How can I call a pointer of an QObject in a QML Fi

2019-05-10 09:14发布

I'm developping an application for Blackberry 10 and I'm using the Momentics IDE (BB native SDK).

I get the following error when I try to execute the code below. Any one have an idea how can I fix this ?

Error

QMetaProperty::read: Unable to handle unregistered datatype 'DataHandler*' for property 'xxx::dataHandler'
asset:///main.qml:104: TypeError: Result of expression 'xxx.dataHandler' [undefined] is not an object.
Process 627863799 (xxx) terminated SIGSEGV code=1 fltno=11 ip=08055b30(/accounts/1000/appdata/com.ddd.xxx.testDev_e_xxx45b0f435/app/native/xxx@main+0x5d63) ref=006e0075

** xxx.hpp **

class xxx: public QObject {
Q_OBJECT

Q_PROPERTY(DataHandler* dataHandler READ getDataHandler)

public:
    xxx(bb::cascades::Application *app);
    virtual ~xxx() {
    }

    Q_INVOKABLE
    DataHandler* getDataHandler() const;

private:
    DataHandler* m_dataHandler;
}

** xxx.cpp **

xxx::xxx(bb::cascades::Application *app) :
        QObject(app) {
m_dataHandler = new DataHandler();
}

** QML file **

Button {
         id: backBtn
         objectName: "backBtnObject"
         text: qsTr("Back") + Retranslate.onLocaleOrLanguageChanged
         preferredWidth: backBtn.text.length
         visible: false
         onClicked: {
              xxx.dataHandler.displayLicencesList();
         }
 }

2条回答
虎瘦雄心在
2楼-- · 2019-05-10 09:51

DataHandler is unknown type.

Q_PROPERTY docu says:

The type can be any type supported by QVariant

I'd try to encapsulate your DataHandler* in QVariant and make the property of type QVariant.

You could create the value containing your DataHandler pointer e.g like this:

const int DATA_HANDLER = QVariant::UserType +1;
//...
QVariant dataHandler( DATA_HANDLER, myDataHandlerPointer );
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放荡不羁爱自由
3楼-- · 2019-05-10 09:53

You need to register your pointer type like this

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

(source)

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