Any way to close popup by button clicking in qml? - qt

Question

I want a button to show/hide popup,like below:
Button{
id:btn
text:"open"
onClicked: {
if(pup.opened)
{
pup.close()
}
else
{
pup.open()
}
}
}
Popup {
id: pup
x: 100
y: 100
width: 200
height: 300
}
Open the popup is ok.
But,when mouse above the button and be clicked, the popup is closed by effects of default closePolicy first.
And then onClicked of button is emitted,so the popup is opened again.
I want it looks like the popup is closed by button clicked(I want keep default closePolicy also).Has any advise?
Qt version:5.15.2

Answer 1

Open the popup is ok. But,when mouse above the button and be clicked,
the popup is closed by effects of default closePolicy first. And then
onClicked of button is emitted,so the popup is opened again. I want it
looks like the popup is closed by button clicked(I want keep default
closePolicy also). Has any advise?
It's impossible because the default policy value is Popup.CloseOnEscape | Popup.CloseOnPressOutside. In your case, the "open" button is outside the popup area and that's why it breaks policy. You have two solutions:
I. Change policy:
Button{
id:btn
text:"open"
z: 100
onClicked: {
if(pup.opened)
{
pup.close()
}
else
{
pup.open()
}
}
}
Popup {
id: pup
x: 100
y: 100
width: 200
height: 300
closePolicy: Popup.NoAutoClose
}
II. Fake behaviour:
Use the "open" button only to open the dialogue and close it by tapping outside the popup. You can disable the button when the popup is opened and it should fake the behaviour which you want. Additionally, you can customise the button and the style can be the same for enabled and disabled states.
Button{
id:btn
text:"open"
z: 100
enabled: !pup.opened
onClicked: {
if(pup.opened)
{
pup.close()
}
else
{
pup.open()
}
}
}
Popup {
id: pup
x: 100
y: 100
width: 200
height: 300
}