i get a error of when i try to get the url about youtube-search HOT 35 CLOSED

`ytdl(results, { filter: function(f) {
return f.container === 'mp4' && !f.encoding; } })
.pipe(fs.createWriteStream(audioOutput))
.on('finish', function() {
ffmpeg()
.input(ytdl(urlm, { filter: function(f) {
return f.container === 'mp4' && !f.audioEncoding; } }))
.videoCodec('copy')
.input(audioOutput)
.audioCodec('copy')
.save(path.resolve(__dirname, 'output.mp4'))
.on('error', console.error)
.on('progress', function(progress) {

  }).on('end', function() {
    console.log();
  });` script which im using

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Hard to tell what is going on from your code snippet. Are you passing the results of the youtube search to ytdl? If so just pass results.link instead and not the entire object. If this does not solve you problem can you please submit a small working example which demonstrates the problem?

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yes is for a discord bot

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@frankhale results.link does not work

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Will you please provide a small example that shows the issue because it's difficult to tell what you are doing from the sample you posted.

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want a bigger sample for more anallising

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@frankhale

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I don't know where 'url' is coming from in your error. 'url' isn't something returned from youtube-search results. The result object looks like this:

[ { id: 'UCYEK6xds6eo-3tr4xRdflmQ',
    link: 'https://www.youtube.com/channel/UCYEK6xds6eo-3tr4xRdflmQ',
    kind: 'youtube#channel',
    publishedAt: '2006-10-19T12:22:09.000Z',
    channelId: 'UCYEK6xds6eo-3tr4xRdflmQ',
    title: 'deadmau5',
    description: 'random carnage a la deadmau5.',
    thumbnails: { default: [Object], medium: [Object], high: [Object] } },
  { id: 'PKFcaXd5G8c',
    link: 'https://www.youtube.com/watch?v=PKFcaXd5G8c',
    kind: 'youtube#video',
    publishedAt: '2016-12-16T14:58:58.000Z',
    channelId: 'UCYEK6xds6eo-3tr4xRdflmQ',
    title: 'deadmau5 - Let Go Feat. Grabbitz (Cube 2.1)',
    description: 'http://www.lotsofshowsinarow.com W:/2016ALBUM/ out now: http://smarturl.it/2016album deadmau5 offers an exclusive Behind The Scenes look at the work that ...',
    thumbnails: { default: [Object], medium: [Object], high: [Object] } } ]

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im trying to make it so people can do a command like this ^play (youtube search term here) @frankhale

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I noticed that you are using the module https://github.com/fent/node-ytdl-core to download the YouTube video. Are you having an issue with that module or the youtube-search module?

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youtube search @frankhale

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If all you are trying to do is play the video you can use the YouTube IFrame API to do that.

https://developers.google.com/youtube/iframe_api_reference

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im tying to get the url from the 1 result of the seacrh term

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Have a look at the object returned by the results. It's an array of objects that look like this:

the URL is the 'link' string (see below)

{ id: 'PKFcaXd5G8c',
    link: 'https://www.youtube.com/watch?v=PKFcaXd5G8c',
    kind: 'youtube#video',
    publishedAt: '2016-12-16T14:58:58.000Z',
    channelId: 'UCYEK6xds6eo-3tr4xRdflmQ',
    title: 'deadmau5 - Let Go Feat. Grabbitz (Cube 2.1)',
    description: 'http://www.lotsofshowsinarow.com W:/2016ALBUM/ out now: http://smarturl.it/2016album deadmau5 offers an exclusive Behind The Scenes look at the work that ...',
    thumbnails: { default: [Object], medium: [Object], high: [Object] } }

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ok one more thing is there any other youtube url grabbing thing

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like youtube-api

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npmjs.com lists a bunch of YouTube modules but still you haven't told us really what the issue is with this module. I use this module in my own YouTube player and it works very well for searching YouTube. I'm trying to understand what issue you are having but so far am not able to figure out what problem you are encountering.

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ok let me put it easly ith the code sample and saying this im making a discord bot with a music function
and im trying to get the url aka link from the search term 1result link automatically here is the script i have in my ftp right now

`if (content.indexOf("^play ") == 0) {
const urlm = content.replace("^play ", "");
channel.sendMessage("debug: url/search term: " + urlm);
let vc = e.message.member.getVoiceChannel()
if(vc){vc.join();
} else {
}

   try{
var audioOutput = path.resolve(__dirname, e.message.guild.name + '.mp3');
ytdl(urlm, { filter: function(f) {
  return f.container === 'mp4' && !f.encoding; } })
    .pipe(fs.createWriteStream(audioOutput))
  .on('finish', function() {
    ffmpeg()
    .input(ytdl(urlm, { filter: function(f) {
      return f.container === 'mp4' && !f.audioEncoding; } }))
      .videoCodec('copy')
      .input(audioOutput)
      .audioCodec('copy')
      .save(path.resolve(__dirname, 'output.mp4'))
      .on('error', console.error)
      .on('progress', function(progress) {
        

      }).on('end', function() {
        console.log();
      });
channel.sendMessage("^^" + urlm + "^^ is downloaded and will now play :arrow_forward: ");
filenamee = e.message.guild.name + ".mp3";


play();


});
} catch(error) { 

var opts = {
  maxResults: 1,
  key: 'AIzaSyD1J1tnUlAxa0WxO9-XY6AwsuoBc7Dso1w'
};

search(urlm, opts, function(err, results) {
var audioOutput = path.resolve(__dirname, e.message.guild.name + '.mp3');
ytdl(results.link(), { filter: function(f) {
  return f.container === 'mp4' && !f.encoding; } })
    .pipe(fs.createWriteStream(audioOutput))
  .on('finish', function() {
    ffmpeg()
    .input(ytdl(results.link, { filter: function(f) {
      return f.container === 'mp4' && !f.audioEncoding; } }))
      .videoCodec('copy')
      .input(audioOutput)
      .audioCodec('copy')
      .save(path.resolve(__dirname, 'output.mp4'))
      .on('error', console.error)
      .on('progress', function(progress) {
        

      }).on('end', function() {
        console.log();
      });
channel.sendMessage("^^" + results + "^^ is downloaded and will now play  from youtube search :arrow_forward: ");
filenamee = e.message.guild.name + ".mp3";


});
});
}`

from youtube-search.

even though you specified maxResults of 1 you'll still get an array with 1 element. Trying to do this results.link will not work like that. Try results[0].link. Additionally it's not a function so you can't do results.link(). Link is a string that you can pass to the ytdl module like you are trying to do in your code sample.

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ok

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i get this error url.js:87
throw new TypeError('Parameter "url" must be a string, not ' + typeof url);
^

TypeError: Parameter "url" must be a string, not undefined
at Url.parse (url.js:87:11)
at Object.urlParse [as parse] (url.js:81:5)
at Object.exports.getVideoID (/var/www/html/bot/node_modules/ytdl-core/lib/util.js:225:20)
at getInfo (/var/www/html/bot/node_modules/ytdl-core/lib/info.js:39:17)
at ytdl (/var/www/html/bot/node_modules/ytdl-core/lib/index.js:18:3)
at WriteStream. (/var/www/html/bot/bot.js:285:12)
at emitNone (events.js:91:20)
at WriteStream.emit (events.js:185:7)
at finishMaybe (_stream_writable.js:515:14)
at afterWrite (_stream_writable.js:389:3)
at onwrite (_stream_writable.js:379:7)

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now got it semi working

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I don't think the issue is on the youtube-search module side. Looks like the error is coming from the ytdl-core module based on your sample code.

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shows string like this ^^[object Object]^^

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var r = results[0].link; string for results

from youtube-search.

I think for me or anyone to help further we'd need to see a small sample code that is complete (and something we could run locally) so we could do some more debugging.

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what it shows in discord and sample code
search(urlm, opts, function(err, results) {
var audioOutput = path.resolve(__dirname, e.message.guild.name + '.mp3');
var r = results[0].link; //string
ytdl(r, { filter: function(f) {
return f.container === 'mp4' && !f.encoding; } })
.pipe(fs.createWriteStream(audioOutput))
.on('finish', function() {
ffmpeg()
.input(ytdl(r, { filter: function(f) {
return f.container === 'mp4' && !f.audioEncoding; } }))
.videoCodec('copy')
.input(audioOutput)
.audioCodec('copy')
.save(path.resolve(__dirname, 'output.mp4'))
.on('error', console.error)
.on('progress', function(progress) {

  }).on('end', function() {
    console.log();
  });

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@frankhale got it working now

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Awesome!

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want to join my discord server and try it @frankhale

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I don't have the discord app installed on this machine =(

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@frankhale ok use online https://discord.gg/XH92D and discord made an improvement it works online

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@frankhale they made it to work online now

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@frankhale non expire link https://discord.gg/gxUkCvw

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Please just use these issues for actual issues with the package. For more general questions you can always resort to sites like stackoverflow. @kessanro

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