如何在QML ApplicationWindow中获取activeFocusControl的className

赋料扬雄敌，诗看子建亲。这篇文章主要讲述如何在QML ApplicationWindow中获取activeFocusControl的className相关的知识，希望能为你提供帮助。
我试试

ApplicationWindow { onActiveFocusControlChanged: { console.log(activeFocusControl) console.log(activeFocusControl.objectName) } }

输出：

qml: QQuickTextField(0xa6ec00)//the 'activeFocusControl' qml://the 'activeFocusControl.objectName' qml: QQuickButton(0xd7ccb0) qml:

我想要

onActiveFocusControlChanged: { if (activeFocusControl.className == "QQuickTextField") { //do something } else if (activeFocusControl.className == "QQuickButton") { //do something }

但是“className”方法不存在，所以我怎么做呢？
对不起，我的英语是游泳池，谢谢
答案没有方法可以从qml访问className，因此可能的解决方案是从c ++创建一个帮助程序，如下所示：

#include < QGuiApplication> #include < QQmlApplicationEngine> #include < QQmlContext> #include < QObject> class Helper : public QObject { Q_OBJECT public: Q_INVOKABLE QString getClassName(QObject *obj) const{ return obj? obj-> metaObject()-> className(): ""; } }; #include "main.moc"int main(int argc, char *argv[]) { #if defined(Q_OS_WIN) QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling); #endifQGuiApplication app(argc, argv); Helper helper; QQmlApplicationEngine engine; engine.rootContext()-> setContextProperty("helper", & helper); engine.load(QUrl(QStringLiteral("qrc:/main.qml"))); if (engine.rootObjects().isEmpty()) return -1; return app.exec(); }

【如何在QML ApplicationWindow中获取activeFocusControl的className】那么它在QML方面使用：

import QtQuick 2.9 import QtQuick.Window 2.2 import QtQuick.Controls 2.3ApplicationWindow { visible: true width: 640 height: 480 title: qsTr("Hello World")Button { id: button x: 270 y: 47 text: qsTr("Button") }TextField { id: textField x: 220 y: 169 text: qsTr("Text Field") }onActiveFocusControlChanged:{ var className = helper.getClassName(activeFocusControl) switch(className){ case "QQuickTextField": console.log("I am QQuickTextField") break case "QQuickButton": console.log("I am QQuickButton") break default: console.log("empty") } } }

完整的例子可以在以下link找到。