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我正在制作一个不和谐的机器人，我使用以下代码，同时安装了所有正确的 npm 东西并且 ffmpeg 工作。这个机器人今天早些时候工作，我把它搞砸了，所以我恢复到旧代码，现在它不工作了。

const Discord = require('discord.js');
const api = require("imageapi.js");
const client = new Discord.Client();
const YouTube = require('simple-youtube-api')
const ytdl = require('ytdl-core')
const prefix = '!';


client.once('ready', () => {
    console.log("Online!");
    client.user.setActivity('!help');
});

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

  const args = message.content.substring(prefix.length).split("")

  if(message.content.startsWith(`${prefix}play`)) {
    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')) return message.channel.send(`I don\'t have permission to connect`)
    if(!permissions.has('SPEAK')) return message.channel.send(`I don\'t have permission to speak`)
     
    try {
        var connection = await voiceChannel.join()
    } catch(error){
        console.log("error")
        return message.channel.send(`There was a error when connection to the voice channel`)
    }

    const dispatcher = connection.play(ytdl(args[1]))
    .on('finish', () => {
      voiceChannel.leave()
    })
    .on('error', error => {
      console.log(error)
    })
    dispatcher.setVolumeLogarithmic(5 / 5)
  } else if (message.content.startsWith(`${prefix}stop`)) {
    if(!message.member.voice.channel) return message.channel.send("You need to be in a voice channel to stop the music")
    message.member.voice.channel.leave()
    return undefined
  }
})```
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1 回答 1

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这意味着这args[1]不是有效的 youtube URL 或 ID。拆分消息时：

const args = message.content.substring(prefix.length).split("")

你分开''而不是' '. 这是每个字符与每个空格的区别。

const str = 'Hello World';

console.log(str.split(''));
console.log(str.split(' '));

因此，您可能会调用ytdl('w')in www.youtube.com/...。即使您解决了这个问题，您也应该添加错误处理以确保：

  1. args[1]存在
  2. args[1]是一个有效的 ID
if (!args[1]) return message.channel.send('...');

try {
  const audio = ytdl(args[1]);
} catch (err) {
  return message.channel.send('...');
}
于 2021-05-12T23:40:06.477 回答