It seems that it impossible to utilize Peer.Call without passing in a user media stream. Peer.Call should be able to call any peer even if the caller decides to not give the browser access to their webcam/mic.
choudeshell
All 14 comments
Good point; we'll look into adding this.
michelle
on 3 Mar 2014
+1
Calling without sending a stream seems a reasonable feature to me. Any update on that ? :)
dervalp
on 23 May 2014
Sorry, haven't had the time recently :(. Happy to review PRs though.
Michelle
On Fri, May 23, 2014 at 2:01 PM, Pierre [email protected] wrote:
+1
Calling without sending a stream seems a reasonable feature to me. Any
update on that ? :)—
Reply to this email directly or view it on GitHubhttps://github.com/peers/peerjs/issues/158#issuecomment-44059854
.
michelle
on 23 May 2014
I solved that by passing new webkitMediaStream() as a media stream to method call(). As my app is being used only in Chrome, it's not an issue for me.
Hope it helps :)
grabbou
on 24 Jun 2014
@grabbou can you paste some example code?
thomasverleye
on 20 Jan 2015
+1
@grabbou that`s so simple solution and works for me tks, do you know how do it in firefox? i try MediaStream but not works :sweat_smile:
jhogoforbroke
on 24 Jan 2016
@michelle, I don't suppose you have any thoughts on how this could be implemented? I know very little about WebRTC, but it would be enormously helpful to me to be able to make calls as a user without a stream.
doppio
on 29 Jan 2016
+1
Rastopyr
on 30 Jan 2016
+1 too
ndac-todoroki
on 24 Feb 2016
+1
mahmed0715
on 24 Oct 2016
@jhogoforbroke Can you post some sample of how you achieved this. I am attaching the client and server code. The server code is just expected to broadcast the video and all clients can just view the video without having to share any media. When I use new webkitMediaStream() the server is not streaming any data.
peer_server.txt
vikramarjun
on 1 Feb 2017
Not sure, if it's still actual, but got able to bypass it using this trick:
const audioCtx = new AudioContext();
const dest = audioCtx.createMediaStreamDestination();
var call = peer.call('testid', dest.stream);
demogoran
on 31 Aug 2017
Not sure, if it's still actual, but got able to bypass it using this trick:
const audioCtx = new AudioContext();
const dest = audioCtx.createMediaStreamDestination();
var call = peer.call('testid', dest.stream);
it's not working
can anyone solve this problem?
yunusberateke
on 11 Feb 2020
@yunusberateke it's work
```
const createMediaStreamFake = () => {
return new MediaStream([createEmptyAudioTrack(), createEmptyVideoTrack({ width:640, height:480 })]);
}
const createEmptyAudioTrack = () => {
const ctx = new AudioContext();
const oscillator = ctx.createOscillator();
const dst = oscillator.connect(ctx.createMediaStreamDestination());
oscillator.start();
const track = dst.stream.getAudioTracks()[0];
return Object.assign(track, { enabled: false });
}
const createEmptyVideoTrack = ({ width, height }) => {
const canvas = Object.assign(document.createElement('canvas'), { width, height });
canvas.getContext('2d').fillRect(0, 0, width, height);
const stream = canvas.captureStream();
const track = stream.getVideoTracks()[0];
return Object.assign(track, { enabled: false });
};
wilywork
on 16 Apr 2020
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Most helpful comment
@yunusberateke it's work
```
const createMediaStreamFake = () => {
return new MediaStream([createEmptyAudioTrack(), createEmptyVideoTrack({ width:640, height:480 })]);
}
const createEmptyAudioTrack = () => {
const ctx = new AudioContext();
const oscillator = ctx.createOscillator();
const dst = oscillator.connect(ctx.createMediaStreamDestination());
oscillator.start();
const track = dst.stream.getAudioTracks()[0];
return Object.assign(track, { enabled: false });
}
const createEmptyVideoTrack = ({ width, height }) => {
const canvas = Object.assign(document.createElement('canvas'), { width, height });
canvas.getContext('2d').fillRect(0, 0, width, height);
};