Is it possible to pass an instance of a derived `QAbstractListModel` from backend to QML without using `setContextProperty`?

Curtwagner1984

Hello,

I'm trying to figure out how to pass an instance of a model to QML from the backend, but I don't want to use setContextProperty to expose said model.
Because I want to dynamically change the model the view is using.

I know that I can register the model as a QML type and then instantiate it on the QML side, but doing this will lock me out of changing the model data from the backend,
because as far as I know, the backend doesn't have references to QML instantiated classes.

I tried registering QAbstractListModel as a QML type and then pass it as a return value of a slot, like so:

    @Slot(result=VideoModel)
    def get_video_model(self):
        return self.video_model

And in QML

    Shortcut {
        id: getModelShortuct
        sequences: ["Ctrl+W"]
        onActivated: {
            console.log("getModelShortuct activated")
            console.log("bridge.get_video_model()= " + bridge.get_video_model())
        }
    }

Where bridge is the class that's exposed through setContextProperty .
When trying this, I get "Error: Unknown method return type: QAbstractListModel* ". This also happens when registering the video model class specifically as a QML type, not just QAbstractListModel

Alternatively, I wonder if it's possible to expose a class through setContextProperty and have one of its class variables be a model and then change the value of the said class variable. For example

class ModelWrapper(QObject):

    def __init__(self):
    super().__init__()
    self.model = SomeModel()

    def switch_model(self):
        self.model = AnotherModel()

And in QML:

GridView{
    id:myGridview
    model:ModelWrapper.model
}

Button{
    id:modelSwitchButton
    onClicked{
        ModelWrapper.switch_model()
    }
}

Curtwagner1984

So, answering my own question:

Do you mean like it's described here?

Yes, it is indeed how it's described here.

Also, do SomeModel and AnotherModel needs to be registered as a QML type, or is it enough to register QAbstractListModel if both of them derive from it?

No. Apparently, you need to specify that the 'return' type is just a QObject neither QAbstractListModel nor its derived classes need to be registered.

This is what I ended up doing to test this:


class CurrentView(QObject):

    def __init__(self):
        super().__init__()
        self.person_model: PersonModel = PersonModel()
        self.video_model: VideoModel = VideoModel()
        self.search_backend: SearchBackend = SearchBackend(self.person_model, self.video_model)
        self._current_model = self.person_model # setting current model to this for test
        self.search_backend.video_search("", "Name", False) #fills models with data

    def _current_model(self):
        return self._current_model

    @Signal
    def current_model_changed(self):
        pass

    @Slot()
    def swap_model(self): #called from QML
        logger.info(f"Swapping Model...")
        if self._current_model == self.video_model:
            self._current_model = self.person_model
        else:
            self._current_model = self.video_model
        self.current_model_changed.emit()

    #the return type needs to be QObject.
    #At first I tried QAbstractListModel or it's derived classes. Didn't work.
    current_model = Property(QObject, _current_model, notify=current_model_changed)

In QML:

Shortcut {
    id: swapModels
    sequences: ["Ctrl+W"]
    onActivated: {
        console.log("Swapping Models...")
        current_view.swap_model()
    }
}

GridView {
    id: gridview
    anchors.fill: parent
    model: current_view.current_model
    delegate: MyDelegate{}

}

CurrentView is exposed through:

    cw = CurrentView()
    engine.rootContext().setContextProperty("current_view", cw)

It works like a charm, when I press Ctrl+W the model instantly switches.

This SO answer helped me figure out the missing pieces. Thank you @raven-worx for pointing me in the right direction.

raven-worx

@Curtwagner1984
sure, if ModelWrapper.model is a property with a notifier signal

Curtwagner1984

Thank you for the reply.

Do you mean like it's described here?

Also, do SomeModel and AnotherModel needs to be registered as a QML type, or is it enough to register QAbstractListModel if both of them derive from it?

Curtwagner1984

So, answering my own question:

Do you mean like it's described here?

Yes, it is indeed how it's described here.

Also, do SomeModel and AnotherModel needs to be registered as a QML type, or is it enough to register QAbstractListModel if both of them derive from it?

No. Apparently, you need to specify that the 'return' type is just a QObject neither QAbstractListModel nor its derived classes need to be registered.

This is what I ended up doing to test this:


class CurrentView(QObject):

    def __init__(self):
        super().__init__()
        self.person_model: PersonModel = PersonModel()
        self.video_model: VideoModel = VideoModel()
        self.search_backend: SearchBackend = SearchBackend(self.person_model, self.video_model)
        self._current_model = self.person_model # setting current model to this for test
        self.search_backend.video_search("", "Name", False) #fills models with data

    def _current_model(self):
        return self._current_model

    @Signal
    def current_model_changed(self):
        pass

    @Slot()
    def swap_model(self): #called from QML
        logger.info(f"Swapping Model...")
        if self._current_model == self.video_model:
            self._current_model = self.person_model
        else:
            self._current_model = self.video_model
        self.current_model_changed.emit()

    #the return type needs to be QObject.
    #At first I tried QAbstractListModel or it's derived classes. Didn't work.
    current_model = Property(QObject, _current_model, notify=current_model_changed)

In QML:

Shortcut {
    id: swapModels
    sequences: ["Ctrl+W"]
    onActivated: {
        console.log("Swapping Models...")
        current_view.swap_model()
    }
}

GridView {
    id: gridview
    anchors.fill: parent
    model: current_view.current_model
    delegate: MyDelegate{}

}

CurrentView is exposed through:

    cw = CurrentView()
    engine.rootContext().setContextProperty("current_view", cw)

It works like a charm, when I press Ctrl+W the model instantly switches.

This SO answer helped me figure out the missing pieces. Thank you @raven-worx for pointing me in the right direction.

GrecKo

QAbstractListModel as a property type or a return type should work, using QObject instead is not the correct solution.
Have you tried calling qRegisterMetaType?

raven-worx

@GrecKo said

using QObject instead is not the correct solution

and why?
Context properties also dont't require a type registration and still can be used as models. Since the cast / type-check anyway happens on the qobject instance.

GrecKo

Because you lose the type information than can be used by the QML engine/compiler and Qt Creator.
Also root context properties should not be used anymore, access lookup is heavy for them.

Curtwagner1984

@GrecKo said in Is it possible to pass an instance of a derived `QAbstractListModel` from backend to QML without using `setContextProperty`?:

Have you tried calling qRegisterMetaType?

No, I have not. I tried

qmlRegisterType(QAbstractListModel, "com.videoModel", 1, 0, "VideoModel")

Curtwagner1984

@GrecKo Could you elaborate on how to use qRegisterMetaType? And on what you mean by

root context properties should not be used anymore, access lookup is heavy for them.