title: C#: OpenFileDialog copy a file
body: <p>Hi I have a button with this code:</p>
  
  <pre><code>     OpenFileDialog ofd = new OpenFileDialog();
          string percorso = Environment.CurrentDirectory;
          percorso += @"\img\Articoli";
          ofd.InitialDirectory = percorso;
          ofd.Filter = "File JPEG/JPG (*.jpeg *.jpg)|*.jpg;*.jpeg|File PNG (*.png)|*.png;|Tutti i files (*.*)|*.*"; 
          if (ofd.ShowDialog() == DialogResult.OK)
          {
              tPercorso.Text = System.IO.Path.GetFileName(ofd.FileName);
              Refresh_Image();
          }       
  </code></pre>
  
  <p>I need that when I select a file in the OpenFileDialog, if his path is not equal to "percorso" a MessageBox will appear: "Do you want to copy file in default path?" YES | NO if DialogResult == YES I need to copy this file to path "percorso" and select filename only like I'm doing.
  Can you help me?
  Regards.</p>