javascript - 如何使基本的 YouTube 音乐机器人能够搜索标题而不是 URL

Question

您好，我已按照本教程操作，并将此代码添加到我当前的机器人中，以使其具有音乐机器人功能。我想知道如何使以下代码与 youtube 搜索功能一起使用，例如现在我必须执行 !play URL 但我也希望能够执行 !play name of歌曲，然后机器人将搜索并播放最匹配的歌曲。

我是 javascript 新手，但我知道我不应该寻找讲义，但如果能提供一些帮助，我将不胜感激。

const Discord = require("discord.js");
const { prefix, token } = require("./config.json");
const ytdl = require("ytdl-core");

const client = new Discord.Client();

const queue = new Map();

client.once("ready", () => {
  console.log("Ready!");
});

client.once("reconnecting", () => {
  console.log("Reconnecting!");
});

client.once("disconnect", () => {
  console.log("Disconnect!");
});

client.on("message", async message => {
  if (message.author.bot) return;
  if (!message.content.startsWith(prefix)) return;

  const serverQueue = queue.get(message.guild.id);

  if (message.content.startsWith(`${prefix}play`)) {
    execute(message, serverQueue);
    return;
  } else if (message.content.startsWith(`${prefix}skip`)) {
    skip(message, serverQueue);
    return;
  } else if (message.content.startsWith(`${prefix}stop`)) {
    stop(message, serverQueue);
    return;
  } else {
    message.channel.send("You need to enter a valid command!");
  }
});

async function execute(message, serverQueue) {
  const args = message.content.split(" ");

  const voiceChannel = message.member.voice.channel;
  if (!voiceChannel)
    return message.channel.send(
      "You need to be in a voice channel to play music!"
    );
  const permissions = voiceChannel.permissionsFor(message.client.user);
  if (!permissions.has("CONNECT") || !permissions.has("SPEAK")) {
    return message.channel.send(
      "I need the permissions to join and speak in your voice channel!"
    );
  }

  const songInfo = await ytdl.getInfo(args[1]);
  const song = {
    title: songInfo.title,
    url: songInfo.video_url
  };

  if (!serverQueue) {
    const queueContruct = {
      textChannel: message.channel,
      voiceChannel: voiceChannel,
      connection: null,
      songs: [],
      volume: 5,
      playing: true
    };

    queue.set(message.guild.id, queueContruct);

    queueContruct.songs.push(song);

    try {
      var connection = await voiceChannel.join();
      queueContruct.connection = connection;
      play(message.guild, queueContruct.songs[0]);
    } catch (err) {
      console.log(err);
      queue.delete(message.guild.id);
      return message.channel.send(err);
    }
  } else {
    serverQueue.songs.push(song);
    return message.channel.send(`${song.title} has been added to the queue!`);
  }
}

function skip(message, serverQueue) {
  if (!message.member.voice.channel)
    return message.channel.send(
      "You have to be in a voice channel to stop the music!"
    );
  if (!serverQueue)
    return message.channel.send("There is no song that I could skip!");
  serverQueue.connection.dispatcher.end();
}

function stop(message, serverQueue) {
  if (!message.member.voice.channel)
    return message.channel.send(
      "You have to be in a voice channel to stop the music!"
    );
  serverQueue.songs = [];
  serverQueue.connection.dispatcher.end();
}

function play(guild, song) {
  const serverQueue = queue.get(guild.id);
  if (!song) {
    serverQueue.voiceChannel.leave();
    queue.delete(guild.id);
    return;
  }

  const dispatcher = serverQueue.connection
    .play(ytdl(song.url))
    .on("finish", () => {
      serverQueue.songs.shift();
      play(guild, serverQueue.songs[0]);
    })
    .on("error", error => console.error(error));
  dispatcher.setVolumeLogarithmic(serverQueue.volume / 5);
  serverQueue.textChannel.send(`Start playing: **${song.title}**`);
}

client.login(token);

Answer 1

您可以使用yt-search :

const yts = require("yt-search");

// Searches YouTube with the message content (this joins the arguments
// together because songs can have spaces)
const {videos} = await yts(args.slice(1).join(" "));
if (!videos.length) return message.channel.send("No songs were found!");
const song = {
  title: videos[0].title,
  url: videos[0].url
};

// rest of code...

如果您想同时支持 URL 和搜索，可以使用 ytdl.validateURL 测试第一个参数是否为有效 URL。 :

let song;
if (ytdl.validateURL(args[1])) {
  const songInfo = await ytdl.getInfo(args[1]);
  song = {
    title: songInfo.title,
    url: songInfo.video_url
  };
} else {
  const {videos} = await yts(args.slice(1).join(" "));
  if (!videos.length) return message.channel.send("No songs were found!");
  song = {
    title: videos[0].title,
    url: videos[0].url
  };
}

// rest of code...

这两个示例都替换

const songInfo = await ytdl.getInfo(args[1]);
const song = {
  title: songInfo.title,
  url: songInfo.video_url
};

除了 const yts = require("yt-search");，它应该与其他 require 一起放置在顶部。