Question

我正在尝试调试以下场景：有两个并发进程，将完全相同的行插入到具有唯一约束的表中。这是在显式事务之外完成的（虽然我假设InnoDB将其作为单个语句处理内部自动提交？）

架构如下：

CREATE TABLE locks (
    id bigint(20) UNSIGNED NOT NULL AUTO_INCREMENT,
    lock_uid varchar(255) NOT NULL,
    count smallint(6) NOT NULL,
    processor_id varchar(255) DEFAULT NULL,
    created_at timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
    updated_at timestamp NOT NULL DEFAULT '0000-00-00 00:00:00',
    PRIMARY KEY (id),
    UNIQUE INDEX locks_lock_uid_unique (lock_uid)
)

正如您所看到的，lock_uid上有唯一索引，以防止表格中有多个具有相同值的行。

正在运行的命令（对于上下文，这些是从通用查询日志中获取的，以获得完整的健全性，在排序规则命令之外的任何一个线程上都没有其他语句）：

主题1：

insert into `locks` (`lock_uid`, `count`, `processor_id`, `created_at`, `updated_at`) 
values ('11161567', '0', NULL, '2017-11-07 10:46:36', '2017-11-07 10:46:36')

主题2：

insert into `locks` (`lock_uid`, `count`, `processor_id`, `created_at`, `updated_at`) 
values ('11161567', '0', NULL, '2017-11-07 10:46:36', '2017-11-07 10:46:36')

这导致以下死锁：

LATEST DETECTED DEADLOCK
------------------------
2017-11-07 10:46:36 0x2ac88f791700
*** (1) TRANSACTION:
TRANSACTION 6089510736, ACTIVE 0 sec inserting
mysql tables in use 1, locked 1
LOCK WAIT 4 lock struct(s), heap size 1136, 4 row lock(s), undo log entries 1
MySQL thread id 177584962, OS thread handle 47059008030464, query id 13109086103 ec2-34-232-58-13.compute-1.amazonaws.com 34.232.58.13 appserver update
insert into `locks` (`lock_uid`, `count`, `processor_id`, `created_at`, `updated_at`) values ('11161567', '0', NULL, '2017-11-07 10:46:36', '2017-11-07 10:46:36')
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 6403 page no 4 n bits 176 index locks_lock_uid_unique of table `core`.`locks` trx id 6089510736 lock_mode X locks gap before rec insert intention waiting
Record lock, heap no 107 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
0: len 30; hex 636865636b6f75745265636f7665727950726f636573735f313131363137; asc 111617; (total 32 bytes);
1: len 8; hex 0000000003266637; asc      &f7;;

*** (2) TRANSACTION:
TRANSACTION 6089510734, ACTIVE 0 sec inserting
mysql tables in use 1, locked 1
4 lock struct(s), heap size 1136, 5 row lock(s), undo log entries 1
MySQL thread id 177584971, OS thread handle 47040888903424, query id 13109086092 ec2-34-237-3-244.compute-1.amazonaws.com 34.237.3.244 appserver update
insert into `locks` (`lock_uid`, `count`, `processor_id`, `created_at`, `updated_at`) values ('11161567', '0', NULL, '2017-11-07 10:46:36', '2017-11-07 10:46:36')
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 6403 page no 4 n bits 176 index locks_lock_uid_unique of table `core`.`locks` trx id 6089510734 lock mode S
Record lock, heap no 104 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
0: len 30; hex 636865636b6f75745265636f7665727950726f636573735f313131363135; asc 111615; (total 32 bytes);
1: len 8; hex 0000000003266632; asc      &f2;;

Record lock, heap no 105 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
0: len 30; hex 636865636b6f75745265636f7665727950726f636573735f313131363135; asc 111615; (total 32 bytes);
1: len 8; hex 0000000003266634; asc      &f4;;

Record lock, heap no 107 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
0: len 30; hex 636865636b6f75745265636f7665727950726f636573735f313131363137; asc 111617; (total 32 bytes);
1: len 8; hex 0000000003266637; asc      &f7;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 6403 page no 4 n bits 176 index locks_lock_uid_unique of table `core`.`locks` trx id 6089510734 lock_mode X locks gap before rec insert intention waiting
Record lock, heap no 107 PHYSICAL RECORD: n_fields 2; compact format; info bits 32
0: len 30; hex 636865636b6f75745265636f7665727950726f636573735f313131363137; asc 111617; (total 32 bytes);
1: len 8; hex 0000000003266637; asc      &f7;;

*** WE ROLL BACK TRANSACTION (2)

我已经阅读了类似的答案（例如MySQL locking in Duplicate Key Error），但我不太清楚在这种情况下发生了什么。解释与死锁输出不匹配。

为什么在事务外的2个插入处发生此死锁？
为什么T2在请求X之前已经拥有S锁，而相同的T1没有？

Answer 1

为什么在事务外的2个插入处发生此死锁？

它仍有交易 https://dev.mysql.com/doc/refman/5.7/en/innodb-autocommit-commit-rollback.html

在InnoDB中，所有用户活动都发生在事务中。如果启用自动提交模式，每个SQL语句形成一个交易本身

因此，您的单个查询可被视为短期交易

为什么T2在请求X之前已经拥有S锁，而在相同的T1没有？

它有。这就是打印最新死锁信息的功能如何工作:-) 它不会打印第一个事务所持有的锁。您可以通过模拟来自2个并行mysql会话的简单死锁来自行检查。这是代码：

https://github.com/mysql/mysql-server/blob/5.7/storage/innobase/lock/lock0lock.cc#L7236

DeadlockChecker::notify(const lock_t* lock) const
{
    ut_ad(lock_mutex_own());

    start_print();

    print("\n*** (1) TRANSACTION:\n");

    print(m_wait_lock->trx, 3000);

    print("*** (1) WAITING FOR THIS LOCK TO BE GRANTED:\n");

    print(m_wait_lock);

    print("*** (2) TRANSACTION:\n");

    print(lock->trx, 3000);

    print("*** (2) HOLDS THE LOCK(S):\n");

    print(lock);

    /* It is possible that the joining transaction was granted its
    lock when we rolled back some other waiting transaction. */

    if (m_start->lock.wait_lock != 0) {
        print("*** (2) WAITING FOR THIS LOCK TO BE GRANTED:\n");

        print(m_start->lock.wait_lock);
    }

    DBUG_PRINT("ib_lock", ("deadlock detected"));
}

说明与死锁输出

不匹配

它与您案件中似乎发生的情况相当接近。这是一个模拟：

的制备：将

create table test (id int primary key, val int not null unique) engine=innodb;
insert into test values (1, 1), (2, 2), (3, 3);

现在让我们打开3个shell并运行以下命令：

1> begin;
2> begin;
3> begin;

然后

1>insert into test values (1, 1);
2>insert into test values (1, 1); (will hang)
3>insert into test values (1, 1); (will hang)

现在

1>rollback
2>would produce: ERROR 1213 (40001): Deadlock found when trying to get lock; try restarting transaction
3>would produce: Query OK, 1 row affected

1>show engine innodb status;
...
------------------------
LATEST DETECTED DEADLOCK
------------------------
2017-11-15 23:21:47 0x700000d95000
*** (1) TRANSACTION:
TRANSACTION 2336, ACTIVE 8 sec inserting
mysql tables in use 1, locked 1
LOCK WAIT 4 lock struct(s), heap size 1136, 2 row lock(s)
MySQL thread id 8, OS thread handle 123145316831232, query id 58 localhost 127.0.0.1 root update
insert into test values (1, 1)
*** (1) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 26 page no 3 n bits 72 index PRIMARY of table `test`.`test` trx id 2336 lock_mode X insert intention waiting
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** (2) TRANSACTION:
TRANSACTION 2335, ACTIVE 13 sec inserting
mysql tables in use 1, locked 1
4 lock struct(s), heap size 1136, 2 row lock(s)
MySQL thread id 5, OS thread handle 123145316552704, query id 57 localhost root update
insert into test values (1, 1)
*** (2) HOLDS THE LOCK(S):
RECORD LOCKS space id 26 page no 3 n bits 72 index PRIMARY of table `test`.`test` trx id 2335 lock mode S
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** (2) WAITING FOR THIS LOCK TO BE GRANTED:
RECORD LOCKS space id 26 page no 3 n bits 72 index PRIMARY of table `test`.`test` trx id 2335 lock_mode X insert intention waiting
Record lock, heap no 1 PHYSICAL RECORD: n_fields 1; compact format; info bits 0
 0: len 8; hex 73757072656d756d; asc supremum;;

*** WE ROLL BACK TRANSACTION (2)

你可以看到它接近你的结果

Answer 2

插入查询在分配共享OR独占锁之前使用插入意图锁。

例如：架构：

 CREATE TABLE `temp_table` (
    `id` INT NOT NULL AUTO_INCREMENT,
    `u_id` INT NULL DEFAULT NULL,
    PRIMARY KEY (`id`),
    INDEX `u_id` (`u_id`)
 )
 COLLATE='latin1_swedish_ci'
 ENGINE=InnoDB;

让我在表中有2,4,7 u_id。现在我正在从另一个会话中执行两个插入操作。

以下是可能的情况。

情况1：

T1：插入temp_table（u_id）值（'5'）;

T2：插入temp_table（u_id）值（'6'）;

此事务首先应用间隙插入意图锁定并检查应分配哪种类型的锁定（S或X）。就像这里5＆amp; 6是适合间隙（4-7）所以专用锁定将分配给T1＆amp; T2。

情况2：

T1：插入temp_table（u_id）值（'5'）;

T2：插入temp_table（u_id）值（'5'）;

这里虽然插入意图锁定T1将适合间隙（4-7）并且它将分配独占锁定但是T2不适合间隙作为间隙更新为（5-7）因为我们已经分配了独占锁定T1因此T2被分配了共享锁并等待T1完成。

如果T1成功完成，那么T2将被视为重复键错误，但如果由于任何原因T1失败，则T2将被独占锁定并尝试插入记录。

有关更多信息，请参阅Mysql指南： https://dev.mysql.com/doc/refman/5.7/en/innodb-locking.html

MySQL并发插入导致（显式）事务之外的死锁

2 个答案: