JavaScript audio.play() error

Question

my code stopped working today (it worked before).

Here is my code:

playMe = new Audio(link);
playMe.play();

Here is the error I get in the console:

Uncaught (in promise) DOMException: Failed to load because no supported source was found.

It is really the audio.play(); that doesn't work because if I only keep the first line I don't get an error (but obviously is dos't play).

I'm using chrome 52.0.2743.82 m (64-bit)

Thanks.

Amm... it started to give me the error again, I didn't change the code and there was not any chore updates... — Zvi Karp, Jul 28 '16 at 20:16
Did you ever figure out this issue, I'm running into the exact same problem. — John Anderson, Jul 14 '17 at 02:58
It might be a problem with Chrome 50+. Have a look at this: [https://stackoverflow.com/a/28812647/4465386](https://stackoverflow.com/a/28812647/4465386) — Zvi Karp, Jul 15 '17 at 19:25

score 1 · Answer 1 · answered Mar 06 '19 at 07:51

1

Before you play audio, window must be focused. For this case you can use:

playMe = new Audio(link);
window.focus();
playMe.play();

answered Mar 06 '19 at 07:51

Nhat Ngo

113
2
9

score 1 · Answer 2 · answered Sep 03 '19 at 07:39

The below answer is for video. The same you can use for audio as well.

 var playPromise = document.querySelector('video').play();
  // In browsers that don’t yet support this functionality,
 // playPromise won’t be defined.
 if (playPromise !== undefined) {
  playPromise.then(function() {
 // Automatic playback started!
 }).catch(function(error) {
  // Automatic playback failed.
 // Show a UI element to let the user manually start playback.
});
}

Working on audio, tnx :) – Jiro Matchonson Feb 11 '22 at 12:26 — Jiro Matchonson, Feb 11 '22 at 12:26

score 0 · Answer 3 · answered Feb 07 '22 at 20:12

I had the exact problem as you.
I tried playing a <audio> html element and it worked!
So I thought that maybe...
The only thing you should do is to append playMe to body...

var play_me = new Audio("music.mp3");
document.body.appendChild(play_me);
play_me.play();

But don't worry the play_me isn't visible.

JavaScript audio.play() error

3 Answers3

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