Windows Phone 8 中的弹出窗口 [英] Popup in windows phone 8

问题描述

我想显示一个带有媒体元素的弹出窗口作为一个控件.当用户单击按钮时，我必须显示此弹出窗口.当用户点击设备的后退按钮时，弹出窗口应该关闭.

I want to show a popup with media element as one control. When user clicks on the button, then I have to show this popup. And when user clicks on the back button of the device then the popup should be closed.

请帮助我如何在 Windows Phone 8 应用程序中执行此操作.

Please help me how to do this in windows phone 8 application.

推荐答案

Popup with MediaElement(视图是PhoneApplicationPage的名称)

<Popup
    x:Name="popup">

    <Grid
        Background="{StaticResource PhoneChromeBrush}"
        Height="{Binding Path=ActualHeight, ElementName=view}"
        Width="{Binding Path=ActualWidth, ElementName=view}">

        <MediaElement />

    </Grid>

</Popup>

应用栏

<phone:PhoneApplicationPage.ApplicationBar>
    <shell:ApplicationBar>

        <shell:ApplicationBarIconButton
            Click="ShowPopup"
            IconUri="/Icons/show.png"
            Text="show" />

    </shell:ApplicationBar>
</phone:PhoneApplicationPage.ApplicationBar>

背后的代码

protected override void OnBackKeyPress(System.ComponentModel.CancelEventArgs e)
{
    if (this.popup.IsOpen)
    {
        this.popup.IsOpen = false; 
        e.Cancel = true;
    }

    base.OnBackKeyPress(e);
}


private void ShowPopup(object sender, EventArgs e)
{
    this.popup.IsOpen = true;
}

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