C# 在窗体之间传递值时出错

在将用户名和密码从form2传递到form3时，当我第一次运行应用程序时，该值不会第一次被传递。然而，当我重新运行应用程序时，它每次都能正常工作，并且添加成功，我对c#和windows编程非常陌生，所以请原谅任何执行不力的情况。流程为form1->form2->form3

更新： 这是表格3的定义

  public partial class Form3 : Form
    {
        MySqlConnection connection;
        public string username;
        public string password;
        public Form3(string user,string pass)
        {
            username = user;
            password = pass;
            InitializeComponent();
        }
        private void button1_Click(object sender, EventArgs e)
        {
            MessageBox.Show(username, password);
            string connstring = "server=localhost;user=" + username + ";database=testdb;port=3306;password=" + password + ";";
            connection = new MySqlConnection(connstring);
            connection.Open();
            string query = "INSERT INTO user_books values('" + textBox1.Text + "','" + textBox2.Text + "','" + textBox3.Text + "');";
            MySqlCommand newcommand = new MySqlCommand(query,connection);
            try
            {
                newcommand.ExecuteNonQuery();
                MessageBox.Show("DONE");
            }
            catch (Exception e1)
            {
                MessageBox.Show(e1.ToString());
            }
            connection.Close();

        }

        private void Form3_Load(object sender, EventArgs e)
        {
            MessageBox.Show(username, password);
        }
    } 

这是表格2：

 public partial class Form2 : Form
    {
        public static MySqlConnection connection;

        public string username;
        public string password;

        public Form2(string user,string pass)
        {
            InitializeComponent();
            this.Text = "MAIN MENU";
            username = user;
            password = pass;
        }

        Form3 form3 = new Form3(username,password);
        private void button1_Click(object sender, EventArgs e)
        {
            MessageBox.Show(username,password);//this lines  no longer gives an empty message box with title Error
            form3.Show();
        }
    }

这是表格一：

public partial class Form1 : Form
    {
        public Form1()
        {
            InitializeComponent();
            this.Text = "LOGIN";
        }

        MySqlConnection connection;
        Form2 form2;
        private void button1_Click(object sender, EventArgs e)
        {
            string username = textBox1.Text;
            string password = textBox2.Text;

            string connstring = "server=localhost;user=" + username + ";database=testdb;port=3306;password=" + password + ";";

            try
            {
                connection = new MySqlConnection(connstring);
                connection.Open();

                form2 = new Form2(username,password);
                form2.Show();
            }
            catch (Exception e1)
            {
                MessageBox.Show(e1.ToString(), "Exception");
            }
            connection.Close();
        }
    }

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使用mysql和visual studio 2012

请为表单指定适当的名称，并删除所有静态变量

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将Form3的创建移到事件处理程序中，这样它将获得在单击时设置的用户名/密码，而不是在构建类时设置的用户名/密码。

如何导航？表格一->表格二->表格三。给出明确的规范。最初它们不是静态的，但后来我得到了这个错误：字段初始值设定项不能引用非静态字段、方法或属性“MySQLDatabaseProject.Form2.password”使用公共属性作为变量，而不是静态的