The SelectionModelis the right approach. You can get the default from your TabPane or assign your own implementation by using setSelectionModel(…). The default model should be good enough for the beginning. SingleSelectionModel<Tab> selectionModel = tabPane.getSelectionModel(); Once you stored it in some local variable, you have different options to select a tab. selectionModel.select(tab); //select by object … Read more
Right, got it sorted. I kept trying to cast it like IList<PuzzleViewModel> collection = (IList<PuzzleViewModel>)param; Which told me it couldn’t convert from SelectedItemCollection to IList… This is in fact what you need to do. System.Collections.IList items = (System.Collections.IList)param; var collection = items.Cast<PuzzleViewModel>();
You can use the SelectedItem property to get the currently selected object, which you can then cast into the correct type. For instance, if your DataGrid is bound to a collection of Customer objects you could do this: Customer customer = (Customer)myDataGrid.SelectedItem; Alternatively you can bind SelectedItem to your source class or ViewModel. <Grid DataContext=”MyViewModel”> … Read more
I realise this has already had an answer accepted, but I put this together to solve the problem. It uses a similar idea to Delta’s solution, but without the need to subclass the TreeView: public class BindableSelectedItemBehavior : Behavior<TreeView> { #region SelectedItem Property public object SelectedItem { get { return (object)GetValue(SelectedItemProperty); } set { SetValue(SelectedItemProperty, … Read more
Their names can be a bit confusing :). Here’s a summary: The SelectedItem property returns the entire object that your list is bound to. So say you’ve bound a list to a collection of Category objects (with each Category object having Name and ID properties). eg. ObservableCollection<Category>. The SelectedItem property will return you the currently … Read more