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MySQL中InnoDB的间隙锁问题

2024-07-24 12:45:50
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 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。

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