Question

我正在开发一个通用的Windows项目。

Windows Phone 8.1 UI包含列表视图。 Listviews的源代码是数据绑定，其datatemplate包含一个按钮。我想在按下按钮时显示一个MenuFlyout MenuFlyout（如wp8工具包中的ContextMenu）。

我的代码是：

    <DataTemplate x:Key="ListTemplate" x:Name="ListTemplate">
        <Button Style="{StaticResource ListButtonStyle}">
            <Button.Flyout> 
                <MenuFlyout>
                        <MenuFlyoutItem Text="delete"/>
                </MenuFlyout>
            </Button.Flyout>
        </Button>
    </DataTemplate>

    <ListView ItemsSource="{Binding List}" ItemTemplate="{StaticResource ListTemplate}">

但是，这会在单击按钮时打开MenuFlyout。如何将此行为更改为按住事件以打开MenuFlyout？

此外，MenuFlyout会放大打开它的按钮。如何禁用此缩放效果？

Answer 1

您可以从Windows Phone Toolkit修改ContextMenuService。

有很好的教程。

http://www.visuallylocated.com/post/2014/05/29/Migrating-from-the-Windows-Phone-Toolkit-ContextMenu-to-the-new-Runtime-MenuFlyout.aspx

简而言之：

替换ContextMenu

MenuFlyout

添加事件处理程序OnElementHolding

private void OnElementHolding(object sender, HoldingRoutedEventArgs args)
{
    // this event is fired multiple times. We do not want to show the menu twice
    if (args.HoldingState != HoldingState.Started) return;

    FrameworkElement element = sender as FrameworkElement;
    if (element == null) return;

    // If the menu was attached properly, we just need to call this handy method
    FlyoutBase.ShowAttachedFlyout(element);
 }

修改OnMenuFlyoutChanged

private static void OnMenuFlyoutChanged(DependencyObject o, DependencyPropertyChangedEventArgs e)
{
    FrameworkElement element = o as FrameworkElement;
    if (null != element)
    {
        // just in case we were here before and there is no new menu
        element.Holding -= OnElementHolding;

        MenuFlyout oldMenuFlyout = e.OldValue as MenuFlyout;
        if (null != oldMenuFlyout)
        {
            element.SetValue(FlyoutBase.AttachedFlyoutProperty, null);
        }
        MenuFlyout newMenuFlyout = e.NewValue as MenuFlyout;
        if (null != newMenuFlyout)
        {
            element.SetValue(FlyoutBase.AttachedFlyoutProperty, newMenuFlyout);
            element.Holding += OnElementHolding;
        }
    }
}

Answer 2

将按钮放在网格中并将menuflyout连接到网格而不是按钮。

像这样：

<DataTemplate x:Key="ListTemplate" x:Name="ListTemplate">
 <Grid Holding="Grid_Holding">
    <Button Style="{StaticResource ListButtonStyle}">
    </Button>
    <FlyoutBase.AttachedFlyout>
      <MenuFlyout>
        <MenuFlyoutItem Text="delete"/>
      </MenuFlyout>
    </FlyoutBase.AttachedFlyout>
 </Grid>
</DataTemplate>

在代码背后：

    private void Grid_Holding(object sender, Windows.UI.Xaml.Input.HoldingRoutedEventArgs e)
    {
        if (e.HoldingState == Windows.UI.Input.HoldingState.Started)
        {
            FrameworkElement frameworkElement = sender as FrameworkElement;
            FlyoutBase flyoutBase = FlyoutBase.GetAttachedFlyout(frameworkElement);
            flyoutBase.ShowAt(frameworkElement);
        }
    }

Windows Phone 8.1中ListView中的MenuFlyout

2 个答案: