Inheritance: System.Windows.Forms.Form
QuickOpenDialog Class Documentation
Exemple #1
0
 private void menuQuickOpen_Click(object sender, EventArgs e)
 {
     using (var dlg = new QuickOpenDialog())
     {
         if (dlg.ShowDialog(this) == System.Windows.Forms.DialogResult.OK)
         {
             OpenDisc(dlg.DiscPath);
         }
     }
 }
Exemple #2
0
 private void menuQuickOpen_Click(object sender, EventArgs e)
 {
     using (var dlg = new QuickOpenDialog())
     {
     if (dlg.ShowDialog(this) == System.Windows.Forms.DialogResult.OK)
     {
         OpenDisc(dlg.DiscPath);
     }
     }
 }