Sql проблемa логина ?

[CyberTramp]

Подскажите что я делаю не правильно ?

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using System;
using Tamir.SharpSsh;
using Tamir.Streams;
using System.Threading;
using MySql.Data.MySqlClient;
using System.Data.Common;

namespace sshconsole
{
    class MainClass
    {   
        public static void Main (string[] args)
        {
            sshConect sshR = new sshConect();
            sshR.conect("uname -a");
            Console.WriteLine ("SSH*******");
            sqlConect sqlC = new sqlConect();
            sqlC.sendCommand();
            Console.WriteLine ("SQL*******");
        }
    }
    public class sshConect
    {
        public static string host = "10.0.0.15";
        public static string user = "root";
        public static string pass = "pass";
        public static int port = 22;
        public static string output;
       
        public void conect (string comand)
        {
            SshShell ssh = new SshShell(host, user, pass);
            ssh.Connect(port);
            //ssh.RedirectToConsole();
            ssh.WriteLine(comand);
            Thread.Sleep(1000);
            output = ssh.Expect();
            Console.WriteLine(output);
            //ssh.Close();
        }
       
    }
   
    public class sqlConect
{
        public static string output;
       
    public void sendCommand()
    {
        const string connectinString = "Server=localhost;Uid=mysql -u mailsql -p mailsql;Pwd='pass';";
        MySqlConnection connection = new MySqlConnection(connectinString);
        const string query =     "update users set clear='Test' where id=2;";
           
        MySqlCommand command = new MySqlCommand(query, connection);
            Console.WriteLine("startConect");
        connection.Open();
        MySqlDataReader reader = command.ExecuteReader();
           
        foreach (DbDataRecord record in reader)
            Console.WriteLine(reader[0] + "   " + record[1]);
        connection.Close();
        Console.ReadKey(true);
    }
}   
}

*****************************************************************************************************************
*********************************** "Result" 01******************************************************************
*****************************************************************************************************************
Last login: Wed Oct 12 13:36:57 EEST 2011 from 10.0.0.10 on pts/2
Linux nonama 2.6.32-33-generic #72-Ubuntu SMP Fri Jul 29 21:08:37 UTC 2011 i686 GNU/Linux
SSH*******

Unhandled Exception: System.ArgumentException: Format of initialization string does not conform to specifications at index 64.
  at System.Data.Common.DbConnectionStringBuilder.ParseConnectionStringNonOdbc (System.String connectionString) [0x00000]
  at System.Data.Common.DbConnectionStringBuilder.ParseConnectionString (System.String connectionString) [0x00000]
  at System.Data.Common.DbConnectionStringBuilder.set_ConnectionString (System.String value) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnectionStringBuilder..ctor (System.String connectionString) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnection.set_ConnectionString (System.String value) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnection..ctor (System.String connectionString) [0x00000]
  at (wrapper remoting-invoke-with-check) MySql.Data.MySqlClient.MySqlConnection:.ctor (string)
  at sshconsole.sqlConect.sendCommand () [0x00000] in /home/NoName/Projects/ssh console/ssh console/Main.cs:83
  at sshconsole.MainClass.Main (System.String[] args) [0x00021] in /home/NoName/Projects/ssh console/ssh console/Main.cs:30
The application was terminated by a signal: SIGHUP


*****************************************************************************************************************
*********************************** "Result" 02******************************************************************
*****************************************************************************************************************
IF  const string connectinString = "Server=localhost;Uid=mysql -u mailsql -p mailsql;Pwd='pass';";
==  const string connectinString = "Server=localhost;Uid='mysql -u mailsql -p mailsql';Pwd='pass';";

Last login: Wed Oct 12 13:39:50 EEST 2011 from 10.0.0.10 on pts/2
Linux nonama 2.6.32-33-generic #72-Ubuntu SMP Fri Jul 29 21:08:37 UTC 2011 i686 GNU/Linux
SSH*******

Unhandled Exception: System.ArgumentException: Keyword not supported.
Parameter name: ;pwd
  at MySql.Data.MySqlClient.MySqlConnectionStringBuilder.GetKey (System.String key) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnectionStringBuilder.SetValue (System.String keyword, System.Object value) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnectionStringBuilder.set_Item (System.String keyword, System.Object value) [0x00000]
  at System.Data.Common.DbConnectionStringBuilder.ParseConnectionStringNonOdbc (System.String connectionString) [0x00000]
  at System.Data.Common.DbConnectionStringBuilder.ParseConnectionString (System.String connectionString) [0x00000]
  at System.Data.Common.DbConnectionStringBuilder.set_ConnectionString (System.String value) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnectionStringBuilder..ctor (System.String connectionString) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnection.set_ConnectionString (System.String value) [0x00000]
  at MySql.Data.MySqlClient.MySqlConnection..ctor (System.String connectionString) [0x00000]
  at (wrapper remoting-invoke-with-check) MySql.Data.MySqlClient.MySqlConnection:.ctor (string)
  at sshconsole.sqlConect.sendCommand () [0x00000] in /home/NoName/Projects/ssh console/ssh console/Main.cs:83
  at sshconsole.MainClass.Main (System.String[] args) [0x00021] in /home/NoName/Projects/ssh console/ssh console/Main.cs:30
The application was terminated by a signal: SIGHUP

Более читаемый -> http://paste.org.ru/?lt0uyv

Как правильно делать запрос ?

const string connectinString = "Server=localhost;Uid='mysql -u mailsql -p mailsql';Pwd='pass';";
Обычно в манах Uid=Логин
В моем случае Uid='mysql -u mailsql -p mailsql'

Чем можно заменить команду mysql -u mailsql -p mailsql ?
на "mysql" ругается

Код: [Выделить]

ERROR 1045 (28000): Access denied for user 'root'@'localhost' (using password: NO)