C＃WPF - 暂停调度程序处理时无法执行此操作

Question

我有ContentDialogHost在ContentDialog中显示Grid。

ShowDialog()方法处理在主机中显示控件，并返回ContentDialogResult。

我希望它像Windows对话框一样工作，你调用ShowDialog()，代码会停止，直到用户按下任何对话框按钮。

这是我的实施：

public ContentDialogResult ShowDialog(ContentDialog dialog)

        {
            // show host
            this.Visibility = Visibility.Visible;

            _dialogResult = null;

            if (_currentDialog == null)
            {
                containerGrid.Children.Add(dialog);
                _currentDialog = dialog;
            }

            dialog.PrimaryButtonClick += (s, ev) => { _dialogResult = ContentDialogResult.Primary; };
            dialog.SecondaryButtonClick += (s, ev) => { _dialogResult = ContentDialogResult.Secondary; };

            while (_dialogResult == null)
            {
                if (this.Dispatcher.HasShutdownStarted ||
                    this.Dispatcher.HasShutdownFinished)
                {
                    break;
                }

                // !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
                // THIS IS WHERE IT THROWS THE EXCEPTION BELOW
                // !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

                this.Dispatcher.Invoke(
                    DispatcherPriority.Background,
                    new ThreadStart(delegate { }));
                Thread.Sleep(20);
            }

            CloseDialog();

            return _dialogResult.Value;
        }

这包含在Popups类中，如下所示：

public ContentDialogHost.ContentDialogResult ShowDialog(ContentDialog dialog)
        {
            return mainwindow.contentdialogHost.ShowDialog(dialog);
        }

仅在从IsVisibleChanged事件调用ShowDialog()时发生异常，例如：

private void Page_IsVisibleChanged(object sender, DependencyPropertyChangedEventArgs e)
        {
            if (this.IsVisible)
                ShowSubviewDebugDialog(); // this calls `ShowDialog()` from the above mentioned Popups class
        }

我该怎么做才能解决或解决这个问题？