我是XAML的新手。我写了下面的程序,将图像旋转一个给定的角度(0到360)。我放入了一个滑块控件,根据滑块值设置角度。效果很好!但是,当运行程序并单击"旋转"按钮时,For/Next循环从0变为360,图像将仅显示最后一个角度旋转(360)。我确实输入了"睡眠"命令来减慢速度,以防我没有赶上之前的更新。如果您能帮助我们解释为什么它不会持续更新,我们将不胜感激。非常感谢。
Imports System.Threading.Thread
Imports System.Windows.Media.Imaging.BitmapImage
Class MainWindow
Private Sub Slider1_ValueChanged(sender As System.Object, e As System.Windows.RoutedPropertyChangedEventArgs(Of System.Double)) Handles Slider1.ValueChanged
' ---- when I adjust manually, this works perfectly
Dim rotateTransform1 As New RotateTransform
rotateTransform1.Angle = Slider1.Value
lblAngle.Content = rotateTransform1.Angle
Image1.RenderTransform = rotateTransform1
End Sub
Private Sub btnSpin_Click(sender As System.Object, e As System.Windows.RoutedEventArgs) Handles btnSpin.Click
Dim spinAngle as Double
For SpinAngle 0 to 360
spinWheel(spinAngle)
Sleep(50)
Next spinAngle
End Sub
Private Sub spinWheel(ByVal spinAngle)
Dim rotateTransform1 As New RotateTransform
rotateTransform1.Angle = SpinAngle 'Slider1.Value
Image1.RenderTransform = rotateTransform1
lblAngle.Content = rotateTransform1.Angle
Image1.InvalidateVisual()
End Sub
Public Sub New()
' This call is required by the designer.
InitializeComponent()
' Add any initialization after the InitializeComponent() call.
' Image1.createOption = BitmapCreateOptions.IgnoreImageCache
End Sub
End Class
XAML
<Window x:Class="MainWindow"
xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml"
Title="MainWindow" Height="677" Width="910">
<Grid Background="#FF006903">
<Grid.RowDefinitions>
<RowDefinition Height="238*" />
<RowDefinition Height="178*" />
</Grid.RowDefinitions>
<Button Content="SPIN!" Height="23" HorizontalAlignment="Left" Margin="521,101,0,0" Name="btnSpin" VerticalAlignment="Top" Width="75" Grid.Row="1" FontFamily="Tahoma" FontSize="15" FontWeight="ExtraBold" />
<Image Height="615" Margin="32,7,0,0" Name="Image1" Stretch="None" VerticalAlignment="Top"
RenderTransformOrigin=" 0.5,0.5" Source="/rotatePicture;component/Images/purp_wheel_cropped.png"
Grid.RowSpan="2" HorizontalAlignment="Left" Width="619" />
<Slider Height="25" HorizontalAlignment="Left" Margin="127,188,0,0" Name="Slider1" VerticalAlignment="Top" Width="350" Maximum="360" Grid.Row="1" />
<Label Content="Label" Height="28" HorizontalAlignment="Left" Margin="521,175,0,0" Name="lblAngle" VerticalAlignment="Top" Width="75" Grid.Row="1" FontFamily="Tahoma" FontSize="15" FontWeight="ExtraBold" />
<Image Height="29" HorizontalAlignment="Left" Margin="535,252,0,0" Name="Image2" Stretch="Fill" VerticalAlignment="Top" Width="55" Source="/rotatePicture;component/Images/wheel_pointer.png" />
</Grid>
</Window>
您的方法的问题是通过在Click处理程序中重复调用Sleep
来阻塞UI线程。WPF为你想要做的事情提供了一个非常优雅的机制,它叫做动画。
为FrameworkElement
的旋转设置动画的方法可能如下C#中所示(对不起,我不会说VB)。
private void RotateElement(
FrameworkElement element, double from, double to, TimeSpan duration)
{
var transform = element.RenderTransform as RotateTransform;
if (transform != null)
{
var animation = new DoubleAnimation(from, to, duration);
transform.BeginAnimation(RotateTransform.AngleProperty, animation);
}
}
注意,RotateTransform
必须已经包含在FrameworkElement
的RenderTransform
属性中。例如,它可以在XAML中这样分配:
<Image RenderTransformOrigin="0.5,0.5" ...>
<Image.RenderTransform>
<RotateTransform/>
</Image.RenderTransform>
</Image>
您可以在按钮点击处理程序中调用RotateElement
方法,如下所示:
private void btnSpin_Click(object sender, RoutedEventArgs e)
{
RotateElement(Image1, 0, 360, TimeSpan.FromMilliseconds(360 * 50));
}
还要注意,在Slider1_ValueChanged
中,也没有必要每次都创建一个新的RotateTransform
。
此外,很少需要像在spinWheel
中那样调用InvalidateVisual
。