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WebRTC call: allow starting call when already one

This commit is contained in:
ganfra 2021-07-13 18:13:31 +02:00
parent 3793df1b36
commit 94981d5f37
1 changed files with 3 additions and 12 deletions
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15
vector/src/main/java/im/vector/app/features/home/room/detail/StartCallActionsHandler.kt
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@ -26,7 +26,6 @@ import im.vector.app.core.utils.PERMISSIONS_FOR_VIDEO_IP_CALL
import im.vector.app.core.utils.checkPermissions
import im.vector.app.features.call.webrtc.WebRtcCallManager
import im.vector.app.features.settings.VectorPreferences
import org.matrix.android.sdk.api.session.widgets.model.WidgetType
class StartCallActionsHandler(
private val roomId: String,
@ -36,7 +35,7 @@ class StartCallActionsHandler(
private val roomDetailViewModel: RoomDetailViewModel,
private val startCallActivityResultLauncher: ActivityResultLauncher<Array<String>>,
private val showDialogWithMessage: (String) -> Unit,
private val onTapToReturnToCall: () -> Unit) {
private val onTapToReturnToCall: () -> Unit) {
fun onVideoCallClicked() {
handleCallRequest(true)
@ -61,16 +60,8 @@ class StartCallActionsHandler(
}
2 -> {
val currentCall = callManager.getCurrentCall()
if (currentCall != null) {
// resume existing if same room, if not prompt to kill and then restart new call?
if (currentCall.signalingRoomId == roomId) {
onTapToReturnToCall()
}
// else {
// TODO might not work well, and should prompt
// webRtcPeerConnectionManager.endCall()
// safeStartCall(it, isVideoCall)
// }
if (currentCall?.signalingRoomId == roomId) {
onTapToReturnToCall()
} else if (!state.isAllowedToStartWebRTCCall) {
showDialogWithMessage(fragment.getString(
if (state.isDm()) {