MySQL 死锁套路：唯一索引下批量插入顺序不一致

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

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`)
);
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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

可以看到目前锁的状态

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'  |
+-------------+-------------+-----------+-----------+------------+------------+------------+-----------+----------+-----------+
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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

------------------------
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)
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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