ToggleButton/CheckBox内容取决于其检查状态？

Question

制作ToggleButton内容的最短的xamly方法取决于其检查状态是什么？

在WPF中,我可能会选择DataTriggerSilverlight中不存在的.

我尝试了以下操作,但它不起作用,只要我包含触发器,就会破坏与源的绑定.触发器无论如何都不会起作用.

<ToggleButton
xmlns:i="http://schemas.microsoft.com/expression/2010/interactivity"
xmlns:ei="http://schemas.microsoft.com/expression/2010/interactions"
IsChecked="{Binding IsArchived, Mode=TwoWay}">
  <i:Interaction.Triggers>
    <i:EventTrigger EventName="Checked">
      <ei:ChangePropertyAction
      TargetObject="{Binding
        RelativeSource={RelativeSource AncestorType=ToggleButton}}" 
      PropertyName="Content" Value="Unarchive project"/>
    </i:EventTrigger>
    <i:EventTrigger EventName="Unchecked">
      <ei:ChangePropertyAction 
      TargetObject="{Binding 
        RelativeSource={RelativeSource AncestorType=ToggleButton}}" 
      PropertyName="Content" Value="Archive project"/>
    </i:EventTrigger>
  </i:Interaction.Triggers>
</ToggleButton>

Answer 1

<ToggleButton Width="50" Height="50">
  <ToggleButton.Content>
        <TextBlock x:Name="obj" Text="Foo"/>
    </ToggleButton.Content>
  <i:Interaction.Triggers>
        <i:EventTrigger EventName="Checked">
            <ei:ChangePropertyAction PropertyName="Text" Value="On" TargetName="obj"/>
        </i:EventTrigger>
        <i:EventTrigger EventName="Unchecked">
            <ei:ChangePropertyAction PropertyName="Text" Value="Off" TargetName="obj"/>
        </i:EventTrigger>
    </i:Interaction.Triggers>
</ToggleButton>