我关于XAML的问题:
我想在WP8中开发滑入/退出菜单
下面是我的代码:
<VisualStateManager.VisualStateGroups>
<VisualStateGroup x:Name="ShowHideMenu">
<VisualStateGroup.Transitions>
<VisualTransition GeneratedDuration="0:0:0.3"/>
</VisualStateGroup.Transitions>
<VisualState x:Name="ShowMenu">
<Storyboard>
<DoubleAnimation Duration="0" To="335" Storyboard.TargetProperty="(UIElement.RenderTransform).(CompositeTransform.TranslateX)" Storyboard.TargetName="stackPanel" d:IsOptimized="True"/>
</Storyboard>
</VisualState>
<VisualState x:Name="HideMenu"/>
</VisualStateGroup>
</VisualStateManager.VisualStateGroups>
我的按钮代码:
<Button Content="Button" Margin="0,0,145,0">
<i:Interaction.Triggers>
<i:EventTrigger EventName="Tap">
<ec:GoToStateAction StateName="ShowMenu"/>
</i:EventTrigger>
</i:Interaction.Triggers>
</Button>
如何使它显示菜单如果它的状态是HideMenu,隐藏如果它是显示的?
实际上看起来是这样的:
发布于 2013-07-08 07:04:54
与使用Button
不同,您可以使用ToggleButton
,这样就可以使用Checked
和Unchecked
事件:
<ToggleButton Margin="0,0,145,0">
<ToggleButton.Content>
<TextBlock Text="Button"/>
</ToggleButton.Content>
<i:Interaction.Triggers>
<i:EventTrigger EventName="Checked">
<ec:GoToStateAction StateName="ShowMenu"/>
</i:EventTrigger>
<i:EventTrigger EventName="Unchecked">
<ec:GoToStateAction StateName="HideMenu"/>
</i:EventTrigger>
</i:Interaction.Triggers>
</ToggleButton>
https://stackoverflow.com/questions/17510515
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