How to make a basic youtube music bot work with searching titles instead of the URL

Question

Hello so i've followed this tutorial and added this code to my current bot to make it have a music bot function. Im wondering how to make the following code work with the youtube search function, for example right now I have to do !play URL but I would also like to be able to do !play name of song then the bot will search and play the most matched song.

I am new to javascript but I know I shouldn't be looking for handouts, but some help would be appreciated.

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);

user_8289918 · Answer

You can use 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...

If you want to support both URLs and searching, you can test if the first argument is a valid URL using ytdl.validateURL:

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...

Both of these examples replace

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

except for const yts = require("yt-search");, which should be placed at the top with the other requires.

How to make a basic youtube music bot work with searching titles instead of the URL

1 Answers