Do not open more than one form at a time in C #

From what I found by searching, it is not possible to open a form mdi child as dialog .

You have two options:

Using singleton:

public partial class Form2 : Form
{
 .....
 private static Form2 inst;
 public static Form2  GetForm
 {
   get
    {
     if (inst == null || inst.IsDisposed)
         inst = new Form2();
     return inst;
     }
 }
 ....
}
Invoke/Show this form,

Form2.GetForm.Show();

Source: link

I found this solution on the net:

frm2 f = Application.OpenForms["NameOfForm2"];
if(f != null)
   f.BringToFront();
else
{
   frm2 f = new frm2();
   f.Show();
}

Source: link

Or see the example below where I check if Outlook is running before starting it. You can do this in the builder of your screen.

//Se não estiver iniciado, inicia uma instância do outlook
if (Process.GetProcessesByName("OUTLOOK").Count().Equals(0))
{
    Process.Start("outlook.exe");//Aqui continue seu código normalmente
}
else
{
   //Cancele o processo
   Window.GetWindow(this).Close();
}

Hermano boa, open the form this way:

DialogResult dialogResult = frmNomeDoSeuForm.ShowDialog();

But I have a hint, as today the needs of multiteam systems would be interesting you just check if that same form is already open ... if yes let it in the foreground and in focus .... so you you can leave another part of the system on another screen ... see it is simple ...

        if (Application.OpenForms.OfType<FrmNomeDoForm>().Count() > 0)
        {
            Application.OpenForms.OfType<FrmNomeDoForm>().First().Focus();
        }
        else
        {
            FrmNomeDoForm frmNomeDoForm = new FrmNomeDoForm();
            frmNomeDoForm.Show();
        }