[solved]How can I register C++ class to QML?

beidaochuan

I want to add just one command
[qmlRegisterType<QmlConnector>("MyLib", 1, 0, "QmlConnector");]
to "main.cpp" for registering C++ to QML.
Class QMLConnector is used to manager(get) another classes object.
But [QMetaProperty::read: Unable to handle unregistered datatype for property] error happend.
how should i deal with it?
please help me ,thanks.

p3c0

Hi @beidaochuan Is that complete error ? Did it print which datatype and which property ?

beidaochuan

@p3c0
I want to make the source form ① to ② as follows.
①
int main(int argc, char *argv[]){
QApplication app(argc, argv);
qmlRegisterType<A>("MyLib", 1, 0, "A");
qmlRegisterType<B>("MyLib", 1, 0, "B");
qmlRegisterType<C>("MyLib", 1, 0, "C");
QQmlApplicationEngine engine;
engine.load(QUrl(QStringLiteral("qrc:/form/main.qml")));
return app.exec();
}
②
int main(int argc, char *argv[]){
QApplication app(argc, argv);
qmlRegisterType<ALL>("MyLib", 1, 0, "ALL");
QQmlApplicationEngine engine;
engine.load(QUrl(QStringLiteral("qrc:/form/main.qml")));
return app.exec();
}

class ALL is used to get objects of class A, B ,C.

p3c0

@beidaochuan Ok. Not sure. But what approach did you follow to get that error you posted earlier ?

beidaochuan

@p3c0
I just modified the class ALL as follows.

// .h
class ALL : public QObject
{
    Q_OBJECT
public:
    explicit ALL(QObject *parent = 0);

    Q_PROPERTY(A* bmA READ bmA NOTIFY bmAChanged)
    Q_INVOKABLE A *bmA();

    Q_PROPERTY(B* bmB READ bmB NOTIFY bmBChanged)
    Q_INVOKABLE B *bmB();

Q_PROPERTY(C* bmC READ bmC NOTIFY bmCChanged)
    Q_INVOKABLE C *bmA();
    
private:
    A *a;
    B * b;
    C *c;

signals:
    void bmAChanged();
    void bmBChanged();
    void bmCChanged();

public slots:
};

//.cpp

ALL::ALL(QObject *parent) : QObject(parent)
{
    a = new A(this);
    b = new B(this);
    c = new C(this);
}

A *ALL::bmA()
{
    return a;
}

B *ALL::bmB()
{
    return b;
}

C *ALL::bmC()
{
    return c;
}

beidaochuan

@p3c0
when I ran the project, the errors happend.

QMetaProperty::read: Unable to handle unregistered datatype 'A*' for property 'ALL::bmA'
QMetaProperty::read: Unable to handle unregistered datatype 'B*' for property 'ALL::bmB'
QMetaProperty::read: Unable to handle unregistered datatype 'C*' for property 'ALL::bmC'

p3c0

@beidaochuan To make pointers available in Q_PROPERTY you will need to register it first as:

qRegisterMetaType<A*>("A*");

Then since it is declared as a property no need to make the accessor Q_INVOKABLE. So

//In cpp
Q_PROPERTY(A* bmA READ bmA NOTIFY bmAChanged)

A *MyClass::bmA() const
{
    return a;
}

private:
   A* a;
     
//In QML
//Access it as usual property
All {
    id: all
}

console.log(all.bmA) //should get A object.

beidaochuan

@p3c0
Thanks a lot.
That means I must register class A,B,C and class ALL to QML, and just manager A,B,C with class ALL.
In QML I can use like this:
ALL {id: all}
all.bmA
all.bmB
all.bmC

that is right?

p3c0

@beidaochuan Class A,B and C as:

qRegisterMetaType<A*>("A*");
qRegisterMetaType<B*>("B*");

to make pointers available
And class ALL as usual using qmlRegisterType

qmlRegisterType<ALL>("ALL", 1, 0, "ALL");

In QML I can use like this:
ALL {id: all}
all.bmA
all.bmB
all.bmC

that is right?

Right.

beidaochuan

@p3c0
It's very kind of you.
Thank you very much.

p3c0

@beidaochuan Glad that I was helpful :)
Please mark the post as [Solved] if done.

vladstelmahovsky

@beidaochuan take a look here https://forum.qt.io/topic/54576/wrote-short-article-about-dynamically-created-qquickitem-s-with-example-code