What condition should i pass, to show the popup again - qml

Question

I implemented a feature where after pressing the back button twice then it will show a popup and that popup consist of tow buttons i.e yes and cancel, so when you press yes it closes the app.
My issue is when i click on cancel button then popup closes, and after that when i again press the back button it closes the app instead of showing the pop-up again. So what can I do? What condition should I pass?( It's in pure qml)
here is what i have done
Popup {
visible: false
id: popup
background: Rectangle {
width: 300
height: 200
}
Row{
topPadding: 10
Button {
text: qsTr("YES")
onClicked:
{
Qt.quit();
}
Button{
text: qsTr("cancel")
height:40
width:100
onClicked:
{
popup.close();
}
}
}
}
modal: true
focus: true
}
Keys.onBackPressed: {
timer.pressBack()
}
Timer{
id: timer
property bool backPressed: false
repeat: false
interval: 300//ms
onTriggered: backPressed = false
function pressBack(){
if(backPressed){
timer.stop()
backPressed = false
popup.open();
}
else{
backPressed = true
timer.start()
}
}
}

Answer 1

I guess this happens, because you're calling the close method directly and qml instead of closing the popup calls the close function of the main window instead (you didn't post the code for the leaveApp function so I can't tell for sure if this is what happens in your case). You also definitely don't need the call to leaveApp() after popup.open()
Here is the sample I used to test this. It should work as expected:
Keys.onReleased: {
if (event.key == Qt.Key_Back) {
event.accepted = true
timer.pressBack()
}
}
Timer{
id: timer
property bool backPressed: false
repeat: false
interval: 300//ms
onTriggered: backPressed = false
function pressBack(){
if(backPressed){
timer.stop()
backPressed = false
popup.open();
//leaveApp()
}
else{
backPressed = true
timer.start()
}
}
}
Popup {
id: popup
x: 100
y: 100
width: 200
height: 300
modal: true
focus: true
Row {
Button {
text: "ok"
onClicked: {
console.log("leave app")
mainWindow.close()
}
}
Button {
text: "cancel"
onClicked: {
console.log("close popup")
popup.close() //you have to call close method of popup explicitly (only close() will cause the issue)
}
}
}
}