我有一个列表视图:
<ListView
VerticalAlignment="Top"
Width="210"
Height="150"
SelectedValuePath="SelectedFile"
SelectionMode="Single"
SelectedIndex="0"
behaviour:CommandsBehaviour.SelectionChanged = "{Binding SelectionFileChange}"
ItemsSource="{Binding files}"
IsSynchronizedWithCurrentItem="True"
atachedProperties:GridViewSort.AutoSort="True"
atachedProperties:GridViewSort.ShowSortGlyph="True">
<ListView.View>
<GridView>
<GridView.Columns>
<GridViewColumn Header="File Name" Width="100" DisplayMemberBinding="{Binding Name}"/>
<GridViewColumn Header="Date" Width="100" DisplayMemberBinding="{Binding Date}"/>
</GridView.Columns>
</GridView>
</ListView.View>
</ListView>
我想在单击要打开的文件的文件名(仅文件名,而不是日期)时(或任何自定义操作),即文件名为链接的列。我怎么能这样做?
答案 0 :(得分:2)
您可以在 DataTemplate
中使用超链接 <ListView VerticalAlignment="Top" Width="210" Height="150" SelectedValuePath="SelectedFile"
SelectionMode="Single" SelectedIndex="0" ItemsSource="{Binding files}" >
<ListView.View>
<GridView>
<GridView.Columns>
<GridViewColumn Width="Auto" Header="URL" >
<GridViewColumn.CellTemplate>
<DataTemplate>
<TextBlock Name="Name" MinWidth="100" Width="Auto">
<Hyperlink NavigateUri="{Binding Path=Name}" Name="NameURl" RequestNavigate="OpenPageRequestNavigate">
<TextBlock Text="{Binding Path=Name}"/>
</Hyperlink>
</TextBlock>
</DataTemplate>
</GridViewColumn.CellTemplate>
</GridViewColumn>
</GridView.Columns>
</GridView>
</ListView.View>
</ListView>