Question

我使用代码创建一个列表视图，我希望第二列包含一个超链接，打开默认浏览器并加载网址。

我创建了类似的列表视图：

GridView myGridView = new GridView();
myGridView.AllowsColumnReorder = false;

ListView l1 = new ListView();

GridViewColumn gvc0 = new GridViewColumn();
gvc0.DisplayMemberBinding = new Binding("AA");
gvc0.Header = "A/A";
gvc0.Width = 30;
myGridView.Columns.Add(gvc0);
GridViewColumn gvc1 = new GridViewColumn();
gvc1.DisplayMemberBinding = new Binding("Description");
gvc1.Header = "Description";
gvc1.Width = 300;
myGridView.Columns.Add(gvc1);

l1.View = myGridView;

我也像这样填写列表视图：

l1.Items.Add(new { AA = aa++, Description = descriptions});

我希望描述文本是一个超链接。它可以吗？

Answer 1

在代码隐藏中，您可以创建CellTemplate：

GridView myGridView = new GridView();
myGridView.AllowsColumnReorder = false;

ListView l1 = new ListView();

GridViewColumn gvc0 = new GridViewColumn();
gvc0.DisplayMemberBinding = new Binding("AA");
gvc0.Header = "A/A";
gvc0.Width = 30;
myGridView.Columns.Add(gvc0);

GridViewColumn gvc1 = new GridViewColumn();
gvc1.Header = "Description";
gvc1.Width = 300;
FrameworkElementFactory fef = new FrameworkElementFactory(typeof(TextBlock));
FrameworkElementFactory hyperlinkHolder = new FrameworkElementFactory(typeof(Hyperlink));
hyperlinkHolder.SetBinding(Hyperlink.NavigateUriProperty, new Binding("Description"));
hyperlinkHolder.AddHandler(Hyperlink.RequestNavigateEvent, new RequestNavigateEventHandler(Hyperlink_RequestNavigate));
FrameworkElementFactory fef2 = new FrameworkElementFactory(typeof(TextBlock));
Binding placeBinding = new Binding();
fef2.SetBinding(TextBlock.TextProperty, placeBinding);
placeBinding.Path = new PropertyPath("Description");
hyperlinkHolder.AppendChild(fef2);
fef.AppendChild(hyperlinkHolder);
var dataTemplate = new DataTemplate();
dataTemplate.VisualTree = fef;
dataTemplate.DataType = typeof(ListViewItem);
gvc1.CellTemplate = dataTemplate;           
myGridView.Columns.Add(gvc1);

l1.View = myGridView;

事件处理程序：

private void Hyperlink_RequestNavigate(object sender, RequestNavigateEventArgs e)
{
    Process.Start(new ProcessStartInfo(e.Uri.AbsoluteUri));
    e.Handled = true;
}

你应该使用XAML来做到这一点，它比你在代码隐藏中定义的所有GUI的第一个解决方案更清晰。

<ListView Name="l1">
    <ListView.View>
        <GridView>
            <GridView.Columns>
                <GridViewColumn Header="A/A" Width="30" DisplayMemberBinding="{Binding AA}" />
                <GridViewColumn Header="Description" Width="300">
                    <GridViewColumn.CellTemplate>
                        <DataTemplate>
                            <TextBlock>
                                <Hyperlink NavigateUri="{Binding Description}" RequestNavigate="Hyperlink_RequestNavigate">
                                    <TextBlock Text="{Binding Description}" />
                                </Hyperlink>
                            </TextBlock>
                        </DataTemplate>
                    </GridViewColumn.CellTemplate>
                </GridViewColumn>
            </GridView.Columns>
        </GridView>
    </ListView.View>
</ListView>

在listview列中添加动态超链接

1 个答案: