Accessing QAbstractListModel item in Qml without using a ListView - qt

Question

I have subclassed QAbstractListModel in order to have a model on the qml side. I can easily use this model in ListViews and other similar components that deal with models, however, I can't access it directly. This is what I am trying without success:
myModel[0].name // TypeError: Cannot read property 'name' of undefined
Is this possible? Am I using the wrong syntax?

Answer 1

You can access generic model (based on QAbstractListModel) easily when you use DelegateModel model as a mediator.
import QtQuick 2.2
import QtQml.Models 2.2
DelegateModel {
id: delegateModel
}
MyModel {
id: myModel
onDataLoaded: {
delegateModel.model = myModel;
for (var row = 0; row < myModel.rowCount(); row++) {
var item = delegateModel.items.get(row).model;
console.log(" name " + row + ":" + item.name);
}
}
}

Answer 2

(1) You're mixing up roles and properties.
Your model implements roles which are used to feed a delegate in a view. In your code, you try to access a property.
(2) A model is not an array, where you can access rows by the [] operator.
But you can write a getter function to achieve exactly that:
class ConversationListModel : public QAbstractListModel
{
Q_OBJECT
public:
enum ConversationRoles {
IdRole = Qt::UserRole, // 256
NameRole,
};
explicit ConversationListModel(QObject *parent = 0);
QHash<int, QByteArray> roleNames() const;
QVariant data(const QModelIndex &index, int role = Qt::DisplayRole) const;
Q_INVOKABLE ConversationModel* get(quint32 conversationId) const;
signals:
// ...
Than you need a data type that represents a row, e.g. ConversationModel. Now you can access a row using
myModel.get(0).name
As stated in (1), you now need to give ConversationModel a property called name
public:
explicit ConversationModel(QObject *parent = 0);
Q_PROPERTY(QString name READ name NOTIFY nameChanged)
QString name() const;

Answer 3

To make the (correct) solution of Simon Warta work, you have to register ConversationModel as QML type. For example in your main you have to state
qmlRegisterType<ConversationModel>("ConversationModel", 1, 0, "ConversationModel");
Then in your qml file you have to import this type with
import ConversationModel 1.0
Afterwards everything works as expected and you can use
myModel.get(0).name
in your qml code, given that myModel is of type ConversationListModel and was registered using QDeclarativeContext::setContextProperty.