用于ListBox项目的DataTemplate中的IsSelected的WPF触发器

问题描述

我有一个列表框，并且它具有以下ItemTemplate:

I have a listbox, and I have the following ItemTemplate for it:

<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的可见性"设置为折叠".

As you can see the ContentControl has Visibility set to collapsed.

我需要定义一个将可见性"设置为可见"的触发器

I need to define a trigger that causes the Visibility to be set to "Visible"

当选择列表项，但我不能弄清楚.

when the ListItem is selected, but I can't figure it out.

有什么想法吗?

更新:当然，我可以简单地复制DataTemplate并添加触发器到有问题的列表框以使用其中一个，但我想避免重复此代码.

UPDATE: Of course I could simply duplicate the DataTemplate and add triggersto the ListBox in question to use either one or the other, but I want to prevent duplicating this code.

推荐答案

您可以设置ContentControl的样式，以便在选择其容器(ListBoxItem)时触发触发器:

You can style your ContentControl such that a trigger fires when its container (the ListBoxItem) becomes selected:

<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.我不太了解这种技术，无法从内存中键入它并认为它会起作用，但这是这样的:

Alternatively, I think you can add the trigger to the template itself and reference the control by name. I don't know this technique well enough to type it from memory and assume it'll work, but it's something like this:

<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>

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