导入到qml中的枚举是否不可以有命名空间 [复制链接]

如下所示定义的枚举 SPC::Klass::Type, 在QML中调用createObject()时，报错不知道参数 SPC::Klass::Type是什么，请问枚举要想导入QML是不是不能带名字空间？
namespace SPC {
     class Klass : public QObject
     {

Q_OBJECT
Q_EMUS(Type)

     public:

enum Type
{
      type1
}

Q_INVOKABLE Q_QObject * createObject(SPC::Klass::Type type) {...}

     }
}

没有人这样用过吗？

普通的类的枚举 是可以在QML中使用的，带有命名空间的没有试过，或者也是可以的

类似
#include <QtCore>

class DeviceModel: public QObject {
    Q_OBJECT
    Q_ENUMS(DeviceModelEnum)

public:
    enum DeviceModelEnum {
        Z10 = 1, Q10 = 2, Q5 = 3
    };
};

加载qml的地方
#include <devicemodel.hpp>
qmlRegisterType<DeviceModel>("ekkes_enums", 1, 0, "DeviceModel");


qml中：
import ekkes_enums 1.0

Page {
    Container {
        DropDown {
            Option {
                text: "Z10"
                value: DeviceModel.Z10
            }
            Option {
                text: "Q10"
                value: DeviceModel.Q10
            }
            Option {
                text: "Q5"
                value: DeviceModel.Q5
            }
            onSelectedIndexChanged: {
                deviceLabel.deviceModel = selectedValue
            }
        }
        Label {
            text: "Hello ENUM"
            textStyle.base: SystemDefaults.TextStyles.BigText
        }
        Label {
            text: "please select from DropDown"
            textStyle.base: SystemDefaults.TextStyles.SmallText
        }
        Label {
            id: deviceLabel
            property int deviceModel
            onDeviceModelChanged: {
                switch (deviceModel){
                    case DeviceModel.Z10:
                        text = "Touch Device"
                        break
                    case DeviceModel.Q10:
                    case DeviceModel.Q5:
                        text = "Keyboard Device"
                        break
                    default:
                        text = "unknown"
                }
            }
        }
    }
    onCreationCompleted: {
        console.debug("Enum value "+ DeviceModel.Q10)
        deviceLabel.deviceModel = -1
    }
}

不加命名空间的确可以，加了qml就认不出来了，用整形数可以处理，不知道qt官方是否会改进