WPF

Question

我是 WPF 和 XAML 的新手。这个问题很直接。

我想在按下按钮时将触发器的椭圆填充属性从默认的“#597E0000”修改为 "Black"。显然我没有正确访问属性因为我无法编译它并在下面的代码指示的行上获得以下内容：

"TargetName property cannot be set on a Style Setter. Line 16"

<ResourceDictionary xmlns="http://schemas.microsoft.com/winfx/2006/xaml/presentation"
                xmlns:x="http://schemas.microsoft.com/winfx/2006/xaml">
<Style x:Key="ExitButton" TargetType="{x:Type Button}">
    <Setter Property="Template">
        <Setter.Value>
            <ControlTemplate x:Name="exitButtonTemplate" TargetType="Button">
                <Grid>
                    <Ellipse x:Name="exitButtonEllipse" Fill="#597E0000"/>
                    <ContentPresenter HorizontalAlignment="Center" VerticalAlignment="Center"/>
                </Grid>
            </ControlTemplate>
        </Setter.Value>
    </Setter>
    <Style.Triggers>
        <Trigger Property="IsPressed" Value="True">
            <Setter TargetName="exitButtonEllipse" Property="Fill" Value="Black" /> <!-- error -->
        </Trigger>
    </Style.Triggers>
</Style>

Answer 1

如果您在ControlTemplate 中定义了Trigger，您就可以访问ControlTemplate 中的元素。

另见 ControlTemplate.Triggers Property。

<Style x:Key="ExitButton" TargetType="{x:Type Button}">
    <Setter Property="Template">
        <Setter.Value>
            <ControlTemplate x:Name="exitButtonTemplate" TargetType="Button">
                <Grid>
                    <Ellipse x:Name="exitButtonEllipse" Fill="#597E0000"/>
                    <ContentPresenter HorizontalAlignment="Center" VerticalAlignment="Center"/>
                </Grid>
                <ControlTemplate.Triggers>
                    <Trigger Property="IsPressed" Value="True">
                        <Setter TargetName="exitButtonEllipse" Property="Fill" Value="Black" />
                    </Trigger>
                </ControlTemplate.Triggers>
            </ControlTemplate>
        </Setter.Value>
    </Setter>
</Style>

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