【发布时间】:2017-01-04 05:55:41
【问题描述】:
我想在左侧面板是菜单(带有切换按钮)和右侧面板是视图时创建窗口。当我单击切换按钮时,视图可见,而另一个视图将被隐藏。但是我的代码尽管他自己也能打开视图。 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;
}
}
【问题讨论】:
标签: wpf togglebutton