wpf4正式测试成功完成了VisualStateManager的功能。在此之前我没有确认wpf3.5是否已经支持此功能。以下是我使用vs2010 rc写的demo程式:

1.xaml部份一般都是用blend来设计的state。这里为了文章篇写的方便，我直接把xaml放出来.在blend中只能用鼠标拖拖放放即可弄出很多state。这功能对很多应用都很有好处。不用每次都自己写动画。还支持动画延时效果。

xaml部分:

<Window

xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"

xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"

xmlns:ei="http://schemas.microsoft.com/expression/2010/interactions" xmlns:ee="http://schemas.microsoft.com/expression/2010/effects" x:Class="WpfApplication1.MainWindow"

Title="MainWindow" Height="350" Width="525">

<Grid x:Name="LayoutRoot">

<VisualStateManager.VisualStateGroups>

<VisualStateGroup x:Name="VisualStateGroup">

<VisualStateGroup.Transitions>

<VisualTransition GeneratedDuration="0:0:0.3"/>

</VisualStateGroup.Transitions>

<VisualState x:Name="nothing">

<Storyboard>

<DoubleAnimationUsingKeyFrames Storyboard.TargetProperty="(UIElement.RenderTransform).(TransformGroup.Children)[3].(TranslateTransform.X)" Storyboard.TargetName="button1">

<EasingDoubleKeyFrame KeyTime="0" Value="0"/>

</DoubleAnimationUsingKeyFrames>

</Storyboard>

</VisualState>

<VisualState x:Name="change">

<Storyboard>

<DoubleAnimationUsingKeyFrames Storyboard.TargetProperty="(UIElement.RenderTransform).(TransformGroup.Children)[2].(RotateTransform.Angle)" Storyboard.TargetName="button1">

<EasingDoubleKeyFrame KeyTime="0" Value="50.596"/>

</DoubleAnimationUsingKeyFrames>

</Storyboard>

</VisualState>

</VisualStateGroup>

</VisualStateManager.VisualStateGroups>

<Button x:Name="button1" Content="Button" HorizontalAlignment="Left" Margin="216.2,176.8,0,0" VerticalAlignment="Top" Width="75" RenderTransformOrigin="0.5,0.5">

<Button.RenderTransform>

<TransformGroup>

<ScaleTransform/>

<SkewTransform/>

<RotateTransform/>

<TranslateTransform/>

</TransformGroup>

</Button.RenderTransform>

</Button>

</Grid>

</Window>

代码部份：

a)使用State:

public partial class MainWindow : Window

{

public MainWindow()

{

InitializeComponent();

button1.MouseEnter += button1_MouseEnter;

button1.MouseLeave += button1_MouseLeave;

}

void button1_MouseLeave(object sender, MouseEventArgs e)

{

VisualStateManager.GoToElementState(this.LayoutRoot, "nothing", true);

}

void button1_MouseEnter(object sender, MouseEventArgs e)

{

VisualStateManager.GoToElementState(this.LayoutRoot, "change", true);

}

}

}

至此，一个简单的state即完成。当鼠标移动到button时,button要翻转，鼠标离开时button会回到原样.