Mysql中mvcc各场景理解应用
作者：王者之峰
这篇文章主要为大家介绍了Mysql中mvcc各场景理解应用，有需要的朋友可以借鉴参考下，希望能够有所帮助，祝大家多多进步，早日升职加薪
前言
- mysql版本为
mysql> select version(); +-----------+ | version() | +-----------+ | 8.0.27 | +-----------+ 1 row in set (0.00 sec)
- 隔离级别
mysql> show variables like '%isola%'; +-----------------------+-----------------+ | Variable_name | Value | +-----------------------+-----------------+ | transaction_isolation | REPEATABLE-READ | +-----------------------+-----------------+ 1 row in set (0.02 sec)
- 表结构
mysql> show create table test; +-------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ | Table | Create Table | +-------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ | test | CREATE TABLE `test` ( `id` bigint unsigned NOT NULL AUTO_INCREMENT COMMENT '主键', `name` char(32) NOT NULL COMMENT '用户姓名', `num` int DEFAULT NULL, `phone` char(11) DEFAULT '' COMMENT '手机号', PRIMARY KEY (`id`), KEY `idx_name_phone` (`name`,`phone`) ) ENGINE=InnoDB AUTO_INCREMENT=108 DEFAULT CHARSET=utf8mb4 COLLATE=utf8mb4_0900_ai_ci COMMENT='test表' | +-------+----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------+ 1 row in set (0.01 sec
- 现有表数据
mysql> select * from test; +-----+---------------+---------+-------+ | id | name | num | phone | +-----+---------------+---------+-------+ | 1 | 执行业 | 1234567 | | | 2 | 执行业务1 | NULL | | | 3 | a | NULL | | | 4 | a | NULL | | | 5 | a | NULL | | | 6 | b | 1 | | | 7 | wdf | NULL | | | 10 | dd | 1 | | | 11 | hello | NULL | | | 15 | df | NULL | | | 16 | e | NULL | | | 20 | e | NULL | | | 21 | 好的 | NULL | | | 25 | g | 1 | | | 106 | hello | NULL | | | 107 | a | NULL | | +-----+---------------+---------+-------+ 16 rows in set (0.00 sec)
场景一
- 事务A:
select * from test where id in (7,15) for update;。 - 事务B:
update test set name='d' where id=10;和insert into test(id,name) values(8,'hello');。 - 事务A:
select * from test where id in (7,8,10,15);。 第二步是否阻塞。第三步是否能读到事务B执行的更新。
试验步骤
事务A第一步
mysql> begin;select * from test where id in (7,15) for update; Query OK, 0 rows affected (0.00 sec) +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 2 rows in set (0.01 sec)
持有锁情况：
mysql> select * from performance_schema.data_locks; +--------+-----------------------------+-----------------------+-----------+----------+---------------+-------------+----------------+-------------------+------------+-----------------------+-----------+---------------+-------------+-----------+ | ENGINE | ENGINE_LOCK_ID | ENGINE_TRANSACTION_ID | THREAD_ID | EVENT_ID | OBJECT_SCHEMA | OBJECT_NAME | PARTITION_NAME | SUBPARTITION_NAME | INDEX_NAME | OBJECT_INSTANCE_BEGIN | LOCK_TYPE | LOCK_MODE | LOCK_STATUS | LOCK_DATA | +--------+-----------------------------+-----------------------+-----------+----------+---------------+-------------+----------------+-------------------+------------+-----------------------+-----------+---------------+-------------+-----------+ | INNODB | 4974808984:1063:4890706744 | 46666 | 50 | 123 | my_test | test | NULL | NULL | NULL | 4890706744 | TABLE | IX | GRANTED | NULL | | INNODB | 4974808984:2:4:7:4915866136 | 46666 | 50 | 123 | my_test | test | NULL | NULL | PRIMARY | 4915866136 | RECORD | X,REC_NOT_GAP | GRANTED | 15 | | INNODB | 4974808984:2:4:9:4915866136 | 46666 | 50 | 123 | my_test | test | NULL | NULL | PRIMARY | 4915866136 | RECORD | X,REC_NOT_GAP | GRANTED | 7 | +--------+-----------------------------+-----------------------+-----------+----------+---------------+-------------+----------------+-------------------+------------+-----------------------+-----------+---------------+-------------+-----------+ 3 rows in set (0.00 sec)
发现7,15持有了行锁。
事务B执行
mysql> update test set name = 'sds' where id=10;insert into test(id,name) values(8,'hello'); Query OK, 1 row affected (0.01 sec) Rows matched: 1 Changed: 1 Warnings: 0 Query OK, 1 row affected (0.00 sec)
事务A执行第二步
mysql> select * from test where id in (7,8,10,15); +----+-------+------+-------+ | id | name | num | phone | +----+-------+------+-------+ | 7 | wdf | NULL | | | 8 | hello | NULL | | | 10 | sds | 1 | | | 15 | df | NULL | | +----+-------+------+-------+ 4 rows in set (0.01 sec)
结果
步骤二执行了，事务A读到了事务B提交的数据。下面我们来看看正常的select;
场景二
还原数据：
mysql> update test set name = 'dd' where id=10;delete from test where id=8; Query OK, 1 row affected (0.01 sec) Rows matched: 1 Changed: 1 Warnings: 0 Query OK, 1 row affected (0.00 sec)
- 事务A:
select * from test where id in (7,15);。 - 事务B:
update test set name='d' where id=10;和insert into test(id,name) values(8,'hello');。 - 事务A:
select * from test where id in (7,8,10,15);。 第二步是否阻塞。第三步是否能读到事务B执行的更新。
试验步骤
事务A第一步
mysql> begin;select * from test where id in (7,15); Query OK, 0 rows affected (0.00 sec) +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 2 rows in set (0.00 sec)
持有锁情况：
mysql> select * from performance_schema.data_locks; Empty set (0.00 sec)
事务B执行
mysql> update test set name = 'sds' where id=10;insert into test(id,name) values(8,'hello'); Query OK, 1 row affected (0.00 sec) Rows matched: 1 Changed: 1 Warnings: 0 Query OK, 1 row affected (0.00 sec)
事务A执行第二步
mysql> select * from test where id in (7,8,10,15); +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 10 | dd | 1 | | | 15 | df | NULL | | +----+------+------+-------+ 3 rows in set (0.00 sec)
结果
步骤二执行了，事务A没读到了事务B提交的数据。笔者猜测for update加锁之后会清除readview或者没开启readview，所以后面会读到事务B的。
所以我们来看看到底是清除还是没开启。
事务A后续步骤
mysql> select * from test where id in (7,15) for update; +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 2 rows in set (0.00 sec) mysql> select * from test where id in (7,8,10,15); +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 10 | dd | 1 | | | 15 | df | NULL | | +----+------+------+-------+ 3 rows in set (0.00 sec)
可以发现重新执行了场景一的步骤后结果没变。
所以应该是没开启，应该是当前读不会开启readview。
笔者找了下资料没找到，找到的笔者可以留言。
不过我们可以使用继续实验验证下。
场景三
- 事务A:
update test set name = 'dgf' where id in (7,15);。 - 事务B:
update test set name='d' where id=10;和insert into test(id,name) values(8,'hello');。 - 事务A:
select * from test where id in (7,8,10,15);。 第二步是否阻塞。第三步是否能读到事务B执行的更新。
这个场景就不搞实验步骤了，结果是和笔者的猜想一样的 ”当前读不会开启readview，第一个快照读才会开启“
场景四
- 事务A:
select * from test where id in (7,15);。 - 事务B:
insert into test(id,name) values(8,'hello');。 - 事务A:
select * from test where id in (7,8,15);。 - 事务A:
update test set name ='cv' where id =8;。 - 事务A:
select * from test where id in (7,8,15);。
事务A第一步
mysql> begin;select * from test where id in (7,15); Query OK, 0 rows affected (0.00 sec) +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 2 rows in set (0.00 sec)
开启了事务，浅读一下。
事务B执行
insert into test(id,name) values(8,'hello');
事务A第二步
mysql> select * from test where id in (7,8,15); +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 2 rows in set (0.00 sec)
检验一下是否读的到，发现读不到。
事务A第三步
mysql> update test set name ='cv' where id =8; Query OK, 1 row affected (0.00 sec) Rows matched: 1 Changed: 1 Warnings: 0
对插入的进行更新。
事务A第四步
mysql> select * from test where id in (7,8,15); +----+------+------+-------+ | id | name | num | phone | +----+------+------+-------+ | 7 | wdf | NULL | | | 8 | cv | NULL | | | 15 | df | NULL | | +----+------+------+-------+ 3 rows in set (0.00 sec)
发现可以读到了。
原因
能读到的原因是因为本事务对版本链内容进行了修改，所以就读到了。
这个场景可能会出现在实际开发中，会比较懵，当然“事务A第三步”是笔者随便模拟的，实际生产中直接拿大不到刚刚插入的id，所以应该是模糊(没有确定行)update。所以在生产中还是要确定行去进行修改，避免出现这种比较难理解的场景。
虽然也可以使用lock in share mode或者for update读当前借助next-key去实现不幻读（第二次读到第一次没有读到的行），还是需要根据具体业务选择。
总结
根据以上的场景，我们可以知道：
- readview是第一个select的时候才会创建的。
- rr级别下读快照如果中间出现修改版本链内容还是会出现幻读（很合理，但是不容易发现这个原因），如果真的要想做到不幻读还是要通过加锁（当然要有索引，没有的话就锁表了）。
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