MySql.Data.dll 不支持输出参数

insert INTO stu(name) VALUES('maimai');
set @ReturnValue=@@IDENTITY;

  

string sql="insert INTO stu(name) VALUES(@name);set @ReturnValue=@@IDENTITY;
 MySqlParameter[] parameters = {
		new MySqlParameter("@name", MySqlDbType.VarChar,100),MySqlDbType.Int32,1),
                new MySqlParameter("@ReturnValue",MySqlDbType.Int32)
                   };

  


public static int ExecuteSql(string SQLString, params MySqlParameter[] cmdParms)
        {
            using (MySqlConnection connection = new MySqlConnection(connectionString))
            {
                using (MySqlCommand cmd = new MySqlCommand())
                {
                    try
                    {
                        PrepareCommand(cmd, connection, null, SQLString, cmdParms);
                        int rows = cmd.ExecuteNonQuery();
                        cmd.Parameters.Clear();
                        return rows;
                    }
                    catch (MySql.Data.MySqlClient.MySqlException e)
                    {
                        throw e;
                    }
                }
            }
        }


     private static void PrepareCommand(MySqlCommand cmd, MySqlConnection conn, MySqlTransaction trans, string cmdText, MySqlParameter[] cmdParms)
        {
            if (conn.State != ConnectionState.Open)
                conn.Open();
            cmd.Connection = conn;
            cmd.CommandText = cmdText;
            if (trans != null)
                cmd.Transaction = trans;
            cmd.CommandType = CommandType.Text;//cmdType;
            if (cmdParms != null)
            {


                foreach (MySqlParameter parameter in cmdParms)
                {
                    if ((parameter.Direction == ParameterDirection.InputOutput || parameter.Direction == ParameterDirection.Input) &&
                        (parameter.Value == null))
                    {
                        parameter.Value = DBNull.Value;
                    }
                    cmd.Parameters.Add(parameter);


                }
            }
        }

 


 然后执行 

ExecuteSql(sql,parameters);

想要返回 return 1(int)parameters[1].Value;

发现报错 : near  NULL=@@IDENTITY;

问题:难道是mysql.data 把输出参数的null值也带入sql语句了?

最后呈现我们的是:insert INTO stu(name) VALUES('maimai');set NULL=@@IDENTITY;

 

根源还是没解决,有大神知道的麻烦解决下,下的mysql.data的源码 调试一直进不去,就没接着调试了...



posted @ 2016-11-18 13:46  败草麦麦  阅读(213)  评论(0编辑  收藏  举报