Show screensharign button only if connected

Signed-off-by: Šimon Brandner <simon.bra.ag@gmail.com>
This commit is contained in:
Šimon Brandner 2021-05-13 18:11:47 +02:00
parent 834579f778
commit adddb0f0d2
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View file

@ -506,10 +506,14 @@ export default class CallView extends React.Component<IProps, IState> {
/> : null;
let screensharingButton;
// Screensharing is possible, if we can send a second stream and identify
// it using SDPStreamMetadata or if we can replace the already existing
// usermedia track by a screensharing track
if (this.props.call.opponentSupportsSDPStreamMetadata() || this.props.call.type === CallType.Video) {
// Screensharing is possible, if we can send a second stream and
// identify it using SDPStreamMetadata or if we can replace the already
// existing usermedia track by a screensharing track. We also need to be
// connected to know the state of the other side
if (
(this.props.call.opponentSupportsSDPStreamMetadata() || this.props.call.type === CallType.Video) &&
this.props.call.state === CallState.Connected
) {
screensharingButton = (
<AccessibleButton
className={screensharingClasses}