WPF ListBox - Getting UIElement instead of of Sele

问题:

I created a ListBox that has a DataTemplate as Itemtemplate. However, is there an easy way to access the generated UIElement instead of the SelectedItem in codebehind?

When I access SelectedItem, I just get the selected object from my ItemsSource collection. Is there a way to access the UIElement (ie. the element generated from the DataTemplate together with the bound object)?

回答1:

You are looking for the ItemContainerGenerator property. Each ItemsSource has an ItemContainerGenerator instance. This class has the following method that might interest you: ContainerFromItem(object instance).

Once you have a handle to the ListBoxItem, you can go ahead and browse the logical and visual tree. Check out Logical Tree Helper and Visual Tree Helper.

Like Andy said in the comments, just because the item exists in your collection doesn't mean a container has been generated for it. Any kind of virtualizing panel scenario will raise this issue; UIElements will be reused across the different items. Be careful with that as well.

回答2:

siz, Andy and Bodeaker are absolutely right.

Here is how I was able to retrieve the textbox of the listbox's selected item using its handle.

var container = listboxSaveList.ItemContainerGenerator.ContainerFromItem(listboxSaveList.SelectedItem) as FrameworkElement;
if (container != null)
{
    ContentPresenter queueListBoxItemCP = VisualTreeWalker.FindVisualChild<ContentPresenter>(container);
    if (queueListBoxItemCP == null)
        return;

    DataTemplate dataTemplate = queueListBoxItemCP.ContentTemplate;

    TextBox tbxTitle = (TextBox)dataTemplate.FindName("tbxTitle", queueListBoxItemCP);
    tbxTitle.Focus();
}

(Note: Here, VisualTreeWalker is my own wrapper over VisualTreeHelper with various useful functions exposed)