发布时间:2025-12-09 12:00:17 浏览次数:12
在依赖事务的项目中,如果SQL语句设计不合理或者执行顺序不合理,就容易引发死锁。本文介绍一个因为Duplicate Key引发的死锁
user CREATE TABLE `user` ( `id` int(11) NOT NULL AUTO_INCREMENT, `name` varchar(64) NOT NULL, `age` int(11) DEFAULT NULL, PRIMARY KEY (`id`), #唯一索引 UNIQUE KEY `uk_name` (`name`)) ENGINE=InnoDB DEFAULT CHARSET=utf8mysql> select * FROM user;+----+------+------+| id | name | age |+----+------+------+| 1 | tenmao | NULL |+----+------+------+1 row in set (0.00 sec)BEGIN;# Duplicate key(所以插入失败)insert INTO user VALUES(null, 'tenmao', 3);# 此时再执行事务2的语句(会等待)# 插入失败则更新update user SET age=3 WHERE name='tenmao';COMMIT;# 单SQL事务(默认)update user SET age =3 WHERE name ='tenmao';在事务1执行insert INTO user VALUES(null, 'tenmao', 3);失败后,执行事务2,事务2等待后,再继续执行事务1,触发死锁。
mysql> BEGIN;Query OK, 0 rows affected (0.00 sec)mysql> insert INTO user VALUES(null, 'tenmao', 3);ERROR 1062 (23000): Duplicate entry 'tenmao' for key 'uk_name'mysql> update user SET age=3 WHERE name='tenmao';Query OK, 1 row affected (0.00 sec)Rows matched: 1 Changed: 1 Warnings: 0mysql> COMMIT;Query OK, 0 rows affected (0.01 sec)mysql> update user SET age =3 WHERE name ='tenmao';ERROR 1213 (40001): Deadlock found when trying to get lock; try restarting transaction当事务2等待时,查看InnoDB锁的情况,分别是INFORMATION_SCHEMA.INNODB_LOCKS, INFORMATION_SCHEMA.INNODB_LOCK_WAITS, INFORMATION_SCHEMA.INNODB_TRX
# 获取锁的情况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 |+---------------+-------------+-----------+-----------+-----------------+------------+------------+-----------+----------+-----------+| 50505:361:4:2 | 50505 | X | RECORD | `tenmao`.`user` | uk_name | 361 | 4 | 2 | 'tenmao' || 50504:361:4:2 | 50504 | S | RECORD | `tenmao`.`user` | uk_name | 361 | 4 | 2 | 'tenmao' |+---------------+-------------+-----------+-----------+-----------------+------------+------------+-----------+----------+-----------+2 rows in set, 1 warning (0.00 sec)# 查看锁等待情况mysql> select * FROM INFORMATION_SCHEMA.INNODB_LOCK_WAITS;+-------------------+-------------------+-----------------+------------------+| requesting_trx_id | requested_lock_id | blocking_trx_id | blocking_lock_id |+-------------------+-------------------+-----------------+------------------+| 50505 | 50505:361:4:2 | 50504 | 50504:361:4:2 |+-------------------+-------------------+-----------------+------------------+1 row in set, 1 warning (0.00 sec)# 查看事务状态mysql> select * FROM INFORMATION_SCHEMA.INNODB_TRXG*************************** 1. row *************************** trx_id: 50505 trx_state: LOCK WAIT trx_started: 2019-12-13 14:55:00 trx_requested_lock_id: 50505:361:4:2 trx_wait_started: 2019-12-13 14:59:49 trx_weight: 2 trx_mysql_thread_id: 2557 trx_query: update user set age =3 where name ='tenmao' trx_operation_state: starting index read trx_tables_in_use: 1 trx_tables_locked: 1 trx_lock_structs: 2 trx_lock_memory_bytes: 1136 trx_rows_locked: 3 trx_rows_modified: 0 trx_concurrency_tickets: 0 trx_isolation_level: REPEATABLE READ trx_unique_checks: 1 trx_foreign_key_checks: 1trx_last_foreign_key_error: NULL trx_adaptive_hash_latched: 0 trx_adaptive_hash_tenmaoeout: 0 trx_is_read_only: 0trx_autocommit_non_locking: 0*************************** 2. row *************************** trx_id: 50504 trx_state: RUNNING trx_started: 2019-12-13 14:54:25 trx_requested_lock_id: NULL trx_wait_started: NULL trx_weight: 4 trx_mysql_thread_id: 2556 trx_query: NULL trx_operation_state: NULL trx_tables_in_use: 0 trx_tables_locked: 1 trx_lock_structs: 4 trx_lock_memory_bytes: 1136 trx_rows_locked: 2 trx_rows_modified: 0 trx_concurrency_tickets: 0 trx_isolation_level: REPEATABLE READ trx_unique_checks: 1 trx_foreign_key_checks: 1trx_last_foreign_key_error: NULL trx_adaptive_hash_latched: 0 trx_adaptive_hash_tenmaoeout: 0 trx_is_read_only: 0trx_autocommit_non_locking: 02 rows in set (0.00 sec)BEGIN;# Duplicate key(所以插入失败)insert INTO user VALUES(null, 'tenmao', 3);# 此时因为重复键,事务拿到记录的`S锁`# 此时再执行事务2的语句(会等待)# 插入失败则更新(获取X锁,但是因为事务2排在前面,需要事务2释放`X锁`,另一方面事务2也在等待事务1释放`S锁`,所以形成死锁)update user SET age=3 WHERE name='tenmao';COMMIT;# 单SQL事务(尝试获取X锁,等待事务1释放`S锁`)update user SET age =3 WHERE name ='tenmao';