UWP - MessageDialog在Windows Phone和平板电脑模式下崩溃应用程序

时间:2015-09-24 20:34:25

标签: c# win-universal-app

在Windows 10 Universal应用程序中,我想在按下后退按钮时显示MessageDialog。

我的页面代码如下:

    protected override void OnNavigatedTo(NavigationEventArgs e)
    {
        base.OnNavigatedTo(e);

        SystemNavigationManager.GetForCurrentView().AppViewBackButtonVisibility = AppViewBackButtonVisibility.Visible;
        SystemNavigationManager.GetForCurrentView().BackRequested += GamePage_BackRequested;
    }

    private async void GamePage_BackRequested(object sender, BackRequestedEventArgs e)
    {
        var dialog = new Windows.UI.Popups.MessageDialog("Are you sure ?");

        dialog.Commands.Add(new Windows.UI.Popups.UICommand("Yes"));
        dialog.Commands.Add(new Windows.UI.Popups.UICommand("No"));

        var result = await dialog.ShowAsync();
    }

当我在“本地机器”中启动应用程序时,对话框会很好地显示。但是当我将Windows转为“平板电脑模式”,或者当我在Windows Phone上试用它时,ShowAsync方法会崩溃应用程序(没有错误)。

为什么应用程序崩溃?

4 个答案:

答案 0 :(得分:2)

问题似乎是" dialog.ShowAsync()"应该从UI线程调用方法。

这就是我解决它的方法:

    private void GamePage_BackRequested(object sender, BackRequestedEventArgs e)
    {
        e.Handled = true;
        Frame rootFrame = Window.Current.Content as Frame;            
        if (rootFrame.CanGoBack)
        {
            var d = dispatcher.RunAsync(CoreDispatcherPriority.Normal, () => ShowConfirmationDialog(rootFrame));
        }
    }

    public async void ShowConfirmationDialog(Frame rootFrame)
    {
        var dialog = new Windows.UI.Popups.MessageDialog("Are you sure ?");

        dialog.Commands.Add(new Windows.UI.Popups.UICommand("Yes") { Id = 0 });
        dialog.Commands.Add(new Windows.UI.Popups.UICommand("No") { Id = 1 });

        var result = await dialog.ShowAsync();

        if (result != null && result.Label == "Yes")
        {
            rootFrame.GoBack();
        }
    }

    protected override void OnNavigatedTo(NavigationEventArgs e)
    {
        base.OnNavigatedTo(e);
        dispatcher = Windows.UI.Core.CoreWindow.GetForCurrentThread().Dispatcher;

        SystemNavigationManager.GetForCurrentView().BackRequested += GamePage_BackRequested;
    }

    protected override void OnNavigatedFrom(NavigationEventArgs e)
    {
        base.OnNavigatedFrom(e);

        SystemNavigationManager.GetForCurrentView().BackRequested -= GamePage_BackRequested;
    }

答案 1 :(得分:1)

You should handle the backrequest ; e.handled = true;

private async void GamePage_BackRequested(object sender, BackRequestedEventArgs e)
        {
            e.handled = true;
            var dialog = new Windows.UI.Popups.MessageDialog("Are you sure ?");

        dialog.Commands.Add(new Windows.UI.Popups.UICommand("Yes"));
        dialog.Commands.Add(new Windows.UI.Popups.UICommand("No"));

        var result = await dialog.ShowAsync();
    }

You should add onnavigatedfrom method to unregister the event, otherwise it will trigger twice !

 protected override void OnNavigatedFrom(NavigationEventArgs e)
        {
            if (gb.DetectPlatform() == Platform.WindowsPhone)
                HardwareButtons.BackPressed -= HardwareButtons_BackPressed;
           elde
                SystemNavigationManager.GetForCurrentView().BackRequested -= GamePage_BackRequested;
};
        }

答案 2 :(得分:0)

没有办法实现你想要的 - 后面的键处理程序需要立即回答(处理与否),但对话框本质上是异步的。

当然,您可以选择在显示对话框之前将事件标记为Handled,但如果用户显示"是"那时你无法走开。您当然可以终止该应用,但这不是一个坏主意(参见最后一段)

也就是说,您通常不需要此对话框,因为在Windows 10中,退出应用程序并不会终止它,它只是切换到之前的应用程序(或“开始”菜单)。用户可以通过任务切换器(或再次启动它)轻松返回到它。

答案 3 :(得分:0)

当我在MessageDialog.ShowAsync事件处理程序中调用Application.UnhandledException时,我的UWP应用程序崩溃了,这是我搜索答案时的最高结果。

我通过设置UnhandledExceptionEventArgs.Handled = true之前 调用MessageDialog.ShowAsync解决了该问题。

another SO question中对此进行了很好的记录,但是直到我找出解决方案后,我才找到答案,因为我没有意识到从Application.UnhandledException调用该方法很重要。