使用FileOpenPicker时出现UnauthorizedAccessException

我的Windows应用商店应用程序中有一个CommandBar，当我单击CommandBar上的Open按钮时，它会运行OpenFile处理程序，如下所示：

private async void OpenFile(object sender, RoutedEventArgs e)
{
    MessageDialog dialog = new MessageDialog("You are about to open a new file. Do you want to save your work first?");
    dialog.Commands.Add(new UICommand("Yes", new UICommandInvokedHandler(SaveAndOpen)));
    dialog.Commands.Add(new UICommand("No", new UICommandInvokedHandler(Open)));
    await dialog.ShowAsync();
}
private async void SaveAndOpen(IUICommand command)
{
    await SaveFile();
    Open(command);
}
private async void Open(IUICommand command)
{
    FileOpenPicker fileOpenPicker = new FileOpenPicker();
    fileOpenPicker.ViewMode = PickerViewMode.List;
    fileOpenPicker.FileTypeFilter.Add(".txt");
    StorageFile file = await fileOpenPicker.PickSingleFileAsync();
    await LoadFile(file);
}

我看到的消息很好，但只有当我点击Yes时，我才会收到FileOpenPicker。当我点击No时，我在以下行得到一个UnauthorizedAccessException: Access is denied.：StorageFile file = await fileOpenPicker.PickSingleFileAsync();

我很困惑。。。有人知道为什么会发生这种事吗？我甚至试着在一个调度器中运行它，以防处理程序在另一个线程上被调用，但是。。。不幸的是，同样的事情：

await CoreApplication.MainView.CoreWindow.Dispatcher.RunAsync(CoreDispatcherPriority.High, async () =>
{
    StorageFile file = await fileOpenPicker.PickSingleFileAsync();
    await LoadFile(file);
});

private async void OpenFile(object sender, RoutedEventArgs e)
{
    MessageDialog dialog = new MessageDialog("You are about to open a new file. Do you want to save your work first?");
    IUICommand result = null;
    dialog.Commands.Add(new UICommand("Yes", (x) =>
    {
        result = x;
    }));
    dialog.Commands.Add(new UICommand("No", (x) =>
    {
        result = x;
    }));
    await dialog.ShowAsync();
    if (result.Label == "Yes")
    {
        await SaveFile();
    }
    FileOpenPicker fileOpenPicker = new FileOpenPicker();
    fileOpenPicker.ViewMode = PickerViewMode.List;
    fileOpenPicker.FileTypeFilter.Add(".txt");
    StorageFile file = await fileOpenPicker.PickSingleFileAsync();
    await LoadFile(file);
}