Question

当我手动更改currentIndex的{{1}}时，ListView在设置为给定值之前设置为0。主要问题是currentIndex被触发2次：值为0，然后值为2.

请参阅以下代码：

onCurrentIndexChanged

我获得的日志：

import QtQuick 2.8
import QtQuick.Controls 2.1

ApplicationWindow {
    visible: true
    width: 100
    height: 100

    ListView {
        id: list
        anchors.fill: parent
        currentIndex: -1
        model: ListModel {}
        delegate: Text {
            text: model.name
        }

        Component.onCompleted: {
            model.append({"name": "plip"})
            model.append({"name": "plop"})
            model.append({"name": "ploup"})
            currentIndex = 2
        }

        onCurrentIndexChanged: {
            console.log("currentIndex", currentIndex)
        }

    }
}

我发现避免此问题的唯一解决方案是拥有另一个属性qml: currentIndex 0 qml: currentIndex 2并将indexToBeSet与实际indexToBeSet进行比较：

currentIndex

日志：

import QtQuick 2.8
import QtQuick.Controls 2.1

ApplicationWindow {
    visible: true
    width: 100
    height: 100

    ListView {
        id: list

        property int indexToBeSet: -1

        anchors.fill: parent
        currentIndex: -1
        model: ListModel {}
        delegate: Text {
            text: model.name
        }

        Component.onCompleted: {
            model.append({"name": "plip"})
            model.append({"name": "plop"})
            model.append({"name": "ploup"})
            setCurrentIndex(2)
        }

        onCurrentIndexChanged: {
            console.log("currentIndex", currentIndex)
            if (currentIndex === indexToBeSet) {
                console.log("=> do some stuff")
            }

        }

        function setCurrentIndex(index) {
            indexToBeSet = index
            currentIndex = index
        }
    }
}

但我认为这是一个非常讨厌的解决方案。

为什么会这样？这是一个错误还是预期的行为？有没有更好的方法来避免这个问题？

更多详情：

看起来问题只发生在我在前一个qml: currentIndex 0 qml: currentIndex 2 qml: => do some stuff等于currentIndex时添加元素后更改currentIndex时。请参阅以下代码：

-1

以及相应的日志：

Component.onCompleted: {
    model.append({"name": ""})
    model.append({"name": ""})
    model.append({"name": ""})
    currentIndex = 1 // WILL BE SET TO 0 BEFORE 1
    model.append({"name": ""})
    currentIndex = 2 // no problem because currentIndex was not -1
    currentIndex = -1 // no problem
    currentIndex = 2 // no problem because we did not add new elements after setting to -1
    currentIndex = -1 // no problem
    model.append({"name": ""})
    currentIndex = 1 // WILL BE SET TO 0 BEFORE 1
}

QML ListView currentIndex错误地设置为0

0 个答案: