Question

I have the code:

private void Button1_Click (object sender, EventArgs e)
(
    Form2 f2 = new Form2 ();
    f2.Show ();
)

how to make so that Form2 when you click on Button1 not opened for the second time, if it is open?

sorry for bad english

Answer 1

You have to make sure you only have 1 instance of Form2 created. One way to do this is to move the object declaration and instantiation to a private variable and in your click event handler simply call the Show() method:

private Form2 f2 = new Form2();
private void Button1_Click (object sender, EventArgs e)
(
    if(f2 == null) 
    {
      f2 = new Form2();
    }
    f2.Show ();
)