c# - MySql Last_Insert_Id() 的并发问题... (C#)

Question

我不太确定为什么以下控制台应用程序没有产生last_insert_id() 的预期行为。我读过，last_insert_id() 返回特定连接的最后一个 auto_incremented 值，但在这段代码中，两个连接返回相同的结果。有人可以解释一下我哪里出了问题吗？

    static void Main(string[] args)
    {
        string ConnectionString = "server=XXXX;database=XXXX;username=XXXX;password=XXXX;pooling=true;max pool size=100;min pool size=0";

        MySqlConnection conn1 = new MySqlConnection(ConnectionString);
        MySqlConnection conn2 = new MySqlConnection(ConnectionString);

        MySqlCommand command1 = new MySqlCommand();
        MySqlCommand command2 = new MySqlCommand();

        command1.Connection = conn1;
        command2.Connection = conn1;

        StringBuilder createTableCommandText = new StringBuilder();
        createTableCommandText.Append("Create Table TestTable (");
        createTableCommandText.Append("Id INT NOT NULL AUTO_INCREMENT, ");
        createTableCommandText.Append("str VARCHAR(20) NOT NULL, ");
        createTableCommandText.Append("PRIMARY KEY (Id));");

        StringBuilder insertCommandText = new StringBuilder();
        insertCommandText.Append("INSERT INTO TestTable (str) VALUES ('what is the dilleo?');");

        StringBuilder getLastInsertId = new StringBuilder();
        getLastInsertId.Append("SELECT LAST_INSERT_ID();");

        conn1.Open();
        conn2.Open();

        command1.CommandText = createTableCommandText.ToString();
        command1.ExecuteNonQuery();

        command1.CommandText = insertCommandText.ToString();
        command2.CommandText = insertCommandText.ToString();

        command1.ExecuteNonQuery();
        command2.ExecuteNonQuery();

        command1.CommandText = getLastInsertId.ToString();
        Console.WriteLine("Command 1:  {0}", command1.ExecuteScalar().ToString());
        command2.CommandText = getLastInsertId.ToString();
        Console.WriteLine("Command 2:  {0}", command2.ExecuteScalar().ToString());

        conn1.Close();
        conn2.Close();

        Console.ReadLine();
    }

我知道这种情况不会发生，因为顶部连接字符串中的池化，因为我尝试在没有该字符串部分的情况下运行程序，并且我仍然得到相同的结果(即命令 1 和命令 2为last_insert_id()显示相同的值。欢迎任何想法!

非常感谢，

安德鲁

Answer 1

看这里:

command1.Connection = conn1;
command2.Connection = conn1;

您对这两个命令使用相同的连接。