MySql e l'inserimento dell'ultimo problema ID rimangono

Chiarimento, il driver ODBC mySQL .net non consente l'esecuzione di più comandi come stai descrivendo. Devi effettuare due chiamate separate e avvolgerle in una transazione.

// open a new connection using a default connection string I have defined elsewhere
using( OdbcConnection connection = new OdbcConnection( s_connectionString ) )
{
      // ODBC command and transaction objects
      OdbcCommand command = new OdbcCommand();
      OdbcTransaction transaction = null;

      // tell the command to use our connection
      command.Connection = connection;

      try
      {
           // open the connection
           connection.Open();

           // start the transaction
           transaction = connection.BeginTransaction();

           // Assign transaction object for a pending local transaction.
           command.Connection = connection;
           command.Transaction = transaction;

           // TODO: Build a SQL INSERT statement
           StringBuilder SQL = new StringBuilder();

           // run the insert using a non query call
           command.CommandText = SQL.ToString();
           command.ExecuteNonQuery();

           /* now we want to make a second call to MYSQL to get the new index 
              value it created for the primary key.  This is called using scalar so it will
               return the value of the SQL  statement.  We convert that to an int for later use.*/
           command.CommandText = "select last_insert_id();";
           id = Convert.ToInt32( command.ExecuteScalar() );

           // Commit the transaction.
           transaction.Commit();
     }
     catch( Exception ex )
     {
          Debug.WriteLine( ex.Message );

          try
          {
               // Attempt to roll back the transaction.
               transaction.Rollback();
            }
            catch
            {
                 // Do nothing here; transaction is not active.
              }
         }
}