Back to previous form with the same data in C#

Question

i was learning to create a window application by using c#. Current i faces a problem that, when i retun from form 2 to form 1, my data get from database is gone. How can i go back to previous page without the losing of my data on form 1

private void AdminMain_Load(object sender, EventArgs e)
{
    conn.Open();
    string getInfo = "SELECT FullName FROM staff WHERE StaffID = '" + Username + "'";
    MySqlCommand cmd = new MySqlCommand(getInfo, conn);
    MySqlDataReader mdr = cmd.ExecuteReader();
    if (mdr.Read())
    {
        userName.Text = mdr["FullName"].ToString();
    }

}

This is my form 1 which will display the username. However, when I am using the code below to return to the form 1, the data in form 1 is gone.

private void backBtn_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    AdminMain am = new AdminMain();
    am.ShowDialog();
    this.Hide();
}

How can I keep the data in the form 1 when I return to it.

UPDATE: how my code open form2 in form1 is

private void profileBtn_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
    {
        MyProfileAdmin pf = new MyProfileAdmin();
        pf.getInformation(Username.ToString());
        this.Hide();
        pf.ShowDialog();
    }

and this is my code on form2 to "try back to form1"

private void backBtn_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
    {
        this.Close();
    }

Answer 1

If you open Form2 with ShowDialog , Form1 still remains open and will continue as soon as Form2 is closed.

So, use this code to close Form2.

private void backBtn_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    this.Close(); // Close, not just hide
}

In order to hide Form1 while Form2 is shown:

private void profileBtn_LinkClicked(object sender, LinkLabelLinkClickedEventArgs e)
{
    MyProfileAdmin pf = new MyProfileAdmin();
    pf.getInformation(Username.ToString());

    this.Hide(); 
    try
    {
        pf.ShowDialog();
    }
    finally
    {
        this.Show();
    }
}

We use try/finally to make sure the first form is shown when there is an exception while showing the other form.