Question

感谢您的光临。如果尝试在其他位置选中复选框，则试图在WPF摇动中创建一个按钮。我尝试仅在XAML中执行此操作，但无法弄清楚如何同时检查单击作为触发器和复选框。这是我的XAML代码：

<Style TargetType="{x:Type Button}">
 <Setter Property="RenderTransformOrigin"
         Value="0.5 0.5" />
 <Setter Property="RenderTransform">
   <Setter.Value>
     <RotateTransform />
   </Setter.Value>
 </Setter>

 <Style.Triggers>

   <MultiTrigger>

     <MultiTrigger.Conditions>
       <Condition Property='IsPressed'
                  Value='True' />
     </MultiTrigger.Conditions>

     <MultiTrigger.EnterActions>
       <BeginStoryboard>
         <Storyboard TargetProperty="RenderTransform.Angle">
           <DoubleAnimation From="0"
                            To="5"
                            Duration="0:0:0.05"
                            AutoReverse="True"
                            FillBehavior="Stop" />
           <DoubleAnimation BeginTime="0:0:0.05"
                            From="5"
                            To="-5"
                            Duration="0:0:0.1"
                            AutoReverse="True"
                            FillBehavior="Stop" />
           <DoubleAnimation BeginTime="0:0:0.2"
                            From="-5"
                            To="0"
                            Duration="0:0:0.1"
                            AutoReverse="False"
                            FillBehavior="Stop" />
         </Storyboard>
       </BeginStoryboard>
     </MultiTrigger.EnterActions>

   </MultiTrigger>
 </Style.Triggers>

这使按钮在按下时总是摇晃。为了使它有条件地工作，我认为我应该在后面的代码中实现它。我认为动画一旦开始工作，就应该像添加if语句一样简单。

MainWindow : Window

{

    private Storyboard myStoryboard;

    public MainWindow()

    {

        InitializeComponent();

        DoubleAnimation myDoubleAnimation = new DoubleAnimation();

        myDoubleAnimation.From = 0.0;
        myDoubleAnimation.To = 20.0;

        myDoubleAnimation.Duration = new Duration(TimeSpan.FromSeconds(5));

        myDoubleAnimation.AutoReverse = true;

        myStoryboard = new Storyboard();

        myStoryboard.Children.Add(myDoubleAnimation);
        Storyboard.SetTargetName(myDoubleAnimation, HelpButton.Name);
        Storyboard.SetTargetProperty(myDoubleAnimation, new 
        PropertyPath(RotateTransform.AngleProperty));
        HelpButton.Click += ShakeObject;

    }

private void ShakeObject(object sender, RoutedEventArgs e)
    {
        myStoryboard.Begin(this);
    }
}

我希望这段代码可以将我的名为“ HelpButton”的按钮（在XAML中）旋转20度，然后向后旋转，但是没有这种运气。

我可以执行一个RotateTransform .BeginAnimation（RotateTransform.AngleProperty，myDoubleAnimation）; 但我想连续进行几次旋转（以模拟晃动）。

感谢您的帮助！谢谢。

Answer 1

我发现的一种解决方案是，当IsChecked触发我的复选框时，将XAML样式应用于按钮，然后取消选中该样式。

Answer 2

您需要在复选框上添加条件，然后在MultiDataTrigger中更改MultiTringger。这是代码

public function store(Request $request)
{

    $requestData = $request->all();

    $requestData['application_id'] = strtolower(trim($requestData['application_id'].'.'.config('app.fqdn_base')));

    $request->replace($requestData);

    $values =  $this->validate($request, [
        'application_type' => 'required|exists:applications,name',
        /* other validations */
    ]);

    $exitCode = Artisan::call('tenant:create', [
        /* passing param here */
    ]);

    $host = Hostname::where('fqdn', $values["application_id"])->first();
    $web = $host->website;

    $web->name = $values["application_name"];
    $web->application_id = Application::where('name', '=', $values["application_type"])->pluck('id')->first();
    $web->owner = auth()->id();
    $web->save();

    $applications = Application::where('id', '!=', '0')->get();
    return view('admin.applications', [ "applications" =>  $applications]);
}

在WPF中有条件地为情节提要动画吗？

2 个答案: