SqlTransaction.Commit()之前的SqlCommand.Dispose()？

Question

是否可以在提交事务之前处理分配给事务的命令？我自己测试了以下代码,它似乎运行良好,但这是一个相当小的例子,所以我正在寻找确认,如果有人肯定知道.

internal static void TestTransaction()
{
    try
    {
        Program.dbConnection.Open();
        using (SqlTransaction transaction = Program.dbConnection.BeginTransaction())
        {
            Boolean doRollback = false;
            for (int i = 0; i < 10; i++)
            {
                using (SqlCommand cmd = new SqlCommand("INSERT INTO [testdb].[dbo].[transactiontest] (testvalcol) VALUES (@index)", Program.dbConnection, transaction))
                {
                    cmd.Parameters.AddWithValue("@index", i);
                    try
                    {
                        cmd.ExecuteNonQuery();
                    }
                    catch (SqlException)
                    {
                        doRollback = true;
                        break;
                    }
                }
            }
            if (doRollback)
                transaction.Rollback();
            else
                transaction.Commit();
        }
    }
    finally
    {
        Program.dbConnection.Close();
    }
}

Answer 1

连接,事务和命令对象只是将命令发送到数据库的工具.执行命令后,数据库已收到该命令.无论你以后如何处理命令对象,将其丢弃,焚烧或射向月球,这一事实都不会改变.(它只能回滚).

您可以在一个范围内SqlConnection(有或没有SqlTransaction)创建和处理任意数量的命令.您可以在一个中启动和处理任意数量的事务SqlConnection.为了证明这一点,请参阅:

using (var conn = new SqlConnection(@"server=(local)\sql2008;database=Junk;Integrated Security=SSPI"))
{
  conn.Open();
  // Repeat this block as often as you like...
  using (var tran = conn.BeginTransaction())
  {
    using (var cmd = new SqlCommand("INSERT INTO Mess VALUES ('mess1')", conn, tran))
    {
      cmd.ExecuteNonQuery(); // Shows in Sql profiler
    }
    tran.Commit(); // Commits
  }
  using (var cmd = new SqlCommand("INSERT INTO Mess VALUES ('mess2')", conn))
  {
    cmd.ExecuteNonQuery(); // Executes and commits (implicit transaction).
  }
}

当然,对于健康的代码,您需要以正确的顺序处理所有对象.在处理a之后处置命令SqlConnection可能导致连接对象留在内存中.