}).toArray(function(err, users) {
            strUserNames = users.map(function(elem){return elem.name;}).join(", ");
            arrRooms.push({_id:rooms[i].roomId, name:strUserNames });
        });

    }

    res.json({rooms:arrRooms});
});

我想要这样的结果：
arrRooms: [ { _id:1:, name:"Alex, Harry" }, { _id:2:, name:"Alex, Tom" } ]

谢谢

Answer 1

您在第二个 .find 回调执行之前返回 res.json，因此它将为空。您需要重组以在响应之前等待所有异步回调，如下所示：

var userId = 99;
var arrRooms = [];
var strUserNames = "";

db.collection("rooms").find({"users.userId":userId}).toArray(function(err, rooms) {

    var completed = 0;
    var complete = function() {
        completed++;
        if (completed === rooms.length - 1) {
            res.json({rooms:arrRooms});
        }
    }


    for (var i=0; i<rooms.length; i++) {
        var arrUserIds = [];
        var roomId = rooms[i].roomId;

        for(var j=0; j<rooms[i].users.length; j++){
            arrUserIds.push(new BSON.ObjectID(rooms[i].users[j].userId));   
        }

        db.collection('users').find({"_id": {$in: arrUserIds}}).toArray(function(err, users) {
            strUserNames = users.map(function(elem){return elem.name;}).join(", ");
            arrRooms.push({_id:roomId, name:strUserNames });

            complete();
        });
    }

});