UnhandledPromiseRejectionWarning：错误：未找到视频 ID

Question

正如标题所说，UnhandledPromiseRejectionWarning: Error: No video id found 是我发送命令后在控制台上显示的内容：!join {link}。 UnhandledPromiseRejectionWarning: Unhandled promise rejection 也出现了。

const commando = require('discord.js-commando'); const YTDL = require('ytdl-core');

function Play(connection, message) {
    var server = servers[message.guild.id];
    server.dispatcher = connection.playStream(YTDL(server.queue[0], {filter: "audioonly"}));
    server.queue.shift();
    server.dispatcher.on("end", function()
    {
        if(server.queue[0])
        {
            Play(connection,message);
        }
        else
        {
            connection.disconnect();
        }
    }); }

class JoinChannelCommand extends commando.Command {
    constructor(client)
    {
        super(client,{
            name: 'join',
            group: 'music',
            memberName: 'join',
            description: 'Join the voice channel'
        });
    }

    async run(message, args)
    {
        if(message.member.voiceChannel)
        {
            if(!message.guild.voiceConnection)
            {
                if(!servers[message.guild.id])
                {
                    servers[message.guild.id] = {queue: []}
                }
                message.member.voiceChannel.join()
                    .then(connection =>{
                        var server = servers[message.guild.id];
                        message.reply("Succesfully Joined!");
                        server.queue.push(args);
                        Play(connection, message);
                    })
            }
        }
        else
        {
            message.reply("You need to be in a voice channel to invite me Baka!")
        }
    }

}

module.exports = JoinChannelCommand;

Answer 1

当你的承诺被拒绝并且你没有捕捉到它产生的错误时，就会发生未处理的 Promiserejection。要获得您想要处理的错误，您可以在.then 语句之后放置一个.catch 语句。例如：

            message.member.voiceChannel.join()
                .then(connection =>{
                    var server = servers[message.guild.id];
                    message.reply("Succesfully Joined!");
                    server.queue.push(args);
                    Play(connection, message);
                }).catch(e => { console.log(e) });

也许这会帮助您处理错误。