C#游戏重置按钮

[grayyy]

做了个游戏，用ado.net链接SQL存储游戏记录，用label显示，如图
重置按钮代码如下

private void button1_Click(object sender, EventArgs e)
        {
            SqlConnection conn = new SqlConnection("Server=(local);Database=Mine;uid=sa;pwd=netbox123*;");
            string sql = "delete from Count";
            SqlCommand comm = new SqlCommand(sql, conn);
            conn.Open();
            comm.ExecuteNonQuery();
            conn.Close();
            Form2 f = new Form2();
            f.Show();
        }

可是这样点击重置按钮后是弹出一个新的Form2窗口，我想要的是点击重置后直接把查询后的内容在原来的窗口是上显示。
请问要怎么做？
PS：之前链接数据库查询的命令都是写在public Form2里的：

 public Form2()
        {
            InitializeComponent();
            SqlConnection sqlConnection1 = new SqlConnection("Server=(local);Database=Mine;uid=sa;pwd=netbox123*;");
            //创建数据连接 连接字符串改成用于读取自己数据库的
            sqlConnection1.Open();
            SqlCommand selectCMD = new SqlCommand("select count(*) from Count;", sqlConnection1);
            //创建并初始化SqlCommand对象 
            SqlDataAdapter sqlDataAdapter1 = new SqlDataAdapter();
            selectCMD.Connection = sqlConnection1;
            SqlDataAdapter custDA = new SqlDataAdapter();
            sqlDataAdapter1.SelectCommand = selectCMD;
            custDA.SelectCommand = selectCMD;

            //创建SqlDataAdapter对象，并根据SelectCommand属性检索数据 
            DataSet dsDataSet1 = new DataSet();
            sqlDataAdapter1.Fill(dsDataSet1, "Customers");
            //使用SqlDataAdapter的Fill方法填充DataSet 
            label5.Text = dsDataSet1.Tables[0].Rows[0][0].ToString();
}

[BitCoffee]

private void button1_Click(object sender, EventArgs e)
        {
            SqlConnection conn = new SqlConnection("Server=(local);Database=Mine;uid=sa;pwd=netbox123*;");
            string sql = "delete from Count";
            SqlCommand comm = new SqlCommand(sql, conn);
            conn.Open();
            comm.ExecuteNonQuery();
            conn.Close();
            //这里给label的text属性直接赋值不就行了.
            label1.Text = "0";
        }