菜单与ToggleButtons

Question

我想创建窗口，当在左面板菜单（与切换按钮），在右侧面板是视图。当我点击切换按钮时，可见视图和其他视图将被隐藏。但我的代码尽管自己打开视图。 这是我的代码：

<StackPanel Grid.Column="0"> 
    <ToggleButton Name="Button1" Checked="MenuItem_Checked"> 
    <ToggleButton Name="Button2" Checked="MenuItem_Checked"> 
</StackPanel> 
<administration:View1 Grid.Column="1" 
           Visibility="{Binding ElementName=Button1, Path=IsChecked, UpdateSourceTrigger=PropertyChanged, 
           Converter={StaticResource BooleanToVisibilityConverter}}"/> 
<administration:View2 Grid.Column="1" 
           Visibility="{Binding ElementName=Button2, Path=IsChecked, UpdateSourceTrigger=PropertyChanged, 
           Converter={StaticResource BooleanToVisibilityConverter}}"/> 

代码背后：

private void MenuItem_Checked(object sender, RoutedEventArgs e) 
    { 
     var el = sender as ToggleButton; 
     if (Equals(_current, el)) 
      return; 
     _current = el; 
     foreach (var menuChild in Menu.Children) 
     { 
      ToggleButton button = menuChild as ToggleButton; 
      if (button == null) continue; 
      if (!Equals(button, _current)) 
       button.IsChecked = false; 
     } 
    } 

Answer 1

钍是我的解决方案：

<View1.Resources> 
    <Style TargetType="View1"> 
       <Setter Property="Visibility" Value="Hidden"/> 
       <Style.Triggers> 
        <DataTrigger Binding="{Binding ElementName=Button1, Path=IsChecked, UpdateSourceTrigger=PropertyChanged}" Value="True"> 
         <Setter Property="Visibility" Value="Visible"/> 
        </DataTrigger> 
       </Style.Triggers> 
      </Style> 
</View1.Resources> 

Answer 2

我猜，设置器isChecked假环路造成这个问题，因为如果你有任何未检查的定义，并设置“_current”那里，第二次迭代_current可以保持选中切换按钮，把一个断点，并检查自己时，你可以通过包括像下面旗避免这种情况，

bool internalChange = false; 
    private void MenuItem_Checked(object sender, RoutedEventArgs e) 
    { 
     if (!internalChange) 
     { 
      var el = sender as ToggleButton; 
      if (Equals(_current, el)) 
       return; 
      _current = el; 
      foreach (var menuChild in Menu.Children) 
      { 
       ToggleButton button = menuChild as ToggleButton; 
       if (button == null) 
        continue; 
       if (!Equals(button, _current)) 
       { 
        internalChange = true; 
        button.IsChecked = false; 
        internalChange = false; 
       } 
      } 
     } 
    } 

Answer 3

我设计了一个自定义窗口镶边，弹出菜单和切换按钮。当切换按钮IsCheced属性设置为true时，弹出式菜单将以良好的动画打开，并在关闭时弹出。

1. 显示状态 flyout menu is open

2. 隐藏状态 flyout menu is closed