WPF: 纯 XAML 当午餐 Storyboard 基础的绑定条件

Question

所以我有这个 Storyboard:

<Storyboard x:Key="animate">
            <DoubleAnimation BeginTime="0:0:0" Storyboard.TargetProperty="Opacity" From="1" To="0" Duration="0:0:2.0"/>
</Storyboard>

我的绑定value:

public bool IsFound
{
    get { return _isFound; }
    set
    {
        _isFound= value;
        NotifyPropertyChanged();
    }
}

我的 Grid 得到这个 Storyboard:

<Grid name="myGrid">
   ....
<Grid>


if(IsFound)
{
    Storyboard storyboard = Resources["animate"] as Storyboard;
        if (storyboard != null)
            storyboard.Begin(myGrid);
}

所以我正在寻找纯粹的东西 XAML 而不是在代码后面检查这个 IsFound。

Answer 1

您可以使用 DataTrigger：

<Style TargetType="Grid" x:Key="MyAnimatedGrid">
    <Style.Triggers>
        <DataTrigger Binding="{Binding IsFound}" Value="True">
            <DataTrigger.EnterActions>
                <BeginStoryboard StoryBoard="{StaticResource animate}" />
            </DataTrigger.EnterActions>
        </DataTrigger>
    </Style.Triggers>
</Style>

WPF: 纯 XAML 当午餐 Storyboard 基础的绑定条件

WPF: pure XAML when lunch Storyboard base of binding condition

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