Question

我有一个列表框，我有以下ItemTemplate：

<DataTemplate x:Key="ScenarioItemTemplate">
    <Border Margin="5,0,5,0"
            Background="#FF3C3B3B"
            BorderBrush="#FF797878"
            BorderThickness="2"
            CornerRadius="5">
        <DockPanel>
            <DockPanel DockPanel.Dock="Top"
                       Margin="0,2,0,0">
                <Button HorizontalAlignment="Left"
                        DockPanel.Dock="Left"
                        FontWeight="Heavy"
                        Foreground="White" />
                <Label Content="{Binding Path=Name}"
                       DockPanel.Dock="Left"
                       FontWeight="Heavy"
                       Foreground="white" />
                <Label HorizontalAlignment="Right"
                       Background="#FF3C3B3B"
                       Content="X"
                       DockPanel.Dock="Left"
                       FontWeight="Heavy"
                       Foreground="White" />
            </DockPanel>
            <ContentControl Name="designerContent"
                            Visibility="Collapsed"
                            MinHeight="100"
                            Margin="2,0,2,2"
                            Content="{Binding Path=DesignerInstance}"
                            Background="#FF999898">
            </ContentControl>
        </DockPanel>
    </Border>
</DataTemplate>

如您所见，ContentControl的Visibility设置为折叠。

我需要定义一个触发器，使Visibility设置为“Visible”

当选择ListItem时，我无法弄明白。

有什么想法吗？

更新：当然我可以简单地复制DataTemplate并添加触发器有问题的ListBox使用其中一个，但我想防止重复此代码。

Answer 1

您可以设置ContentControl的样式，以便在其容器（ListBoxItem）被选中时触发器触发：

<ContentControl 
    x:Name="designerContent"
    MinHeight="100"
    Margin="2,0,2,2"
    Content="{Binding Path=DesignerInstance}"
    Background="#FF999898">
    <ContentControl.Style>
        <Style TargetType="{x:Type ContentControl}">
            <Setter Property="Visibility" Value="Collapsed"/>
            <Style.Triggers>
                <DataTrigger
                        Binding="{Binding
                            RelativeSource={RelativeSource
                                Mode=FindAncestor,
                                AncestorType={x:Type ListBoxItem}},
                                Path=IsSelected}"
                        Value="True">
                    <Setter Property="Visibility" Value="Visible"/>
                </DataTrigger>
            </Style.Triggers>
        </Style>
    </ContentControl.Style>
</ContentControl>

或者，我认为您可以将触发器添加到模板本身并按名称引用控件。我不太清楚这种技术是否可以从内存中输入并假设它可以工作，但它是这样的：

<DataTemplate x:Key="ScenarioItemTemplate">
    <DataTemplate.Triggers>
        <DataTrigger
                Binding="{Binding
                    RelativeSource={RelativeSource
                        Mode=FindAncestor,
                        AncestorType={x:Type ListBoxItem}},
                        Path=IsSelected}"
                Value="True">
            <Setter
                TargetName="designerContent"
                Property="Visibility"
                Value="Visible"/>
        </DataTrigger>
    </DataTemplate.Triggers>

    ...
</DataTemplate>

Answer 2

@Matt，谢谢!!!

只需添加IsSelected == false的触发器，现在它就像一个魅力！

<ContentControl.Style>
<Style TargetType="{x:Type ContentControl}">
    <Setter Property="Visibility" Value="Collapsed"/>
    <Style.Triggers>
        <DataTrigger Binding="{Binding RelativeSource={RelativeSource Mode=FindAncestor,AncestorType={x:Type ListBoxItem}},Path=IsSelected}" Value="True">
            <Setter Property="Visibility" Value="Visible"/>
        </DataTrigger>
        <DataTrigger Binding="{Binding RelativeSource={RelativeSource Mode=FindAncestor,AncestorType={x:Type ListBoxItem}},Path=IsSelected}" Value="False">
            <Setter Property="Visibility" Value="Collapsed"/>
        </DataTrigger>
    </Style.Triggers>
</Style>

用于ListBox项的DataTemplate中的IsSelected的WPF触发器

2 个答案: