曲线silverlight列表框控件的边框

Question

曲线silverlight列表框控件的边框:

我只想弯曲silverlight列表框边框的两端.

我做的事情有什么不对,因为我无法获得边界曲线:

  <Style TargetType="ListBox" x:Key="listboxStyle">
    <Setter Property="Padding" Value="1"/>
    <Setter Property="Background" Value="Transparent" />
    <Setter Property="HorizontalContentAlignment" Value="Left" />
    <Setter Property="VerticalContentAlignment" Value="Top" />
    <Setter Property="IsTabStop" Value="False" />
    <Setter Property="BorderThickness" Value="0" />
    <Setter Property="TabNavigation" Value="Once" />
    <Setter Property="ScrollViewer.HorizontalScrollBarVisibility" Value="Disabled"/>
    <Setter Property="ScrollViewer.VerticalScrollBarVisibility" Value="Auto"/>
    <Setter Property="Template">
      <Setter.Value>
        <ControlTemplate TargetType="ListBox">
          <Grid Background="White">
              <Border Background="White" BorderThickness="0" CornerRadius="10">
                <ScrollViewer Background="White" x:Name="ScrollViewerElement" Padding="{TemplateBinding Padding}">
                  <ItemsPresenter />
                </ScrollViewer>
              </Border>
          </Grid>
        </ControlTemplate>
      </Setter.Value>
    </Setter>
  </Style>

Answer 1

好的,我完成了它,没有必要设置列表框控件的样式.

    <Border BorderBrush="White" BorderThickness="0" CornerRadius="5" Background="White" >
       <ListBox x:Name="lstEnities" BorderThickness="0" Margin="5" Grid.Row="0"></ListBox>
    </Border>