我将如何使用数据库创建登录表单以进行检查？

Question

我有一个数据库，其中有一个名为person的表，其中有两行称为personID和firstName

现在，我有一个带有2个文本框的表单，要求提供ID和名字。

问题是，我不知道如何检查数据库（如果ID和firstName来自同一记录），并且如果记录匹配并且如果它们不仅仅给出错误标签，则将您带到另一种称为admin形式。

我希望对此有任何帮助，谢谢。

编辑：这是到目前为止的代码， connection already open, connection must be open错误

private void btnLoginScreen_Click(object sender, EventArgs e)
{
    {
        switch (dbAuth(txtAdminLogID.Text, txtAdminLogName.Text))
        {
            case true:
                //Open Admin form
                break;
            case false:
                //Show Errorlabel here
                break;
            case null:
                //An error occured while fetching data
                break;
            default:
                break;
        }
    }
}

public bool? dbAuth(string personID, string firstName)
{
    try
    {
        MySqlCommand command = new MySqlCommand();
        command.CommandText = "SELECT * FROM person";
        MySqlDataReader Reader;
        conn.Open();
        Reader = command.ExecuteReader();
        while (Reader.Read())
        {
            if (Reader[0].ToString() == personID) //first check if the ID is equal
            {
                if (Reader[1].ToString() == firstName) //if ID is equal, check if firstName is equal
                {
                    conn.Close();
                    return true;
                }
            }
        }
        conn.Close();
        return false;
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.Message);
        return null;
    }
}

编辑：这是我的连接字符串：

string connstring = "server=localhost;user Id=root;database=collegesystem;Convert Zero Datetime=True ";
        MySqlConnection conn;

Answer 1

这就是我要走的样子（使用MySQL-Connector）：

public bool? Authenticate(string personID, string firstName)
{
    try
    {
        MySqlCommand command = connection.CreateCommand();
        command.CommandText = "SELECT * FROM yourTable";
        MySqlDataReader Reader;
        connection.Open();
        Reader = command.ExecuteReader();
        while (Reader.Read())
        {
            if (Reader[0].ToString() == personID) //first check if the ID is equal
            {
                if (Reader[1].ToString() == firstName) //if ID is equal, check if firstName is equal
                {
                    connection.Close();
                    return true; 
                }
            }
        }
        connection.Close();
        return false;
    }
    catch (Exception ex)
    {
        MessageBox.Show(ex.Message);
        return null;
    }
}

在您的表单中，登录按钮所在的位置使用以下代码：

private void MySQL_Auth_button_Click(object sender, EventArgs e)
        {
        switch (dbAuth.AuthenticatePlain(personID_textBox.Text, firstName_textBox.Text))
                {
                    case true:
                        //Open Admin form
                        break;
                    case false:
                        //Show Errorlabel here
                        break;
                    case null:
                        //An error occured while fetching data
                        break;
                    default:
                        break;
                }
         }

Answer 2

“这是您必须在“登录”按钮后面编写的代码

SqlConnection CN =新的SqlConnection（ConnectionString）;

            string query= "Select * from tblUsers Where (user_id=@id AND user_password=@pwd)";

            CN.Open();
            SqlCommand myCommand = new SqlCommand(txt, CN);
            myCommand.Parameters.Add(new SqlParameter("id", SqlDbType.NVarChar)).Value = this.txtUserID.Text;
            myCommand.Parameters.Add(new SqlParameter("pwd", SqlDbType.NVarChar)).Value = this.txtPassword.Text;
            SqlDataReader myReader;
            myReader = myCommand.ExecuteReader();
            myReader.Read();
            if (myReader.HasRows)
            {

                CN.Close();
                AdminForm mf = new AdminForm();
                mf.Show();
                this.Hide();
            }
            else
            {

                MessageBox.Show("Invalid User Name or Password", "Error", MessageBoxButtons.OK, MessageBoxIcon.Error);

            }