我有一个通过ItemsControl显示的超链接列表,如下所示:
<ItemsControl x:Name="SubMenu" Visibility="Collapsed">
<ItemsControl.ItemTemplate>
<DataTemplate>
<HyperlinkButton Content="{Binding Name}"
NavigateUri="{Binding Url}"
TargetName="ContentFrame"
Style="{StaticResource LinkStyle}"
/>
</DataTemplate>
</ItemsControl.ItemTemplate>
<ItemsControl.ItemsPanel>
<ItemsPanelTemplate>
<StackPanel Style="{StaticResource LinksStackPanelStyle}"
VerticalAlignment="Center"
HorizontalAlignment="Left" />
</ItemsPanelTemplate>
</ItemsControl.ItemsPanel>
</ItemsControl>我需要做的是枚举子菜单中的实际超链接,如下所示:
foreach (UIElement child in SubMenu.Items) // this does not work!
{
HyperlinkButton hb = child as HyperlinkButton;
if (hb != null && hb.NavigateUri != null)
{
if (hb.NavigateUri.ToString().Equals(e.Uri.ToString()))
{
VisualStateManager.GoToState(hb, "ActiveLink", true);
}
else
{
VisualStateManager.GoToState(hb, "InactiveLink", true);
}
}
}问题是,我似乎找不到一种方法来枚举ItemsCollection.Items中的实际UI元素。
有人知道怎么做或者一个可能的解决办法吗?
我可以提到,我想要做的是构建一个菜单和子菜单,将超链接显示为一种面包屑。
更新:最好的方法是,如果我能够以某种方式进入堆栈面板,因为这段代码似乎有效:
foreach (UIElement child in LinksStackPanel.Children)
{
HyperlinkButton hb = child as HyperlinkButton;
if (hb != null && hb.NavigateUri != null)
{
if (hb.NavigateUri.ToString().Equals(e.Uri.ToString()))
{
VisualStateManager.GoToState(hb, "ActiveLink", true);
}
else
{
VisualStateManager.GoToState(hb, "InactiveLink", true);
}
}
}发布于 2010-02-26 08:37:54
解决方案如下:
foreach (var item in SubMenu.Items)
{
var hb = SubMenu.ItemContainerGenerator.ContainerFromItem(item).FindVisualChild<HyperlinkButton>();
if (hb.NavigateUri.ToString().Equals(e.Uri.ToString()))
{
VisualStateManager.GoToState(hb, "ActiveLink", true);
}
else
{
VisualStateManager.GoToState(hb, "InactiveLink", true);
}
}扩展方法FindVisualChild:
public static T FindVisualChild<T>(this DependencyObject instance) where T : DependencyObject
{
T control = default(T);
if (instance != null)
{
for (int i = 0; i < VisualTreeHelper.GetChildrenCount(instance); i++)
{
if ((control = VisualTreeHelper.GetChild(instance, i) as T) != null)
{
break;
}
control = FindVisualChild<T>(VisualTreeHelper.GetChild(instance, i));
}
}
return control;
}发布于 2010-02-26 01:46:53
尝试使用ItemContainerGenerator.ContainerFromItem方法
foreach (var item in SubMenu.Items)
{
var child = SubMenu.ItemContainerGenerator.ContainerFromItem(item);
HyperlinkButton hb = child as HyperlinkButton;
// use hb
}发布于 2013-06-17 15:21:10
来自Johan答案的FindVisualChild存在缺陷:遍历控件层次结构中的较低级别不会产生任何影响,因为他不检查递归调用的结果。
那是固定版本。
public static T FindVisualChild<T>(this DependencyObject instance) where T : DependencyObject
{
T control = default(T);
if (instance != null)
{
for (int i = 0; i < VisualTreeHelper.GetChildrenCount(instance); i++)
{
if ((control = VisualTreeHelper.GetChild(instance, i) as T) != null)
{
break;
}
if ((control = FindVisualChild<T>(VisualTreeHelper.GetChild(instance, i))) != null)
{
break;
}
}
}
return control;
}https://stackoverflow.com/questions/2336989
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