Question

我的应用中有以下代码：

const response = {
    created: await eventModel.count({ author: uid }).exec(),
    participated: await eventModel.count({ participants: uid }).exec(),
    interested: await eventModel.count({ interested: uid }).exec()
};

我想知道如何只使用一个请求获得相同的输出。

以下是模型：

const eventModel = mongoose.model("Event", new mongoose.Schema({
    author: mongoose.Schema.Types.ObjectId,
    interested: [mongoose.Schema.Types.ObjectId],
    participants: [mongoose.Schema.Types.ObjectId],
    title: String
}));

有没有办法正确地做到这一点？

Answer 1

如果您的问题是how do I get data from mongo and only make 1 query for a given criteria?，那么如果没有更明确的解释，那么就是这样。

const eventModelCounts = eventModel.aggregate(
                     [

                         { $group : { _id : { author: uid, participants: uid, interested: uid },  uid: { $sum: 1 }  }}

                     ]

                    ).exec(function(err,data){
                        count = data;
                        console.log("this is data aggregate:",data);

                        callback();
                    });

基本上你在感兴趣的字段上应用聚合。

一次请求中的多个计数

1 个答案: