前言
死锁的本质是资源竞争,批量插入如果顺序不一致很容易导致死锁,我们来分析一下这个情况。为了方便演示,把批量插入改写为了多条 insert。
先来做几个小实验,简化的表结构如下
1
2
3
4
5
6
7
|
CREATE TABLE `t1` (
`id` int (11) NOT NULL AUTO_INCREMENT,
`a` varchar (5),
`b` varchar (5),
PRIMARY KEY (`id`),
UNIQUE KEY `uk_name` (`a`,`b`)
);
|
实验1:
在记录不存在的情况下,两个同样顺序的批量 insert 同时执行,第二个会进行锁等待状态
t1 | t2 | |
---|---|---|
begin; | begin; | |
insert ignore into t1(a, b)values("1", "1"); | 成功 | |
insert ignore into t1(a, b)values("1", "1"); | 锁等待状态 |
可以看到目前锁的状态
1
2
3
4
5
6
7
|
mysql> select * from information_schema.innodb_locks;
+ -------------+-------------+-----------+-----------+------------+------------+------------+-----------+----------+-----------+
| lock_id | lock_trx_id | lock_mode | lock_type | lock_table | lock_index | lock_space | lock_page | lock_rec | lock_data |
+ -------------+-------------+-----------+-----------+------------+------------+------------+-----------+----------+-----------+
| 31AE:54:4:2 | 31AE | S | RECORD | `d1`.`t1` | `uk_name` | 54 | 4 | 2 | '1' , '1' |
| 31AD:54:4:2 | 31AD | X | RECORD | `d1`.`t1` | `uk_name` | 54 | 4 | 2 | '1' , '1' |
+ -------------+-------------+-----------+-----------+------------+------------+------------+-----------+----------+-----------+
|
在我们执行事务t1的 insert 时,没有在任何锁的断点处出现,这跟 MySQL 插入的原理有关系
insert 加的是隐式锁。什么是隐式锁?隐式锁的意思就是没有锁
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 |
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
|
------------------------
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
以上就是这篇文章的全部内容了,希望本文的内容对大家的学习或者工作具有一定的参考学习价值,谢谢大家对服务器之家的支持。
原文链接:https://juejin.im/post/5ce63f8ee51d45775d516ee9