MySQL死锁分析及解决的方法--例子

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5、死锁举例分析

  在MySQL中,行级锁并不是直接锁记录,而是锁索引。索引分为主键索引和非主键索引两种,如果一条sql语句操作了主键索引,MySQL就会锁定这条主键索引;如果一条语句操作了非主键索引,MySQL会先锁定该非主键索引,再锁定相关的主键索引。

  在UPDATE、DELETE操作时,MySQL不仅锁定WHERE条件扫描过的所有索引记录,而且会锁定相邻的键值,即所谓的next-key locking。

  例如,一个表db。tab_test,结构如下:

  id:主键;

  state:状态;

  time:时间;

  索引:idx_1(state,time)

  出现死锁日志如下:

  ?***(1) TRANSACTION:

  ?TRANSACTION 0 677833455, ACTIVE 0 sec, process no 11393, OSthread id 278546 starting index read

  ?mysql tables in use 1, locked 1

  ?LOCK WAIT 3 lock struct(s), heap size 320

  ?MySQL thread id 83, query id 162348740 dcnet03 dcnet Searching rows for update

  ?update tab_test set state=1064,time=now() where state=1061 and time < date_sub(now(), INTERVAL 30 minute) (任务1的sql语句)

  ?***(1) WAITING FOR THIS LOCK TO BE GRANTED: (任务1等待的索引记录)

  ?RECORD LOCKS space id 0 page no 849384 n bits 208 index `PRIMARY` of table `db/tab_test` trx id 0 677833455 _mode X locks rec but not gap waiting

  ?Record lock, heap no 92 PHYSICAL RECORD: n_fields 11; compact format; info bits 0

  ?0: len 8; hex 800000000097629c; asc b ;; 1: len 6; hex 00002866eaee; asc (f ;; 2: len 7; hex 00000d40040110; asc @ ;; 3: len 8; hex 80000000000050b2; asc P ;; 4: len 8; hex 800000000000502a; asc P*;; 5: len 8; hex 8000000000005426; asc T&;; 6: len 8; hex 800012412c66d29c; asc A,f ;; 7: len 23; hex 75706c6f6164666972652e636f6d2f6 8616e642e706870; asc xxx.com/;; 8: len 8; hex 800000000000042b; asc +;; 9: len 4; hex 474bfa2b; asc GK +;; 10: len 8; hex 8000000000004e24; asc N$;;

  ?*** (2) TRANSACTION:

  ?TRANSACTION 0 677833454, ACTIVE 0 sec, process no 11397, OS thread id 344086 updating or deleting, thread declared inside InnoDB 499

  ?mysql tables in use 1, locked 1

  ?3 lock struct(s), heap size 320, undo log entries 1

  ?MySQL thread id 84, query id 162348739 dcnet03 dcnet Updating update tab_test set state=1067,time=now () where id in (9921180) (任务2的sql语句)

  ?*** (2) HOLDS THE LOCK(S): (任务2已获得的锁)

  ?RECORD LOCKS space id 0 page no 849384 n bits 208 index `PRIMARY` of table `db/tab_test` trx id 0 677833454 lock_mode X locks rec but not gap

  ?Record lock, heap no 92 PHYSICAL RECORD: n_fields 11; compact format; info bits 0

  ?0: len 8; hex 800000000097629c; asc b ;; 1: len 6; hex 00002866eaee; asc (f ;; 2: len 7; hex 00000d40040110; asc @ ;; 3: len 8; hex 80000000000050b2; asc P ;; 4: len 8; hex 800000000000502a; asc P*;; 5: len 8; hex 8000000000005426; asc T&;; 6: len 8; hex 800012412c66d29c; asc A,f ;; 7: len 23; hex 75706c6f6164666972652e636f6d2f6 8616e642e706870; asc uploadfire.com/hand.php;; 8: len 8; hex 800000000000042b; asc +;; 9: len 4; hex 474bfa2b; asc GK +;; 10: len 8; hex 8000000000004e24; asc N$;;

  ?*** (2) WAITING FOR THIS LOCK TO BE GRANTED: (任务2等待的锁)

  ?RECORD LOCKS space id 0 page no 843102 n bits 600 index `idx_1` of table `db/tab_test` trx id 0 677833454 lock_mode X locks rec but not gap waiting

  ?Record lock, heap no 395 PHYSICAL RECORD: n_fields 3; compact format; info bits 0

  ?0: len 8; hex 8000000000000425; asc %;; 1: len 8; hex 800012412c66d29c; asc A,f ;; 2: len 8; hex 800000000097629c; asc b ;;

  ?*** WE ROLL BACK TRANSACTION (1)

  ?(回滚了任务1,以解除死锁)

  原因分析:

  当“update tab_test set state=1064,time=now() where state=1061 and time < date_sub(now(), INTERVAL 30 minute)”执行时,MySQL会使用idx_1索引,因此首先锁定相关的索引记录,因为idx_1是非主键索引,为执行该语句,MySQL还会锁定主键索引。

  假设“update tab_test set state=1067,time=now () where id in (9921180)”几乎同时执行时,本语句首先锁定主键索引,由于需要更新state的值,所以还需要锁定idx_1的某些索引记录。

  这样第一条语句锁定了idx_1的记录,等待主键索引,而第二条语句则锁定了主键索引记录,而等待idx_1的记录,这样死锁就产生了。

  6、解决办法

  拆分第一条sql,先查出符合条件的主键值,再按照主键更新记录:

  ?select id from tab_test where state=1061 and time < date_sub(now(), INTERVAL 30 minute);

  ?update tab_test state=1064,time=now() where id in(......);

  关于MySQL死锁问题的实例分析及解决方法就介绍到这里了,希望本次的介绍能够对您有所收获

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