QmlDesigner: Do not require visual node for states

StateGroup is no visual node.

Change-Id: Ia0ff297efe2b73f42eb24e193d8e832250d77237
Reviewed-by: Qt CI Bot <qt_ci_bot@qt-project.org>
Reviewed-by: Henning Gründl <henning.gruendl@qt.io>
This commit is contained in:
Thomas Hartmann
2022-08-22 17:11:48 +02:00
parent 179713ed5c
commit 3682ff76d2

View File

@@ -274,9 +274,6 @@ QmlModelState QmlModelState::duplicate(const QString &name) const
if (!isValid()) if (!isValid())
throw new InvalidModelNodeException(__LINE__, __FUNCTION__, __FILE__); throw new InvalidModelNodeException(__LINE__, __FUNCTION__, __FILE__);
if (!QmlVisualNode::isValidQmlVisualNode(modelNode().parentProperty().parentModelNode()))
throw new InvalidModelNodeException(__LINE__, __FUNCTION__, __FILE__);
// QmlModelState newState(stateGroup().addState(name)); // QmlModelState newState(stateGroup().addState(name));
QmlModelState newState(createQmlState(view(), {{PropertyName("name"), QVariant(name)}})); QmlModelState newState(createQmlState(view(), {{PropertyName("name"), QVariant(name)}}));
const QList<ModelNode> nodes = modelNode().nodeListProperty("changes").toModelNodeList(); const QList<ModelNode> nodes = modelNode().nodeListProperty("changes").toModelNodeList();