C# MySqlCommand Insert 显示 fatal error 命令

Question

我有一个具有以下类型的 table (Horario):

id (int, auto increment)
codMedico (int)
dia (Date)
horarioInicio (Time)
horarioFim (Time)
status_id (int)

我在下面的代码中看不到任何问题，但它只是向我显示了 mysqlException 上的“ fatal error ”(在 ExecuteNonQuerry() 方法中抛出。我在文本字段中收到了两次时间 (11:10) 和 (12 :10)，以及日期选择器中的日期。

代码如下:

public bool agendaMedico_Novo(String codMedico, DateTime dia, String hrInicio, String hrFim, int status_id)
{
    if (bdConn.State == ConnectionState.Open)
    {               
        MySqlCommand cmd = new MySqlCommand();
        cmd.Connection = bdConn;

        cmd.CommandText = "INSERT INTO horario (id, codMedico, dia, horarioInicio, horarioFim, status_id) VALUES (1, ?codMedico, ?dia, ?horarioInicio, ?horarioFim, ?status_id)";
        // cmd.CommandText = "INSERT INTO horario (codMedico, dia, horarioInicio, horarioFim, status_id) VALUES ("+codMedico+", "+dia.ToDate()+", "+"11:12"+ ", "+ "12:13"+ ", 1"+")";                                       
        try 
        {

            cmd.Parameters.Add("?codMedico", MySqlDbType.Int16).Value = Int16.Parse(codMedico);

        }

        catch (Exception e)
        {
            MessageBox.Show("Cod invalido!");

        }
        try
        {
            MessageBox.Show(dia.ToDate().ToString());
            cmd.Parameters.Add("?dia", MySqlDbType.Date).Value = dia.ToDate();
        }

        catch (Exception e)
        {
            MessageBox.Show("Dia invalido!");

        }
        try
        {
            cmd.Parameters.Add("?horarioInicio", MySqlDbType.Time).Value = DateTime.Parse(hrInicio);

        }
        catch (Exception e)
        {
            MessageBox.Show("Horario Invalido");

        }                  
        try
        {
            cmd.Parameters.Add("?horarioFim", MySqlDbType.Time).Value = DateTime.Parse(hrFim);

        }
        catch (Exception e)
        {
            MessageBox.Show("Horario Invalido");

        }
        try
        {
            cmd.Parameters.Add("?status_id", MySqlDbType.Int16).Value = status_id;
        }

        catch (Exception e)
        {
            MessageBox.Show("dasasdasd");
        }

        MessageBox.Show(cmd.CommandText);

        //try
        //{

        try
        {
            cmd.ExecuteNonQuery();
        }

        catch (MySqlException m)
        {
            MessageBox.Show(m.StackTrace);
            MessageBox.Show(m.Message);
        }

        return true;
    }
    else
    {
        MessageBox.Show("Erro de banco de dados. Tente novamente mais tarde");
        return false;
    }
    //return 0;
}

Answer 1

我认为这是因为您尝试将值 (1) 插入到 id 自动增量列中:

cmd.CommandText = "INSERT INTO horario (id, codMedico, dia, horarioInicio, horarioFim, status_id) VALUES (1, ?codMedico, ?dia, ?horarioInicio, ?horarioFim, ?status_id)";

请在您的 CommandText 上尝试此操作:

cmd.CommandText = "INSERT INTO horario (codMedico, dia, horarioInicio, horarioFim, status_id) VALUES (?codMedico, ?dia, ?horarioInicio, ?horarioFim, ?status_id)";

数据库将为id生成它自己的值，因此您不需要!