Question

我想自动在列表视图中显示所选项目（如果不滚动则无法显示所有项目。）

this.listView.SelectedIndex = 999;当然选择一个项目，但它没有显示它。

我可以用什么来自动显示它？

亲切的问候，杰夫

Answer 1

看看这一个：
Scroll WPF Listview to specific line

Answer 2

这可能会对您有所帮助，我不确定您是否正在寻找它，但它会将所选项目放入视图中并在必要时滚动到您的位置。

 int selectedIndex = listView.Items.IndexOf((listView.SelectedItems[0]))

 listView.Items[selectedIndex].EnsureVisible();

Answer 3

安装nuget软件包System.Windows.Interactivity.WPF，创建一个如下所示的类：

public class ScrollToSelectedListBoxItemBehaviour: Behavior<ListBox>
{
    protected override void OnAttached()
    {
        base.OnAttached();
        AssociatedObject.SelectionChanged += AssociatedObjectOnSelectionChanged;
        AssociatedObject.IsVisibleChanged += AssociatedObjectOnIsVisibleChanged;
    }

    protected override void OnDetaching()
    {
        AssociatedObject.SelectionChanged -= AssociatedObjectOnSelectionChanged;
        AssociatedObject.IsVisibleChanged -= AssociatedObjectOnIsVisibleChanged;
        base.OnDetaching();
    }

    private static void AssociatedObjectOnIsVisibleChanged(object sender, DependencyPropertyChangedEventArgs e)
    {
        ScrollIntoFirstSelectedItem(sender);
    }

    private static void AssociatedObjectOnSelectionChanged(object sender, SelectionChangedEventArgs e)
    {
        ScrollIntoFirstSelectedItem(sender);
    }

    private static void ScrollIntoFirstSelectedItem(object sender)
    {
        if (!(sender is ListBox listBox)) 
            return;
        var selectedItems = listBox.SelectedItems;
        if (selectedItems.Count > 0)
            listBox.ScrollIntoView(selectedItems[0]);
    }
}

将此行为类添加到xaml：

<ListView ItemsSource="{Binding Items}">
    <i:Interaction.Behaviors>
        <behaviors:ScrollToSelectedListBoxItemBehaviour />
    </i:Interaction.Behaviors>
</ListView>

WPF ListView显示所选项目

3 个答案: