wpf 虚拟树获取listbox的usercontrol
作者:互联网
<Window x:Class="WpfExpand.MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
xmlns:d="http://schemas.microsoft.com/expression/blend/2008"
xmlns:mc="http://schemas.openxmlformats.org/markup-compatibility/2006"
xmlns:local="clr-namespace:WpfExpand"
mc:Ignorable="d"
Title="MainWindow" Height="450" Width="800">
<Grid>
<StackPanel Orientation="Horizontal">
<ListBox x:Name="d2" >
<ListBox.ItemTemplate>
<DataTemplate>
<StackPanel Name="p1" Margin="10" DockPanel.Dock="Right">
<TextBlock Text="{Binding Name}" FontWeight="Bold" FontSize="18" />
<TextBlock Text="{Binding age}" Margin="0,4,0,0" FontSize="14"
Foreground="#c6de96" TextWrapping="WrapWithOverflow" />
</StackPanel>
</DataTemplate>
</ListBox.ItemTemplate>
</ListBox>
</StackPanel>
</Grid>
</Window>
this.Dispatcher.InvokeAsync(new Action(delegate
{
ListBoxItem myListBoxItem = (ListBoxItem)(d2.ItemContainerGenerator.ContainerFromIndex(0));
ContentPresenter myContentPresenter = FindVisualChild<ContentPresenter>(myListBoxItem);
DataTemplate myDataTemplate = myContentPresenter.ContentTemplate;
StackPanel target = (StackPanel)myDataTemplate.FindName("p1", myContentPresenter);
}));
FindVisualChild方法
private childItem FindVisualChild<childItem>(DependencyObject obj)
where childItem : DependencyObject
{
for (int i = 0; i < VisualTreeHelper.GetChildrenCount(obj); i++)
{
DependencyObject child = VisualTreeHelper.GetChild(obj, i);
if (child != null && child is childItem)
return (childItem)child;
else
{
childItem childOfChild = FindVisualChild<childItem>(child);
if (childOfChild != null)
return childOfChild;
}
}
return null;
}
标签:childOfChild,FindVisualChild,return,usercontrol,child,wpf,null,listbox,childItem 来源: https://blog.csdn.net/weixin_43632687/article/details/120223020