[solved]Testing ApplicationWindow.visibility enum values?

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Very basic question:

I am playing around with making a routine to toggle a QtQuick app fullscreen (code below) and find that I can't test the enum QWindow::Visibility by the const name but only by the value. i.e. I can't do this: if ( window.visibility == QWindow.FullScreen)

I must be doing it wrong but I've tried a few syntax variations.

(alternately, is there a cleaner way of doing what I am trying to do?)

import QtQuick.Controls 1.4

ApplicationWindow {
    id: window
    visible: true
    visibility: "Windowed"
    width: 1024
    height: 768

    Button {
         id: button
        text: "FullScreen"
        onClicked: toggleFullScreen()
    }

    function toggleFullScreen(){
          console.log(window.visibility)

       if ( window.visibility == 5){
           window.visibility = "Windowed"
           button.text = "FullScreen"

       } else if ( window.visibility == 2){
           window.visibility = "FullScreen"
           button.text = "Windowed"
       }
    }
}

p3c0

Hi @TOMATO_QT
You can use Window.FullScreen or Window.Windowed. Just import QtQuick.Window 2.0 or greater.

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@p3c0

Hey thanks! Works as expected.
Another baby step on the road to writing more interesting bugs. . . ;-)