mouse area pin mode

MehrdadPilevar

import QtQuick 2.12
import QtQuick.Window 2.12

Window {
	width: 640
	height: 480
	visible: true
	title: qsTr("Hello World")

	Rectangle{
		color: "red"

		width: 300
		height: 300

		x:10
		y:10

		MouseArea{
			anchors.fill: parent
			hoverEnabled: true
			


			onExited: {
				console.log("exit red")
			}
			onEntered: {
				console.log("enter red")
			}
			onMouseXChanged: {

				console.log("x mouse red"+mouseX)
			}
			onMouseYChanged: {
				console.log("y mouse red"+mouseY)

			}
		}

	}

	Rectangle{
		color: "green"

		width: 100
		height: 100

		x:130
		y:10

		MouseArea{
			anchors.fill: parent
			hoverEnabled: true



			onExited: {
				console.log("exit green")
			}
			onEntered: {
				console.log("enter green")
			}

		}
	}

}

In this part, I have two rectangles and inside each rectangle, there is a mouse area
In this part, I want to do the pin mode with the Aria mouse, but the problem is that when the next rectangle, which is located on the lower rectangle, is removed from the Aria mouse, the window can not be opened at the top.

Exactly so that you can use the mouse's position in mouse area when there is a button on mouse area, for example, and that the mouse area is on the button does not allow us to exit the mouse area.

Thanks for the support

flowery

I am not getting your problem. Are you looking for something like this to print x and y when there is a position change.
onPositionChanged: {
console.log("x:"+ mouse.x+ " y:"+mouse.y)
}