如何用Mongoose获取played_frequency最高的前10条文档？

要实现这个需求，只需对played_frequency字段做降序排序，再限制返回结果数量为10即可，以下是具体实现代码：

1. 确认Mongoose模型定义（基于ITrack接口） #

import mongoose, { Model, Schema } from 'mongoose';

export interface ITrack {
  genre: string;
  title: string;
  artist: string;
  featuring?: string;
  duration: string;
  coverImage?: string;
  image: string;
  url: string;
  trending?: number;
  played_frequency: number;
  promoted?: boolean;
  likes?: number;
}

const trackSchema = new Schema<ITrack>({
  genre: { type: String, required: true },
  title: { type: String, required: true },
  artist: { type: String, required: true },
  featuring: String,
  duration: { type: String, required: true },
  coverImage: String,
  image: { type: String, required: true },
  url: { type: String, required: true },
  trending: Number,
  played_frequency: { type: Number, required: true },
  promoted: Boolean,
  likes: Number
});

const Track: Model<ITrack> = mongoose.model<ITrack>('Track', trackSchema);

2. 编写查询函数 #

async function getTop10MostPlayedTracks() {
  try {
    // 按played_frequency降序排序，取前10条
    const topTracks = await Track.find()
      .sort({ played_frequency: -1 }) // -1代表降序，1代表升序
      .limit(10); // 限制返回结果数量为10
    return topTracks;
  } catch (error) {
    console.error('查询失败:', error);
    throw error;
  }
}

// 调用示例
getTop10MostPlayedTracks()
  .then(tracks => console.log('播放量Top10曲目:', tracks))
  .catch(err => console.error(err));

补充说明 #

• 如果需要排除某些不需要的字段，可以在find()后添加.select()方法，比如.select('-_id -__v')表示不返回_id和__v字段
• 若要只返回特定字段，可写为.select('title artist played_frequency')
• 确保played_frequency字段在Schema中定义为Number类型，否则排序可能出现异常

内容的提问来源于stack exchange，提问作者user3118363