background.qml
import QtQuick 1.1
Item {
MouseArea {
id: backgroundMouseArea
anchors.fill: parent
hoverEnabled: true
onPositionChanged: {
console.log("Background")
}
}
}
foreground.qml
import QtQuick 1.1
Item {
Background {
width: 1920
height: 1080
}
MouseArea {
anchors.fill: parent
hoverEnabled: true
onPositionChanged: {
console.log("Foreground")
[mouse.accepted = false] - Not working (as the docs say)
[backgroundMouseArea.onPositionChanged(mouse)] - Not working
}
}
}
我需要在背景和前景项目上执行onPositionChanged
事件。
f.ex。对于onPressed
,我会通过在前景项目中设置mouse.accepted = false
来做到这一点。
我可以手动调用背景项目的onPositionChanged
吗?如果是,我该怎么办?
我不完全确定您在这里要实现的目标。MouseArea
旨在从硬件中获取鼠标事件。如果您真的想从不同的MouseArea
传播鼠标事件到背景中,也许您实际想做的就是给Background
一个简单的property mousePosition
而不是MouseArea
,然后从前景onPositionChanged
处理程序设置该位置。
此外,您的前景代码依赖于背景内部的内部id
参数。这真的很糟糕。考虑背景和前景"类"的"公共API"通常更有用。如果我上面描述的确实是您想做的,那么这就是IMHO的样子:
// Background.qml
import QtQuick 1.1
Rectangle {
// an object with just x and y properties
// or the complete mouseevent, whatever you want
// Use variant for QtQuick 1/Qt4, var for QtQuick 2.0 / Qt5
property variant mousePosition
onMousePositionChanged: console.log(
"Background " + mousePosition.x + " " + mousePosition.y
)
}
//Foreground.qml
import QtQuick 1.1
Item {
// use only ids defined in the same file
// else, someone might change it and not know you use it
Background { id: background }
MouseArea {
anchors.fill: parent
hoverEnabled: true
onPositionChanged: {
console.log("Foreground")
background.mousePosition = {x: mouse.x, y: mouse.y}
}
}
}
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