Question

我想在左侧面板中创建窗口（带有切换按钮），右侧面板中是视图。当我单击切换按钮时，可以看到视图，并隐藏另一个视图。但是我的代码尽管他自己也开放了。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

我想，在循环中将IsChecked设置为false会导致此问题，因为如果您有任何UnChecked定义并在那里设置“_current”，则在第二次迭代期间_current可能会保留未经检查的ToggleButton，在那里放置一个断点并自行检查，你可以通过包括下面的标志来避免这种情况，

 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 2

这是我的解决方案：

<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 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，弹出菜单会打开漂亮的动画，并在关闭弹出窗口时关闭。

显示状态 flyout menu is open
隐藏状态 flyout menu is closed

带有ToggleButtons的菜单

3 个答案: