Question

对编程仍然很陌生，并试图让我的头脑围绕将项目连接到数据库。我已经使用 XAMPP/phpmyAdmin 创建了一个数据库，并使用以下代码将其成功连接到我的 Windows 窗体应用程序：

const string connectInfo = @"data source = localhost; Port = 3307; username = 'root'; password = ''; database = caesar_cypher;";
public static void AttemptLogin(string username, string password)
        {
            try
            {
               
                var connect = new MySqlConnection(connectInfo);
                var LoginCommand = new MySqlCommand($"SELECT * FROM login WHERE userName = '{username}' AND password = '{password}';", connect);
                MySqlDataReader reader;
                connect.Open();
                reader = LoginCommand.ExecuteReader();
                int count = 0;
                while (reader.Read())
                { count += 1; }
                if (count == 1)
                {
                    MessageBox.Show("Log in successful");
                }
                else if (count > 1)
                {
                    MessageBox.Show("Incorrect username or password");
                }
                else
                {
                    MessageBox.Show("Duplicate username and password - access denied.");
                }
                MessageBox.Show("Connected");
                connect.Close();
                if (count == 1)
                {
                    var form2 = new CypherForm();
                    form2.Activate();
                    form2.Show();
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
            }
        }

此（以上）代码运行良好，但容易受到 SQL 注入攻击。因此，我按照涵盖 SqlCommand.Prepare() 方法的 this Microsoft Docs 页面跟踪并修改了我的代码。现在，当我运行我的代码时，出现以下异常：

Keyword not supported 'port'

这是新代码：

const string connectInfo = @"data source = localhost; Port = 3307; username = 'root'; password = ''; database = caesar_cypher;";
public static bool AttemptSecureLogin(string username, string password)
        {
            try
            {
                using (var connection = new SqlConnection(connectInfo))
                {
                    connection.Open();
                    var loginCommand = new SqlCommand(null, connection)
                    {
                        CommandText =
                        "SELECT * FROM login WHERE userName = @uName AND password = @pWord"
                    };
                    var uNameParam = new SqlParameter("@uName", SqlDbType.VarChar, 30);
                    var pWordParam = new SqlParameter("@pWord", SqlDbType.VarChar, 30);
                    uNameParam.Value = username;
                    pWordParam.Value = password;
                    loginCommand.Parameters.Add(uNameParam);
                    loginCommand.Parameters.Add(pWordParam);
                    loginCommand.Prepare();
                    loginCommand.ExecuteNonQuery();

                    SqlDataReader reader = loginCommand.ExecuteReader();
                    int count = 0;
                    while(reader.Read())
                    { count += 1; }
                    if(count == 1) { 
                        MessageBox.Show("Log in successful!");
                        var cypherForm = new CypherForm();
                        cypherForm.Activate();
                        cypherForm.Show();
                        return true;
                    }
                    else if(count > 1) { 
                        MessageBox.Show("Incorrect username/password.");
                        return false;
                    }
                    else { 
                        MessageBox.Show("Duplicate username/password.");
                        return false;
                    }
                    MessageBox.Show("Connected");
                    connection.Close();
                }
            }
            catch (Exception ex)
            {
                MessageBox.Show(ex.Message);
                return false;
            }

我在这里找到了一些类似的帖子，但真的不明白问题出在哪里，有人可以帮忙吗？

他

“不支持关键字‘端口’”

0 个答案: