[c#] How to open the second form?

I have Form1 and Form2 in my project. Form2 is just a form with settings for Form1. What is the command to open the Form2 from the Form1 and also what's the command to close it please?

Form1 OpenNewForm = new Form1();
OpenNewForm.Show();

"OpenNewForm" is the name of the Form. In the second the form opens.

If you want to close the previous form:

this.Close();

In a single line it would be:

(new Form2()).Show();

Hope it helps.

//To open the form

Form2 form2 = new Form2();

form2.Show();
// And to close
form2.Close();

Hope this helps

On any click event (or other one):

Form2 frm2 = new Form2();
frm2.Show();

I assume your talking about windows forms:

To display your form use the Show() method:

Form form2 = new Form();
form2.Show();

to close the form use Close():

form2.Close();

If you need to show Form2 as a modal dialog, from within Form1 do:

var form2 = new Form2();
if (form2.ShowDialog() == DialogResult.OK) 
{
    // process results here
}

A modal dialog will retain focus while it is open; it will set the parent windows (Form1) "in the background" until it is closed, which is quite a common practice for settings windows.

Start with this:

var dlg = new Form2();
dlg.ShowDialog();

If you want to open Form2 modally (meaning you can't click on Form1 while Form2 is open), you can do this:

using (Form2 f2 = new Form2()) 
{
    f2.ShowDialog(this);
}

If you want to open Form2 non-modally (meaning you can still click on Form1 while Form2 is open), you can create a form-level reference to Form2 like this:

private Form2 _f2;

public void openForm2()
{
    _f2 = new Form2();
    _f2.Show(this); // the "this" is important, as this will keep Form2 open above 
                    // Form1.
}

public void closeForm2()
{
    _f2.Close();
    _f2.Dispose();
}

Respectively Form.Show() (or Form.ShowDialog() if you want the second form to be modal), and Form.Hide() (or Form.Close(), depending on what you mean by close it).