从QML访问一个属性c++对象



我在理解如何使用qml中的c++单例对象时遇到问题。我知道我必须从QObject类继承我的类,并且我必须公开这些属性。为了使它们在qml中可用,我必须执行setContextProperty("ClassName","类名"(。

然而,如果我们承认这个类包含另一个对象,并且我希望能够从qml中使用它,我会得到类似"的错误;不能调用方法名称";来自未定义的对象。

示例:

APi.h

class API : public QObject {
Q_OBJECT
Q_PROPERTY(User *user READ getUser WRITE setUser NOTIFY userChanged)
public:
Q_INVOKABLE inline User *getUser() const {
qDebug() << "get called";
return user;
};
inline void setUser(User *new_user) { this->user = new_user; };
signals:
void userChanged();
private:
User *user;
};

用户.h

#ifndef USER_H
#define USER_H
#include <QObject>
#include <QString>
class User: public QObject{
Q_OBJECT
public:
Q_INVOKABLE inline QString &getName(){return name;}; // qrc:/main.qml:63: TypeError: Cannot call method 'getName' of undefined
private:
QString name;
};
#endif // USER_H

main.cpp

#include <QQmlContext>
#include <QGuiApplication>
#include <QQmlApplicationEngine>
#include <QtQuick3D/qquick3d.h>
#include "API.h"
#include "User.h"
int main(int argc, char *argv[]) {
QGuiApplication app(argc, argv);
QQmlApplicationEngine engine;
API api;
User user;
api.setUser(&user);
engine.rootContext()->setContextProperty("API", &api);
qmlRegisterType<User>("User", 1, 0, "User");
engine.load(QUrl(QStringLiteral("qrc:/main.qml")));
if (engine.rootObjects().isEmpty())
return -1;
return app.exec();
}

main.qml

import QtQuick
import QtQuick3D
import QtQuick.Controls
import QtQuick.Layouts
Window {
Item{
id: test
Component.onCompleted: {
console.log("Completed")
API.getUser() // no error, getUser is called without error
console.log(API.getUser().getName())
}
}
}

我有什么可能通过API从qml访问User对象?

您可以执行以下任何操作:

在API类中添加公共插槽:

API.cpp:
QString API::getUserName()
{
return user.getName();
}
main.qml:
Component.onCompleted: console.log( API.getUserName() )

在API类中将用户设为Q_PROPERTY:

API.h:
Q_PROPERTY( User* user READ user NOTIFY userChanged)
main.qml:
Component.onCompleted: console.log( API.user.getName() )

使用QML:注册用户

main.cpp:
qmlRegisterType<User>("MyUser", 1, 0, "MyUser");
main.qml:
Component.onCompleted: console.log( API.getUser().getName() )

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