Connection to MySql error

[DPaul1994]

Hi guys. I have a form where I have some labels, checkboxes and a button. For connection to MySql I have a class named ConnConfig and functions for open and closing connection(if needed, closeConn(), openConn()). So, I have a function which extracts from database some data:

private void select()
        {
            using (ConnConfig.getConnection())
            {
                MySqlCommand cmd = new MySqlCommand(dataA, ConnConfig.getConnection());
                cmd.CommandType = CommandType.Text;
[indent]            // Connection must be valid and open.[/indent]
                using (MySqlDataReader rdra = cmd.ExecuteReader())
                {
                    try
                    {
                        while (rdra.Read())
                        {
                            label2.Text = rdra["question"].ToString();
                            label3.Text = rdra["answer1"].ToString();
                            label4.Text = rdra["answer2"].ToString();
                            label5.Text = rdra["answer3"].ToString();
                            r1 = (int)rdra["option1"];
                            r2 = (int)rdra["option2"];
                            r3 = (int)rdra["option3"];
                        }
                    }
                    catch (InvalidOperationException ex)
                    {
                        MessageBox.Show(ex.Message);
                    }
                    finally
                    {
                        ConnConfig.closeConn();
                    }
                }
            }
        }

This function is called first on form load. I also call her in button1_Click function.
So, the problem is this: If I acces form, it tells me that 'Connection must be valid and open.', at DataReader from select() function. Then, if I open the connection above 'using (MySqlDataReader rdra = cmd.ExecuteReader())' line and run the script, everything is ok, but if I click on button1 I receive: 'The connection is already open', at the line where I opened connection.(ConnConfig.connOpen())

private void select()
        {
            using (ConnConfig.getConnection())
            {
                MySqlCommand cmd = new MySqlCommand(dataA, ConnConfig.getConnection());
                cmd.CommandType = CommandType.Text;
[indent]              ConnConfig.connOpen(); // error: The connection is already open.[/indent]
                using (MySqlDataReader rdra = cmd.ExecuteReader())
                {
                    try
                    {
                        while (rdra.Read())
                        {
                            label2.Text = rdra["question"].ToString();
                            label3.Text = rdra["answer1"].ToString();
                            label4.Text = rdra["answer2"].ToString();
                            label5.Text = rdra["answer3"].ToString();
                            r1 = (int)rdra["option1"];
                            r2 = (int)rdra["option2"];
                            r3 = (int)rdra["option3"];
                        }
                    }
                    catch (InvalidOperationException ex)
                    {
                        MessageBox.Show(ex.Message);
                    }
                    finally
                    {
                        ConnConfig.closeConn();
                    }
                }
            }
        }

 

[DPaul1994]

Anybody?

 

[JustinEzequiel]

I have no idea what your returns but have you tried

using (var conn = ConnConfig.getConnection())
{
	conn.Open(); // or .open()
	...
}

 

[DPaul1994]

getConnection() returns true or false. If connection is on, returns true, else is trying to connect.

 

[JustinEzequiel]

getConnection() returns true or false

so basically, you are creating an instance of MySqlCommand via

MySqlCommand cmd = new MySqlCommand(dataA, [COLOR=#CC0000]true[/color]);

I think you need to get the MySqlConnection object and pass that instead of true to the MySqlCommand constructor

 

[DPaul1994]

I solved it, thank you