将QML中的对象的QAbstractListModel派生的listmodel成员绑定为Q_PROPERTY

Question

我想出了如何在QML中公开和绑定QAbstractListModel派生的listmodel的实例.

但我真正想做的是将一个对象暴露给QML并将一个QAbstractListModel派生的listmodel成员绑定为Q_PROPERTY.

我试过这种方式:

class MyObject : public QObject
{
    Q_OBJECT
    Q_PROPERTY(MyListModel myListModel READ myListModel NOTIFY myListModelChanged)

    public:
        explicit MyObject(QObject *parent = 0);
        MyListModel *myListModel();

    signals:
        void myListModelChanged();

    public slots:

    private:
        MyListModel *m_myListModel;

};


MyObject::MyObject(QObject *parent) :
    QObject(parent)
{
    m_myListModel = new MyListModel(this);
}

MyListModel *MyObject::myListModel()
{
    return m_myListModel;
}

class MyListModel : public QAbstractListModel {
    Q_OBJECT
//...   
//...
}

int main(int argc, char *argv[])
{
    QGuiApplication a(argc, argv);

    QQuickView *view = new QQuickView();
    MyObject *myObject = new MyObject();

    view->engine()->rootContext()->setContextProperty("myObject", myObject);    
    view->setSource(QUrl::fromLocalFile("main.qml"));   
    view->show();

    return a.exec();
}

Rectangle {
    width: 200
    height: 200

//...
//...

    ListView {
        id: myListView
        anchors.fill: parent
        delegate: myDelegate
        model: myObject.myListModel
    }
}

但我收到编译错误:

E:\ Qt\Qt5\5.1.1\mingw48_32\include\QtCore\qglobal.h:946:错误:'QAbstractListModel&QAbstractListModel :: operator =(const QAbstractListModel&)'是私有Class&operator =(const Class&)Q_DECL_EQ_DELETE; ^

如何正确地做到这一点？

Answer 1

像QAbstractItemModels这样的QObjects无法复制,必须使用指针.我用的是:

Q_PROPERTY(MyListModel* myListModel READ myListModel CONSTANT)

由于您不替换模型本身,仅替换其内容,因此您不需要myListModelChanged()信号并将其标记为CONSTANT.

你的getter已经有了正确的类型,虽然它应该是const:

MyListModel *MyObject::myListModel() const