在我的listview中,我创建了一个控件模板。在这个模板中,我创建了一个按钮,因此mit listview中的每个项目都有一个按钮。此按钮表示一个链接,用于打开显示的目录。当我单击按钮时,我的程序会按照我的预期打开资源管理器。但是当我单击未选中项目的按钮时,它会打开所选项目的路径。所以我的问题是,当我点击列表视图中的按钮时,如何更改所选项目。
以下是它的样子:
如您所见,选择“R1”,但我点击“R3”的链接。会发生什么,C:\ Temp \ Folder1被打开,因为“R1”仍然是selecteditem。我希望C:\ Temp \ Folder3获得opend。我认为诀窍应该是单击代表链接的按钮时选择“R3”。有谁知道怎么做?
这是我的XAML代码:
<ListView Grid.Row="1" ItemsSource="{Binding MyCollection}" FontSize="20" SelectedItem="{Binding SelectedMember, Mode=TwoWay, UpdateSourceTrigger=PropertyChanged}" Margin="7,10,7,0" x:Name="ListView1" SelectionChanged="ListView1_SelectionChanged">
<ListView.View>
<GridView>
<GridViewColumn Header="Header 1" Width="150"/>
<GridViewColumn Header="Header 2" Width="120"/>
<GridViewColumn Header="Header 3" Width="120"/>
<GridViewColumn Header="Header 4" Width="560"/>
<GridViewColumn Header="Header 5" Width="100"/>
</GridView>
</ListView.View>
<ListView.ItemContainerStyle>
<Style TargetType="{x:Type ListViewItem}" BasedOn="{StaticResource {x:Type ListBoxItem}}">
<EventSetter Event="MouseDoubleClick" Handler="ListViewItem_MouseDoubleClick"/>
<Setter Property="Template">
<Setter.Value>
<ControlTemplate>
<Grid>
<Grid.ColumnDefinitions>
<ColumnDefinition Width="150"/>
<ColumnDefinition Width="120"/>
<ColumnDefinition Width="120"/>
<ColumnDefinition Width="560"/>
<ColumnDefinition Width="100"/>
</Grid.ColumnDefinitions>
<ContentPresenter Grid.Column="0" Content="{Binding ID}" HorizontalAlignment="Center"/>
<ContentPresenter Grid.Column="1" Content="{Binding Name}" HorizontalAlignment="Center"/>
<ContentPresenter Grid.Column="2" Content="{Binding Description}" HorizontalAlignment="Center"/>
<Button Grid.Column="3" Content="{Binding Path}" Style="{StaticResource Link}" Command="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type Window}}, Path=DataContext.GoToPathCommand }" HorizontalAlignment="Left" Margin="5,0,0,0" IsEnabled="{Binding ButtonEnabled}"/>
<ContentPresenter Grid.Column="4" Content="{Binding Owner}" HorizontalAlignment="Center"/>
</Grid>
</ControlTemplate>
</Setter.Value>
</Setter>
<Style.Triggers>
<Trigger Property="IsSelected" Value="True">
<Setter Property="FontWeight" Value="Bold"/>
<Setter Property="Foreground" Value="DarkBlue"/>
</Trigger>
</Style.Triggers>
</Style>
</ListView.ItemContainerStyle>
</ListView>
答案 0 :(得分:1)
尝试在ListViewItem样式中添加以下触发器
<Style.Triggers>
<Trigger Property="IsKeyboardFocusWithin" Value="True">
<Setter Property="IsSelected" Value="True"/>
</Trigger>
</Style.Triggers>
答案 1 :(得分:1)
Command="{Binding RelativeSource={RelativeSource FindAncestor, AncestorType={x:Type Window}}, Path=DataContext.GoToPathCommand }"
此命令将跳转到ViewModel
的Cmd。我假设Command将检查选择了哪个项目,它将访问它的属性Path
。
避免此行为,您可以创建一个命令,该命令将接受参数并从Binding设置它:
<Button Content="{Binding Path}"
Command="{Binding RelativeSource={RelativeSource FindAncestor,
AncestorType={x:Type Window}}, Path=DataContext.GoToPathCommand }"
CommandParameter="{Binding Path}"/>
现在必须调整命令本身:
public MyCmd : ICommand
{
public void Execute(object parameter)
{
string path = (string) parameter;
//Open the path via explorer
}
}