Create 2 new session in one Start button click

[Goji]

Hi,
Can we create 2 new session in the same time (with one Start button click) for WebRTC as RTMP and Stream Recording? How should we handle the front end display as I would like to only display one stream video? Assuming I only want to record the video at the backend for storage purpose.
Thanks.

 

[Max]

Can we create 2 new session in the same time (with one Start button click) for WebRTC as RTMP and Stream Recording?

Do you mean two streams from two different web cams?
You can check the Media Devices example:
https://wcs5-eu.flashphoner.com/demo2/media-devices
You can select a web cam device in the Send Video block.
This block is filled out using this code:
https://github.com/flashphoner/flas...treaming/media_devices_manager/manager.js#L40

Flashphoner.getMediaDevices(null, true).then(function (list) {
        list.audio.forEach(function (device) {
            var audio = document.getElementById("audioInput");
            var i;
            var deviceInList = false;
            for (i = 0; i < audio.options.length; i++) {
                if (audio.options[i].value == device.id) {
                    deviceInList = true;
                    break;
                }
            }
...
}

You can set particular device using this code:
https://github.com/flashphoner/flas...reaming/media_devices_manager/manager.js#L214

constraints.video = {
            deviceId: $('#videoInput').val(),
            width: parseInt($('#sendWidth').val()),
            height: parseInt($('#sendHeight').val())
        };

Therefore you can send two different streams to the server

session.createStream({name:'stream1', constraints:{audio: true, video:{deviceId:'camera1'}}}).publish();
session.createStream({name:'stream2', constraints:{audio: true, video:{deviceId:'camera2'}}}).publish();

So you have to pass deviceId - id of camera to get this working.
You can get the list of cams from code:

var video = document.getElementById("videoInput");
for (i = 0; i < video.options.length; i++) {
...

 

[Max]

How should we handle the front end display as I would like to only display one stream video?

Let's say you have two div's:

<div id="display1"></div>
<div id="display2"></div>

And two streams

session.createStream({display: document.getElementById('display1'), name:'stream1', constraints:{audio: true, video:{deviceId:'camera1'}}}).publish();
session.createStream({display: document.getElementById('display2'), name:'stream2', constraints:{audio: true, video:{deviceId:'camera2'}}}).publish();

If you want to see one display only, you can hide second display like this

<div id="display2" style="visibility:hidden"></div>

 

[Max]

Assuming I only want to record the video at the backend for storage purpose.

If you want to record, you pass parameter record: true

session.createStream({record: true, display: document.getElementById('display1'), name:'stream1', constraints:{audio: true, video:{deviceId:'camera1'}}}).publish();
session.createStream({record: true, display: document.getElementById('display2'), name:'stream2', constraints:{audio: true, video:{deviceId:'camera2'}}}).publish();

Two files will be recorded. One file per stream.