如何从tabitem wpf的数据模板中找到控件 [英] How to find control from datatemplate of tabitem wpf
本文介绍了如何从tabitem wpf的数据模板中找到控件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有 TabControl:
I have TabControl:
<TabControl Name="tabControl"
VerticalAlignment="Top"
HorizontalAlignment="Stretch">
<TabControl.Items>
<TabItem x:Name="tab1" Header="ABC">
<TabItem.ContentTemplate>
<DataTemplate>
<ScrollViewer Name="ScrollViewer">
<StackPanel Orientation="Vertical">
<TextBox Name="txt1" HorizontalAlignment="Center" Margin="0,26,0,0" />
<ListBox Name="listBox" DataContext="{Binding Items, Mode=TwoWay}" />
</StackPanel>
</ScrollViewer>
</DataTemplate>
</TabItem.ContentTemplate>
</TabItem>
</TabControl.Items>
</TabControl>
如何在 C# 代码中以编程方式获取列表框?
How I can get listbox programmatically in C# code?
我尝试了以下代码,myContentPresenter.ContentTemplate
显示为空.
I have tried below code and myContentPresenter.ContentTemplate
shows null.
TabItem myListBoxItem = (TabItem)(tabControl.ItemContainerGenerator.ContainerFromItem(tabControl.SelectedItem));
ContentPresenter myContentPresenter = FindVisualChild<ContentPresenter>(myListBoxItem);
DataTemplate myDataTemplate = myContentPresenter.ContentTemplate;
ListBox listBox = (ListBox)myDataTemplate.FindName("listBox", myContentPresenter);
推荐答案
基于@mm8 方法,以下解决方案将按名称而不是按类型查找 ListBox
:
Building on @mm8 approach, the following solution will find the ListBox
by name instead of by type:
XAML
<TabControl x:Name="tabControl1" SelectionChanged="tabControl1_SelectionChanged">
<TabItem x:Name="tab1" Header="ABC">
<TabItem.ContentTemplate>
...
代码
private void tabControl1_SelectionChanged(object sender, SelectionChangedEventArgs e)
{
Dispatcher.BeginInvoke(new Action(() => TabItem_UpdateHandler()));
}
void TabItem_UpdateHandler()
{
ContentPresenter myContentPresenter = tabControl1.Template.FindName("PART_SelectedContentHost", tabControl1) as ContentPresenter;
if (myContentPresenter.ContentTemplate == tab1.ContentTemplate)
{
myContentPresenter.ApplyTemplate();
var lb1 = myContentPresenter.ContentTemplate.FindName("listBox", myContentPresenter) as ListBox;
}
}
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