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Show screensharign button only if connected

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Signed-off-by: Šimon Brandner <simon.bra.ag@gmail.com>
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Šimon Brandner 2021-05-13 18:11:47 +02:00
parent 834579f778
commit adddb0f0d2
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1 changed files with 8 additions and 4 deletions
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12
src/components/views/voip/CallView.tsx
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@ -506,10 +506,14 @@ export default class CallView extends React.Component<IProps, IState> {
/> : null; /> : null;
let screensharingButton; let screensharingButton;
// Screensharing is possible, if we can send a second stream and identify // Screensharing is possible, if we can send a second stream and
// it using SDPStreamMetadata or if we can replace the already existing // identify it using SDPStreamMetadata or if we can replace the already
// usermedia track by a screensharing track // existing usermedia track by a screensharing track. We also need to be
if (this.props.call.opponentSupportsSDPStreamMetadata() || this.props.call.type === CallType.Video) { // 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 = ( screensharingButton = (
<AccessibleButton <AccessibleButton
className={screensharingClasses} className={screensharingClasses}