弹出式菜单Windows Phone 8.1暂停列表视图

Question

如何flyout在使用时按住某个项目时显示菜单listview？我试过但是hold方法引用了listview而不是它自己的项目.

Answer 1

您可以订阅Item的Template Holding事件.例如这样:

<ListView.ItemTemplate>
    <DataTemplate>
        <Grid Holding="Grid_Holding" VerticalAlignment="Stretch">
            <FlyoutBase.AttachedFlyout>
                <MenuFlyout>
                    <MenuFlyoutItem x:Name="EditButton"
                            Text="Edit"
                            Click="EditButton_Click"/>
                    <MenuFlyoutItem x:Name="DeleteButton"
                            Text="Delete"
                            Click="DeleteButton_Click"/>
                </MenuFlyout>
            </FlyoutBase.AttachedFlyout>
            <TextBlock Text="{Binding}" VerticalAlignment="Center"/>
        </Grid>
    </DataTemplate>
</ListView.ItemTemplate>

在后面的代码中,显示弹出窗口:

private void Grid_Holding(object sender, HoldingRoutedEventArgs e)
{
    FrameworkElement senderElement = sender as FrameworkElement;
    // If you need the clicked element:
    // Item whichOne = senderElement.DataContext as Item;
    FlyoutBase flyoutBase = FlyoutBase.GetAttachedFlyout(senderElement);
    flyoutBase.ShowAt(senderElement);
}

private async void EditButton_Click(object sender, RoutedEventArgs e)
{
    // get the clicked element:
    Item datacontext = (e.OriginalSource as FrameworkElement).DataContext as Item;
    await new MessageDialog("Edit").ShowAsync();
}

如果您愿意,也可以使用" 行为"来执行此操作.您还可以在Igrali的博客文章中找到一些帮助.