C ++和QML：将QML信号连接到C ++插槽

Question

我无法在以下代码中使用信号连接。 我特别想通过将信号连接到cpp插槽而不设置上下文来做到这一点。 我想问题是

 item->findChild<QObject*>("signalItem");

没有找到合适的对象？ 这里是相关的代码文件：

main.cpp中：

#include <QGuiApplication>
#include <QQmlApplicationEngine>
#include <QQmlContext>
#include "include/myclass.h"

int main(int argc, char *argv[])
{
    QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling);
    QGuiApplication app(argc, argv);

    QQmlApplicationEngine engine;
    engine.load(QUrl(QLatin1String("qrc:/main.qml")));
    QObject * item = engine.rootObjects().value(0);
    QObject * myObject= item->findChild<QObject*>("signalItem");
    MyClass myClass;
    QObject::connect(item, SIGNAL(testSignal()),&myClass,SLOT(cppSlot()));

    return app.exec();
}

main.qml：

import QtQuick 2.7
import QtQuick.Controls 2.0
import QtQuick.Layouts 1.0

ApplicationWindow {
    visible: true
    width: 800
    height: 460

    Page1 {
        id: page1
        visible: true
    }
}

Page1.qml：

import QtQuick 2.7
import QtQuick.Window 2.2

Item {
    width: 800
    height: 460
    id: signalItem
    objectName: "signalItem"
    signal testSignal()

    CustomButton {
        id: cppSignalButton
        x: 14
        y: 55
        buttonText: "Test CPP Signal"
        onButtonClicked: {
            signalItem.testSignal();
        }
     }
}

Answer 1

因为您要连接item （main.qml）而不是myObject

如果这样做，它将起作用：

QObject::connect(myObject, SIGNAL(testSignal()),&myClass,SLOT(cppSlot()));

实际上，您还应该添加检查，以确保从该函数返回的值不为null：

QObject * item = engine.rootObjects().value(0);
QObject * myObject= item->findChild<QObject*>("signalItem");