在影响另一个对象的一个​​对象上创建一个简单的 wpf 触发器 [英] Create a simple wpf trigger on one object that affects another

问题描述

这是我最接近创建一个简单触发器的方法.我只希望数据网格的 IsMouseOver == true 显示按钮.问题在于 Setter 的 TargetName 说:属性TargetName"不代表Setter"的有效目标，因为找不到名为ButtonExpand"的元素.确保在任何使用它的 Setter、触发器或条件之前声明目标.我做错了什么?

This is the closest that I have come to creating a simple trigger on this. I just want the datagrid's IsMouseOver == true to show the button. The problem is that the Setter's TargetName says: The property 'TargetName' does not represent a valid target for the 'Setter' because an element named 'ButtonExpand' was not found. Make sure that the target is declared before any Setters, Triggers or Conditions that use it. What am I doing wrong?

<UserControl.Resources>
    <Style TargetType="DataGrid">
        <Style.Triggers>
            <Trigger Property="IsMouseOver" Value="True">
                <Setter TargetName="ButtonExpand" Property="Visibility" Value="Visible" />
            </Trigger>
        </Style.Triggers>
    </Style>
</UserControl.Resources>
<Grid>



    <DataGrid Name="MainDataGrid" ItemsSource="{Binding Programs}" IsReadOnly="True" AutoGenerateColumns="false" >
        <DataGrid.Columns>
            <DataGridTextColumn Header="Name" Binding="{Binding Name}"/>
            <DataGridTextColumn Header="Version" Binding="{Binding Version}"/>
            <DataGridTextColumn Header="Publisher" Binding="{Binding Publisher}"/>
        </DataGrid.Columns>
    </DataGrid>

    <Button Name="ButtonExpand" Height="25" Width="25" HorizontalAlignment="Right" VerticalAlignment="Bottom" Visibility="Hidden">+</Button>
</Grid>

推荐答案

TargetName 不适合使用在 Triggers 集合中一种风格.一种风格没有名称范围，所以没有意义在那里按名称引用元素.但是模板(DataTemplate 或ControlTemplate) 确实有一个名称范围.

TargetName is not intended for use within the Triggers collection of a Style. A style does not have a namescope, so it does not make sense to refer to elements by name there. But a template (either DataTemplate or ControlTemplate) does have a namescope.

请参阅此链接.

您可以通过为 Button 设置 DataTrigger 来反其道而行之.请注意，您必须在 Style 中设置 Property Visibility 才能使 DataTrigger 工作.

You can do it the other way around with a DataTrigger for the Button. Note that you must set the Property Visibility within the Style for the DataTrigger to work.

<Grid Name="MainGrid"> 

    <DataGrid ItemsSource="{Binding Programs}"
              IsReadOnly="True"
              AutoGenerateColumns="false" > 
      <DataGrid.Columns> 
        <DataGridTextColumn Header="Name" Binding="{Binding Name}"/> 
        <DataGridTextColumn Header="Version" Binding="{Binding Version}"/> 
        <DataGridTextColumn Header="Publisher" Binding="{Binding Publisher}"/> 
      </DataGrid.Columns> 
    </DataGrid> 

    <Button Name="ButtonExpand"
            Height="25"
            Width="25"
            HorizontalAlignment="Right"
            VerticalAlignment="Bottom"
            Content="+">
        <Button.Style>
            <Style TargetType="Button">
                <Setter Property="Visibility" Value="Hidden"/>
                <Style.Triggers>
                    <DataTrigger Binding="{Binding ElementName=MainGrid,
                                                   Path=IsMouseOver}" 
                                 Value="True">
                        <Setter Property="Visibility" Value="Visible" />
                    </DataTrigger>
                </Style.Triggers>
            </Style>
        </Button.Style>
    </Button>
</Grid>

另一种方法是使用转换器将 ButtonExpand 的 Visibilty 绑定到 DataGrid 的 IsMouseOver 属性.

Another way to do it would be to bind Visibilty of ButtonExpand to the IsMouseOver property of the DataGrid with a converter.

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