爲什麼在WPF中顯示一個OpenFileDialog會阻止我的WPF彈出窗口？

Question

我有一個WPF應用程序，我打開彈出mennu（彈出控件）使用轉義鍵。在該彈出菜單中，當按下按鈕時，我打開文件對話框，按下按鈕時關閉彈出窗口。當我下次按下esc按鈕時，它不會彈出，直到我已經集中另一個程序，例如。重置焦點。有誰知道什麼可能導致這種情況？

編輯

//called when pushing esc 
private void ShowSettingsMenu() 
{ 
    SettingsMenu.IsOpen = true; 
} 

//clicking my button, subsequent presses on my esc, doesnt pop it up (the code is run) 
private void ImportLicenseButton_Click(object sender, RoutedEventArgs e) 
{ 
    SettingsMenu.IsOpen = false; //<- hiding it again 

    OpenFileDialog filedialog = new OpenFileDialog(); 
    filedialog.Filter = "Xml Files|*.xml"; 
    if ((bool)filedialog.ShowDialog()) 
    { 
     string fileName = "license.xml"; 
     string destinationFolder = new FileInfo(Assembly.GetExecutingAssembly().Location).Directory.FullName; 
     if (!string.IsNullOrEmpty(filedialog.FileName)) 
     { 
     File.Copy(filedialog.FileName, System.IO.Path.Combine(destinationFolder, fileName), true); 
     } 
     else 
     { 
     MessageBox.Show("Please select a file name"); 
     } 
    } 
    this.Cursor = Cursors.None; 
} 

Answer 1

通過重新顯示彈出式固定。