MySQL中InnoDB的间隙锁问题

数据库  /  管理员 发布于 6年前   152

 在为一个客户排除死锁问题时我遇到了一个有趣的包括InnoDB间隙锁的情形。对于一个WHERE子句不匹配任何行的非插入的写操作中，我预期事务应该不会有锁，但我错了。让我们看一下这张表及示例UPDATE。
 

mysql> SHOW CREATE TABLE preferences \G*************************** 1. row ***************************    Table: preferencesCreate Table: CREATE TABLE `preferences` ( `numericId` int(10) unsigned NOT NULL, `receiveNotifications` tinyint(1) DEFAULT NULL, PRIMARY KEY (`numericId`)) ENGINE=InnoDB DEFAULT CHARSET=latin11 row in set (0.00 sec)mysql> BEGIN;Query OK, 0 rows affected (0.00 sec)mysql> SELECT COUNT(*) FROM preferences;+----------+| COUNT(*) |+----------+|    0 |+----------+1 row in set (0.01 sec)mysql> UPDATE preferences SET receiveNotifications='1' WHERE numericId = '2';Query OK, 0 rows affected (0.01 sec)Rows matched: 0 Changed: 0 Warnings: 0

InnoDB状态显示这个UPDATE在主索引记录上持有了一个X锁：
 

---TRANSACTION 4A18101, ACTIVE 12 sec2 lock struct(s), heap size 376, 1 row lock(s)MySQL thread id 3, OS thread handle 0x7ff2200cd700, query id 35 localhost msandboxTrx read view will not see trx with id >= 4A18102, sees < 4A18102TABLE LOCK table `test`.`preferences` trx id 4A18101 lock mode IXRECORD LOCKS space id 31766 page no 3 n bits 72 index `PRIMARY` of table `test`.`preferences` trx id 4A18101 lock_mode X

这是为什么呢，Heikki在其bug报告中做了解释，这很有意义，我知道修复起来很困难，但略带厌恶地我又希望它能被差异化处理。为完成这篇文章，让我证明下上面说到的死锁情况，下面中mysql1是第一个会话，mysql2是另一个，查询的顺序如下：
 

mysql1> BEGIN;Query OK, 0 rows affected (0.00 sec)mysql1> UPDATE preferences SET receiveNotifications='1' WHERE numericId = '1';Query OK, 0 rows affected (0.00 sec)Rows matched: 0 Changed: 0 Warnings: 0mysql2> BEGIN;Query OK, 0 rows affected (0.00 sec)mysql2> UPDATE preferences SET receiveNotifications='1' WHERE numericId = '2';Query OK, 0 rows affected (0.00 sec)Rows matched: 0 Changed: 0 Warnings: 0mysql1> INSERT INTO preferences (numericId, receiveNotifications) VALUES ('1', '1'); -- This one goes into LOCK WAITmysql2> INSERT INTO preferences (numericId, receiveNotifications) VALUES ('2', '1');ERROR 1213 (40001): Deadlock found when trying to get lock; try restarting transaction

现在你看到导致死锁是多么的容易，因此一定要避免这种情况――如果来自于事务的INSERT部分导致非插入的写操作可能不匹配任何行的话，不要这样做，使用REPLACE INTO或使用READ-COMMITTED事务隔离。