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MySQL死锁套路之唯一索引下批量插入顺序不一致

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在 t1 插入记录时,是不加锁的。这个时候事务 t1 还未提交的情况下,事务 t2 尝试插入的时候,发现有这条记录,t2 尝试获取 S 锁,会判定记录上的事务 id 是否活跃,如果活跃的话,说明事务未结束,会帮 t1 把它的隐式锁提升为显式锁( X 锁)

源码如下

t2 获取S锁的结果:DB_LOCK_WAIT

实验2:

批量插入顺序不一致的导致的死锁

t1 t2
begin
insert into t1(a, b)values("1", "1"); 成功
insert into t1(a, b)values("2", "2"); 成功
insert into t1(a, b)values("2", "2"); t1 尝试获取 S 锁,把 t2 的隐式锁提升为显式 X 锁,进入 DB_LOCK_WAIT
insert into t1(a, b)values("1", "1"); t2 尝试获取 S 锁,把 t1 的隐式锁提升为显式 X 锁,产生死锁
------------------------
LATEST DETECTED DEADLOCK
------------------------
181101 9:48:36
*** (1) TRANSACTION:
TRANSACTION 3309, ACTIVE 215 sec inserting
mysql tables in use 1, locked 1
LOCK WAIT 3 lock struct(s), heap size 376, 2 row lock(s), undo log entries 2
MySQL thread id 2, OS thread handle 0x70000a845000, query id 58 localhost root update
insert into t1(a, b)values("2", "2")
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 55 page no 4 n bits 72 index `uk_name` of table `d1`.`t1` trx id 3309 lock mode S waiting
Record lock, heap no 3 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 1; hex 32; asc 2;;
 1: len 1; hex 32; asc 2;;
 2: len 4; hex 80000002; asc ;;

*** (2) TRANSACTION:
TRANSACTION 330A, ACTIVE 163 sec inserting
mysql tables in use 1, locked 1
3 lock struct(s), heap size 376, 2 row lock(s), undo log entries 2
MySQL thread id 3, OS thread handle 0x70000a888000, query id 59 localhost root update
insert into t1(a, b)values("1", "1")
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 55 page no 4 n bits 72 index `uk_name` of table `d1`.`t1` trx id 330A lock_mode X locks rec but not gap
Record lock, heap no 3 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 1; hex 32; asc 2;;
 1: len 1; hex 32; asc 2;;
 2: len 4; hex 80000002; asc ;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 55 page no 4 n bits 72 index `uk_name` of table `d1`.`t1` trx id 330A lock mode S waiting
Record lock, heap no 2 PHYSICAL RECORD: n_fields 3; compact format; info bits 0
 0: len 1; hex 31; asc 1;;
 1: len 1; hex 31; asc 1;;
 2: len 4; hex 80000001; asc ;;

*** WE ROLL BACK TRANSACTION (2)

怎么样解决这样的问题呢?

一个可行的办法是在应用层排序以后再插入

总结

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