Currently Kuix works so that regardless whether another system menu is showing, it will show/execute the other menu when that soft key is pressed. I'd prefer a way to hide the other menu when the soft key is first pressed and then execute the action when it is pressed again..
What I mean by this is that (at least on Nokia phones) the normal way of operation is that the right soft key always cancels the current action. So if a left system menu is showing and I push the right soft key, the left menu closes. And only after I press it a second time, the action associated with the right menu is executed.
It's very distracting in my application, because the right soft key is bound to quit on the main screen and when I'm showing the left system menu I automatically press right soft key to get rid of it. But instead now it executes the quit - action.
I managed to get the behavior I wanted by making the following modifications to FocusManager:
// Soft key right
case KuixConstants.KUIX_KEY_SOFT_RIGHT: {
Screen screen = rootWidget.getDesktop().getCurrentScreen();
if (screen != null) {
Menu firstmenu = screen.getFirstMenu();
boolean process = true;
if (firstmenu.getPopup() != null && firstmenu.getPopup().isInWidgetTree()) {
firstmenu.hidePopup();
process = false;
}
if (process) {
Menu secondmenu = screen.getSecondMenu();
if (secondmenu != null) {
secondmenu.processActionEvent();
}
}
return true;
}
break;
}
But I guess it would be prefereable to make this behavior optional.. So any chance of getting this feature to Kuix ?

