不和机器人youtube搜索，等待异步功能的回答

Question

我已经建立了一个不和谐，但是使用javascript，我有一个命令，我希望能够搜索youtube视频，并在语音频道中播放第一个结果。

我正在使用discordjs和discord-youtup-api库。

此代码查找要搜索的命令。args数组是搜索查询

else if (command === 'search') {
      isReady = false;
      if (message.channel.type !== 'text') return;

      const { voiceChannel } = message.member;

      if (!voiceChannel) {
        return message.reply('please join a voice channel first!');
      }

      voiceChannel.join().then(connection => {
        const stream = ytdl(searchYouTubeAsync(args), { filter: 'audioonly' });
        const dispatcher = connection.playStream(stream);

        dispatcher.on('end', () => voiceChannel.leave());
        isReady = true;
      })

};

这就是使用youtube api搜索视频并返回其网址的功能。

async function searchYouTubeAsync(args) {
  var video = await youtube.searchVideos(args.toString().replace(/,/g,' '));
  console.log(video.url);
  console.log(typeof String(video.url));
  return String(video.url);
}

尝试该命令时，我确实收到以下错误消息。

(node:13141) UnhandledPromiseRejectionWarning: TypeError [ERR_INVALID_ARG_TYPE]: The "url" argument must be of type string. Received type object
    at Url.parse (url.js:146:11)
    at Object.urlParse [as parse] (url.js:140:13)
    at Object.exports.getURLVideoID (/Users/alexanderhoerl/Developer/discord-music-bot/node_modules/ytdl-core/lib/util.js:248:20)
    at Object.exports.getVideoID (/Users/alexanderhoerl/Developer/discord-music-bot/node_modules/ytdl-core/lib/util.js:279:20)
    at getInfo (/Users/alexanderhoerl/Developer/discord-music-bot/node_modules/ytdl-core/lib/info.js:46:17)
    at ytdl (/Users/alexanderhoerl/Developer/discord-music-bot/node_modules/ytdl-core/lib/index.js:17:3)
    at voiceChannel.join.then.connection (/Users/alexanderhoerl/Developer/discord-music-bot/index.js:89:24)
    at process._tickCallback (internal/process/next_tick.js:68:7)
(node:13141) UnhandledPromiseRejectionWarning: Unhandled promise rejection. This error originated either by throwing inside of an async function without a catch block, or by rejecting a promise which was not handled with .catch(). (rejection id: 1)
(node:13141) [DEP0018] DeprecationWarning: Unhandled promise rejections are deprecated. In the future, promise rejections that are not handled will terminate the Node.js process with a non-zero exit code.

我猜问题在于，音乐机器人试图在searchYouTube函数找到链接之前加载流，因此没有提供有效的url。虽然我不知道如何解决这个问题，因为函数需要异步才能等待youtube搜索结果。

Answer 1

我几乎可以肯定现在这是不真实的。然而，以下是解决方案：

const stream = ytdl(searchYouTubeAsync(args).toString(), { filter: 'audioonly' });

xxx。当我和ytdl有同样的问题时，toString（）让我很沮丧。

Answer 2

您可以在async函数中运行：

voiceChannel.join().then(async connection => {
  let url = await searchYouTubeAsync(args);
  let stream = ytdl(url, { filter: 'audioonly' });
  let dispatcher = connection.playStream(stream);

  dispatcher.on('end', () => voiceChannel.leave());
  isReady = true;
})