Question

每当我添加这行时：SET @last_id_owner = LAST_INSERT_ID（）;.它给了我警告; MysqlException未在行中处理：cmd.ExecuteNonQuery（）; 如果我删除它就行了。

String sql = "INSERT INTO owner (name,email,reward)"
             + " VALUES('" + owner.Name + "','" + owner.EMail + "'," + owner.Reward + ");SET @last_id_owner = LAST_INSERT_ID();";
connect();
MySqlCommand cmd = new MySqlCommand(sql, conexion);
cmd.ExecuteNonQuery();
disconnect();

我需要获取最后一个插入ID并在另一个表的插入中使用它。怎么了？

Answer 1

您必须注意的第一件事是，您的查询将为SqLInjection打开一扇大门。要避免这种情况，您应该使用参数化查询。然后，您可以使用LastInsertedId类的MySqlCommand来获取最后插入的ID。您可以使用以下代码段来实现这两个目标：

// Parameterization implementation
String sql = "INSERT INTO owner (name,email,reward)"
    + " VALUES(@ownerName,@ownerMail,@ownerReward)";

MySqlCommand cmd = new MySqlCommand(sql, conexion);
cmd.Parameters.Add("@ownerName",MySqlDbType.VarChar).Value= owner.Name;
cmd.Parameters.Add("@ownerMail", MySqlDbType.VarChar).Value = owner.EMail;
cmd.Parameters.Add("@ownerReward", MySqlDbType.VarChar).Value = owner.Reward;

// Executing Query and getting last Inserted ID
cmd.ExecuteNonQuery();
long lastInsertedID = cmd.LastInsertedId;

Answer 2

尝试使用此代码它将帮助您......如您所愿。

 String sql = "INSERT INTO owner (name,email,reward)  VALUES('" + owner.Name + "','" + owner.EMail + "'," + owner.Reward + "); select LAST_INSERT_ID();";
            connect();
            MySqlCommand cmd = new MySqlCommand(sql, conexion);
            cmd.ExecuteScalar();
            disconnect();

MysqlException未处理C＃LAST_INSERT_ID（）

2 个答案: