private void pictureBox4_Click(object sender, EventArgs e)
{
menu m = new menu();
m.Show();
this.Hide();
}
private void button1_Click(object sender, EventArgs e)
{
/* try
* {
* MySqlConnection cn = new MySqlConnection("server=localhost;user id=root;database=projectdatabase;pwd=rohan!@#");
* cn.Open();
*
*
* }
* catch (Exception ex)
* { MessageBox.Show("Connection failed\n" + ex.Message); }*/
menu a = new menu();
a.Show();
this.Visible = false;
}