如何通过Silverlight中的嵌入式wince6.0 R3中的C ++编程获取ControlTemplate中的子对象 [英] How to get a child object in ControlTemplate by C++ programming in silverlight for embedded wince6.0 R3

问题描述

我在嵌入式wince6,0 R3的银色灯光列表框控件设计中遇到了问题，我想要滑动列表框操作，所以我需要获取列表框模板stackpanel对象，以便我可以通过
鼠标控制其Y位置，但我不知道如何获取对象的stackpanel指针。以下代码XAML，

I encountered a problem in listbox control design of silverlight for embedded wince6,0 R3 , I want to slide listbox operation, so I need to get the listbox template stackpanel object, so that I can control its Y position by mouse, but I don't know how to get the stackpanel pointer of the object. The following code XAML,

< UserControl.Resources>

<UserControl.Resources>

  < ItemsPanelTemplate x：Key =" ItemsPanelTemplate1">

   < StackPanel x：Name =" sp"高度= QUOT;自动"宽度= QUOT;自动"&NBSP;的Horizo​​ntalAlignment = QUOT;拉伸" VerticalAlignment = QUOT;拉伸"&NBSP;&NBSP; >

    < StackPanel.Background>

     < LinearGradientBrush EndPoint =" 0.5 ，1" StartPoint =" 0.5,0">

      < GradientStop Color ="＃FF000000" Offset =" 0.549" />

      < GradientStop Color =" #FFFFFFF" Offset =" 1">

      < GradientStop Color ="＃FFF3E9E9" Offset =" 0.004" />

     < / LinearGradientBrush>

    < ; /StackPanel.Background>

   < / StackPanel>

  < / ItemsPanelTemplate>

  <ItemsPanelTemplate x:Key="ItemsPanelTemplate1">
   <StackPanel x:Name="sp" Height="Auto" Width="Auto"  HorizontalAlignment="Stretch" VerticalAlignment="Stretch"   >
    <StackPanel.Background>
     <LinearGradientBrush EndPoint="0.5,1" StartPoint="0.5,0">
      <GradientStop Color="#FF000000" Offset="0.549"/>
      <GradientStop Color="#FFFFFFFF" Offset="1"/>
      <GradientStop Color="#FFF3E9E9" Offset="0.004"/>
     </LinearGradientBrush>
    </StackPanel.Background>
   </StackPanel>
  </ItemsPanelTemplate>

 < /UserControl.Resources>

 </UserControl.Resources>

 

 < Grid x：Name =" LayoutRoot" Background =" White">

  < ListBox x：Name =" listbox"的Horizo​​ntalAlignment = QUOT;左"余量= QUOT; 109,126,0,158"宽度= QUOT; 142" ItemsPanel =" {StaticResource ItemsPanelTemplate1}" >

 <Grid x:Name="LayoutRoot" Background="White">
  <ListBox x:Name="listbox" HorizontalAlignment="Left" Margin="109,126,0,158" Width="142" ItemsPanel="{StaticResource ItemsPanelTemplate1}" >

   < ListBoxItem Content =" ListBoxItem1" />

   < ListBoxItem Content = " ListBoxItem2" />

   < ListBoxItem Content =" ListBoxItem3" />

  < / ListBox>

  < Button 高度= QUOT; 61"的Horizo​​ntalAlignment = QUOT;右"余量= QUOT; 0,126,70,0" VerticalAlignment = QUOT;陀螺"宽度= QUOT; 216"含量="按钮和QUOT;点击= QUOT; Button_Click" x：姓名=" bu" />

 < / Grid>

   <ListBoxItem Content="ListBoxItem1"/>
   <ListBoxItem Content="ListBoxItem2"/>
   <ListBoxItem Content="ListBoxItem3"/>
  </ListBox>
  <Button  Height="61" HorizontalAlignment="Right" Margin="0,126,70,0" VerticalAlignment="Top" Width="216" Content="Button" Click="Button_Click" x:Name="bu"/>
 </Grid>

我想要获取StackPanel（x：Name =" sp"）对象指针，如何获取它？

I want to get StackPanel (x:Name="sp") object pointer,how to get it?

谢谢，

wayinfo

推荐答案

你已发布到Windows Embedded CE的托管代码（C＃，VB.NET）论坛。我认为本机代码论坛可能更好。对不起，我不知道答案......

You've posted to the managed code (C#, VB.NET) forum for Windows Embedded CE. I'm thinking that the native code forum is probably better. Sorry I don't know the answer...

Paul T。

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