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QUuid cannot be constructed from string in Qt6
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nyalldawson committed Jul 20, 2021
1 parent 87e2653 commit ea43f6bfd4185dc4a2869d878ab4d9c959078dcd
Showing with 2 additions and 2 deletions.
  1. +1 −1 src/gui/attributetable/qgsattributetableview.cpp
  2. +1 −1 src/gui/attributetable/qgsdualview.cpp
@@ -456,7 +456,7 @@ void QgsAttributeTableView::actionTriggered()

if ( action->data().toString() == QLatin1String( "user_action" ) )
{
mFilterModel->layer()->actions()->doAction( action->property( "action_id" ).toString(), f );
mFilterModel->layer()->actions()->doAction( action->property( "action_id" ).toUuid(), f );
}
else if ( action->data().toString() == QLatin1String( "map_layer_action" ) )
{
@@ -897,7 +897,7 @@ void QgsDualView::viewWillShowContextMenu( QMenu *menu, const QModelIndex &maste
}

menu->addSeparator();
QgsAttributeTableAction *a = new QgsAttributeTableAction( tr( "Open Form" ), this, QString(), rowSourceIndex );
QgsAttributeTableAction *a = new QgsAttributeTableAction( tr( "Open Form" ), this, QUuid(), rowSourceIndex );
menu->addAction( tr( "Open Form…" ), a, &QgsAttributeTableAction::featureForm );
}

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