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怎么掌握PostgreSQL Locks的基础知识

发表于:2025-01-21 作者:千家信息网编辑
千家信息网最后更新 2025年01月21日,这篇文章主要讲解了"怎么掌握PostgreSQL Locks的基础知识",文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习"怎么掌握PostgreSQL Lo
千家信息网最后更新 2025年01月21日怎么掌握PostgreSQL Locks的基础知识

这篇文章主要讲解了"怎么掌握PostgreSQL Locks的基础知识",文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习"怎么掌握PostgreSQL Locks的基础知识"吧!

如下例所示,session 1执行update语句,session 2 update相同的rows,session 3查询locktype为transactionid的信息.
session 1

[local]:5432 pg12@testdb=# begin;ere relation=295053;BEGINTime: 1.430 ms[local]:5432 pg12@testdb=#* -- SELECT * from t_lock where id < 10 FOR UPDATE;[local]:5432 pg12@testdb=#* select pg_backend_pid(); pg_backend_pid ----------------           2475(1 row)Time: 2.619 ms[local]:5432 pg12@testdb=#* update t_lock set id = 3000 where id = 3;UPDATE 4Time: 7.892 ms[local]:5432 pg12@testdb=#* select pid,locktype,relation::regclass,mode,page,tuple,virtualxid,transactionid,virtualtransaction,granted,fastpath from pg_locks where relation=295053;-[ RECORD 1 ]------+-----------------pid                | 2475locktype           | relationrelation           | t_lockmode               | RowExclusiveLockpage               | tuple              | virtualxid         | transactionid      | virtualtransaction | 3/2granted            | tfastpath           | tTime: 9.013 ms

session 2

[local]:5432 pg12@testdb=# ---- session 2[local]:5432 pg12@testdb=# begin;BEGINTime: 1.117 ms[local]:5432 pg12@testdb=#* select pg_backend_pid(); pg_backend_pid ----------------           2480(1 row)Time: 1.825 ms[local]:5432 pg12@testdb=#* update t_lock set id = 3000 where id = 3;-- 阻塞/挂起

session 3

[local]:5432 pg12@testdb=# select * from pg_locks where pid <> pg_backend_pid() and locktype = 'transactionid';   locktype    | database | relation | page | tuple | virtualxid | transactionid | classid | objid | objsubid | virtualtransaction | pid  |     mode      | granted | fastpath ---------------+----------+----------+------+-------+------------+---------------+---------+-------+----------+--------------------+------+---------------+---------+---------- transactionid |          |          |      |       |            |        669310 |         |       |          | 3/2                | 2475 | ExclusiveLock | t       | f transactionid |          |          |      |       |            |        669312 |         |       |          | 4/4                | 2480 | ExclusiveLock | t       | f transactionid |          |          |      |       |            |        669310 |         |       |          | 4/4                | 2480 | ShareLock     | f       | f(3 rows)Time: 1.243 ms

可以看到,进程2475中的事务669310和进程2480中的669312分别持有transactionid的ExclusiveLock,进程2480在等待事务ID=669310的lock(granted=f).
为什么会等待669310的ShareLock呢?回过头来查看t_lock表的xmax信息:

[local]:5432 pg12@testdb=# select xmin,xmax,ctid from t_lock where id = 3;  xmin  |  xmax  |  ctid   --------+--------+--------- 669246 | 669310 | (0,3) 669247 | 669310 | (4,99) 669248 | 669310 | (8,195) 669252 | 669310 | (13,65)(4 rows)Time: 4.715 ms

可以看到 : 待更新的tuple.xmax = 669310.
回滚事务669310,再次查看xmax:

[local]:5432 pg12@testdb=# select xmin,xmax,ctid from t_lock where id = 3;  xmin  |  xmax  |  ctid   --------+--------+--------- 669246 | 669312 | (0,3) 669247 | 669312 | (4,99) 669248 | 669312 | (8,195) 669252 | 669312 | (13,65)(4 rows)Time: 1.182 ms[local]:5432 pg12@testdb=# SELECT pid,backend_xid,wait_event_type,wait_event,state,query FROM pg_stat_activity WHERE pid IN (2475,2480);-[ RECORD 1 ]---+------------------------------------------pid             | 2475backend_xid     | wait_event_type | Clientwait_event      | ClientReadstate           | idlequery           | rollback;-[ RECORD 2 ]---+------------------------------------------pid             | 2480backend_xid     | 669312wait_event_type | Clientwait_event      | ClientReadstate           | idle in transactionquery           | update t_lock set id = 3000 where id = 3;Time: 5.434 ms

xmax被更新为669312.

感谢各位的阅读,以上就是"怎么掌握PostgreSQL Locks的基础知识"的内容了,经过本文的学习后,相信大家对怎么掌握PostgreSQL Locks的基础知识这一问题有了更深刻的体会,具体使用情况还需要大家实践验证。这里是,小编将为大家推送更多相关知识点的文章,欢迎关注!

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