Quit Application if another Window hides it (X11)

benmao

I have a small qml application which displays an animation on the screen if there is an inactivity by the user for some time.
This application waits for "onClick" (we have a touchscreen only) and exits.
But if another application is started, the application screen is not visible anymore and it never receives any mouseclick. In this case, it should also exit.

But I don't know how how. I have tried to use some onXXX functions. The problem starts already that I cannot find a documentation which "onXXX" functions are already available.

The application:

import QtQuick 2.0

Item {
width: 1024
height: 600
MouseArea {
onClicked: {
console.log("onClicked");
Qt.quit();
}
anchors.fill: parent
}
}

AnimatedSprite {
id: sprite
width: 1024
height: 600
anchors.centerIn: parent
source: "content/awibuben_spritesheet.png"
frameCount: 5
frameSync: true
frameWidth: 1024
frameHeight: 600
}
}

I have also tried to use "Window" where I can receive "onActiveChanged", but the event comes only at the beginning but not when another fullscreen window is placed above this one.
I use X11 with matchbox display manager.

Any solutions?

Not so nice, but it would also help to keep the application always in the foreground.

benmao

I have found a solution:

The app exits, if the touch screen is pressed or if another application is started. Just as I wanted.

import QtQuick 2.2
import QtQuick.Window 2.1

Item {

Window { id: mainwindow
width: 1024
height: 600
visible: true
visibility: Window.Maximized

modality: Qt.ApplicationModal
// if it should stay on top:   flags: Qt.SplashScreen


MouseArea {
    onClicked: Qt.quit();
    anchors.fill: parent
}

onActiveChanged: {
     console.log("onActiveChanged");
     if (active == 0) Qt.quit();
}

AnimatedSprite {
id: sprite
width: 1024
height: 600
anchors.centerIn: parent
source: "content/awibuben_spritesheet.png"
frameCount: 5
frameSync: true
frameWidth: 1024
frameHeight: 600
}
}
}