Question

我正在尝试访问包含这些Person对象的QQmlListProperty的类中的Person对象的名称。我正在努力的部分是从QML文件访问变量。

所有方法和变量与原始人和生日聚会c ++文件相同。使用指向段落下面示例的链接。


我尝试的方法：我添加了Q_Property使其可读，我尝试了多种方法在qml中调用它，包括party1.guests.guest（0）.name。我删除了static和const并将其公开。另外，我在主c ++文件中添加了#include QGuiApplication，QQmlApplicationEngine和QQmlContext，并使用了默认快速应用程序项目提供的第二个引擎。通常，我认为这可能是我对某些地方的指针缺乏了解。

Link to the example！

Link to QQmlListProperty doc！

我的代码将Text对象实现到窗口内部的Main：

import QtQuick 2.12
import QtQuick.Window 2.12
import People 1.0

Window{
    visible: true
    width: 640
    height: 480

BirthdayParty {
        id : party1
        host: Person {
            name: "Bob Jones"
            shoeSize: 12
        }
        guests: [
            Person { name: "Leo Hodges" },
            Person { name: "Jack Smith" },
            Person { name: "Anne Brown" }
        ]


    }

    Text {
        id: name
        text: qsTr(party1.guest(0).name)
    }

}

主要c ++文件的代码：

#include <QQmlComponent>
#include <QDebug>
#include "birthdayparty.h"
#include "person.h"

int main(int argc, char ** argv)
{
    QCoreApplication::setAttribute(Qt::AA_EnableHighDpiScaling);
    QGuiApplication app(argc,argv);
//![register list]
    qmlRegisterType<BirthdayParty>("People", 1,0, "BirthdayParty");
    qmlRegisterType<Person>("People", 1,0, "Person");
//![register list]

//qml engine1
    QQmlApplicationEngine engine1;  
    const QUrl url(QStringLiteral("qrc:/main.qml"));
    QObject::connect(&engine1, &QQmlApplicationEngine::objectCreated,
                     &app, [url](QObject *obj, const QUrl &objUrl) {
        if (!obj && url == objUrl)
            QCoreApplication::exit(-1);
    }, Qt::QueuedConnection);
    engine1.load(url);

//qml engine2
    QQmlEngine engine2;
    QQmlComponent component(&engine2, QUrl("qrc:example.qml"));
    BirthdayParty *party = qobject_cast<BirthdayParty *>(component.create());
    if (party && party->host()) {
        qWarning() << party->host()->name() << "is having a birthday!";
        qWarning() << "They are inviting:";
        for (int ii = 0; ii < party->guestCount(); ++ii)
            qWarning() << "   " << party->guest(ii)->name();
    } else {
        qWarning() << component.errors();
    }



    return app.exec();
}

错误：没有匹配的函数来调用“ BirthdayParty :: guest（）”            情况2：* reinterpret_cast （_ v）= _t-> guest（）;休息;

候选：Person * BirthdayParty :: guest（int）const        人* guest（int）const;                ^ ~~~~   预期有1个arg并提供了0个

候选者：静态Person * BirthdayParty :: guest（QQmlListProperty ，int）        静态Person guest（QQmlListProperty *，int）;                       ^ ~~~~   预期有2个参数，并提供了0个                                                                 ^

如何访问对象的QQmlListProperty并返回其属性

0 个答案: