Pau*_*one 7 listview signals-slots qml
假设我有一个ListView可点击的委托组件(或GridView或Repeater).这些委托组件需要在触发时发出信号以及自定义数据,这些信息可以由父项获取ListView.如何实现这种信号绑定?
例如,以下代码是我的尝试,但我不知道如何将trigger委托组件的componentTriggered信号绑定到root项目中的信号?
Item {
id: root
anchors.fill: parent
signal componentTriggered(string name)
onComponentTriggered: {
console.log(name + ' component was triggered')
}
ListModel {
id: myModel
ListElement { name: "alpha" }
ListElement { name: "beta" }
ListElement { name: "gamma" }
ListElement { name: "delta" }
}
ListView {
id: myListView
width: 100
height: 600
model: myModel
delegate: TheDelegate { name: model.name }
}
}
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访问 TheDelegate.qml
import QtQuick 2.0
Rectangle {
id: root
width: 100
height: 50
color: "steelblue"
border.color: "white"
border.width: 2
property string name
signal trigger(string name)
Text {
anchors.centerIn: parent
text: model.name
}
MouseArea {
anchors.fill: parent
onClicked: {
console.log(root.name + ' clicked')
root.trigger(root.name)
}
}
}
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您可以在Component.onCompleted处理程序中连接两个信号.
使用您的代码将是这样的:
ListView {
id: myListView
width: 100
height: 600
model: myModel
delegate: TheDelegate {
name: model.name
Component.onCompleted: {
trigger.connect(root.componentTriggered)
}
}
}
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componentTriggered您也可以实现一个功能,而不是调用信号,但这取决于您的要求.无论如何,信号都可以.
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