MySQL 死锁与存储过程生成 UID

Question

我有一个从“票证”表生成 UID 的存储过程，但在负载下我遇到了很多死锁。每当我的任务需要新的 UID 时，我都会从多个并发连接中多次调用此过程。

BEGIN
    DECLARE a_uid BIGINT(20) UNSIGNED;
    START TRANSACTION;
    SELECT uid INTO a_uid FROM uid_data FOR UPDATE; # Lock
    INSERT INTO uid_data (stub) VALUES ('a') ON DUPLICATE KEY UPDATE uid=uid+1;
    SELECT a_uid+1 AS `uid`;
    COMMIT;
END

我确实考虑过使用：

BEGIN
    REPLACE INTO uid_data (stub) VALUES ('a');
    SELECT LAST_INSERT_ID();
END

但是我不确定这对于并发连接是否安全，因为没有锁定，这与 SELECT FOR UPDATE 的第一个过程不同。

这是表格：

mysql> DESCRIBE uid_data;
+-------+---------------------+------+-----+---------+----------------+
| Field | Type                | Null | Key | Default | Extra          |
+-------+---------------------+------+-----+---------+----------------+
| uid   | bigint(20) unsigned | NO   | PRI | NULL    | auto_increment |
| stub  | char(1)             | NO   | UNI | NULL    |                |
+-------+---------------------+------+-----+---------+----------------+

我已经设置了读取提交的事务隔离：

mysql> SHOW VARIABLES LIKE 'tx_isolation';
+---------------+-----------------+
| Variable_name | Value           |
+---------------+-----------------+
| tx_isolation  | READ-COMMITTED  |
+---------------+-----------------+

这是我从SHOW ENGINE INNODB STATUS; 回复的内容

...
... dozens and dozens of the following record locks...

Record lock, heap no 1046 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
 0: len 1; hex 61; asc a;;
 1: len 8; hex 00000000000335f2; asc       5 ;;

Record lock, heap no 1047 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
 0: len 1; hex 61; asc a;;
 1: len 8; hex 00000000000335f1; asc       5 ;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 13 page no 4 n bits 1120 index `stub` of table `my_db`.`uid_data` trx id 13AA89 lock_mode X waiting
Record lock, heap no 583 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
 0: len 1; hex 61; asc a;;
 1: len 8; hex 00000000000334a8; asc       4 ;;

*** WE ROLL BACK TRANSACTION (1)

如果有人能解释发生了什么以及如何避免它们，我将不胜感激。

Answer 1

在这种情况下会发生死锁：

事务 1：请求锁 (SELECT...FOR UPDATE) 并获取它

事务 2：请求锁 (SELECT...FOR UPDATE) 并且必须等待

事务 1：尝试插入，遇到重复，因此更新 (INSERT...ON DUPLICATE KEY UPDATE) => 死锁

我不太确定原因，我怀疑它与ON DUPLICATE KEY UPDATE 有关。我还在调查中，如果我发现了会回来的。

[编辑] 即使在以下情况下也会发生死锁：

BEGIN
    START TRANSACTION;
    SELECT uid FROM uid_data FOR UPDATE;
    UPDATE uid_data SET uid = uid +1; -- here, a deadlock would be detected in a blocked, concurrent connection
    COMMIT;
END

这个呢：

BEGIN
    START TRANSACTION;    
    UPDATE uid_data SET uid = uid +1;
    SELECT uid FROM uid_data;
    COMMIT;
END

您可以完全放弃您的 stub 列。唯一的缺点是你必须用一行来初始化你的 uid_data。

Answer 2

你可以试试

UPDATE uid_data SET uid = LAST_INSERT_ID(uid+1);
SELECT LAST_INSERT_ID();

在桌子上

CREATE TABLE `uid_data` (
    `uid` BIGINT(20) UNSIGNED NOT NULL
)
COLLATE='utf8_general_ci'
ENGINE=MyISAM;

这是线程安全的，如果是 MyISAM 则不会锁定表（实际更新语句期间除外）。

Answer 3

这样做：

CREATE TABLE tickets
(
    uid serial
)

然后获取下一个uid：

BEGIN
  INSERT INTO tickets VALUES (NULL);
  SELECT LAST_INSERT_ID();
END

uid 序列号等价于

uid BIGINT(20) UNSIGNED NOT NULL PRIMARY KEY auto_increment

您不应该在使用这种方法时遇到任何死锁，并且可以向其抛出任意数量的连接。