13518219792
这篇文章主要讲解了“怎么理解PostgreSQL全表扫描问题”，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“怎么理解PostgreSQL全表扫描问题”吧！
北关网站建设公司创新互联,北关网站设计制作，有大型网站制作公司丰富经验。已为北关1000+提供企业网站建设服务。企业网站搭建\成都外贸网站建设要多少钱，请找那个售后服务好的北关做网站的公司定做！
本节内容来源于PGer的一个问题:
Q:
由于多版本的存在，那么全表扫描是不是需要更长的时间了呢？
A:
关于全表扫描,不妨考虑2种极端的情况:
1.insert数据(事务已提交,下同),没有执行update/delete,没有dead tuple,全表扫描效率没有影响;
2.insert数据,执行了大量的update/delete,同时禁用了autovacuum也没有手工执行vacuum,那么存在大量的dead tuple,性能上一是需要更多的IO操作,二是需要执行额外的CPU判断(对于所有的tuple都要执行可见性判断).
其判断逻辑如下:
((Xmin == my-transaction && inserted by the current transaction Cmin < my-command && before this command, and (Xmax is null || the row has not been deleted, or (Xmax == my-transaction && it was deleted by the current transaction Cmax >= my-command))) but not before this command, || or (Xmin is committed && the row was inserted by a committed transaction, and (Xmax is null || the row has not been deleted, or (Xmax == my-transaction && the row is being deleted by this transaction Cmax >= my-command) || but it’s not deleted "yet", or (Xmax != my-transaction && the row was deleted by another transaction Xmax is not committed)))) that has not been committed
简单做个实验,创建一张表t_fts,
1.插入数据,大小为s1,执行全表扫描,时间为m秒;
2.update所有行,大小为s2,执行全表扫描,时间为n秒.
理论上来说,n应为m的s2/s1倍左右(相对于IO时间,如果tuple数不多,CPU时间可以忽略不计).
创建数据表,插入数据:
testdb=# drop table if exists t_fts;
DROP TABLE
testdb=# create table t_fts(id int,c1 varchar(200),c2 varchar(200));
CREATE TABLE
testdb=#
testdb=# insert into t_fts select x,lpad('c1'||x,200,'x'),lpad('c1'||x,200,'x') from generate_series(1,2000000) as x;
INSERT 0 2000000
testdb=# select pg_size_pretty(pg_table_size('t_fts'));
pg_size_pretty
----------------
868 MB
(1 row)禁用autovacuum,执行查询:
testdb=# alter system set autovacuum=off; ALTER SYSTEM testdb=# show autovacuum; autovacuum ------------ off (1 row) testdb=# explain analyze verbose select * from t_fts; QUERY PLAN ---------------------------------------------------------------------------------------------------------------------------- Seq Scan on public.t_fts (cost=0.00..131112.16 rows=2000016 width=412) (actual time=0.048..1086.289 rows=2000000 loops=1) Output: id, c1, c2 Planning Time: 30.762 ms Execution Time: 1181.360 ms (4 rows)
执行update:
testdb=# update t_fts set c1 = lpad('c1'||(id+1),200,id+1||''),c2 = lpad('c1'||(id+1),200,id+1||'');
UPDATE 2000000
testdb=# select pg_size_pretty(pg_table_size('t_fts'));
pg_size_pretty
----------------
1737 MB
(1 row)
testdb=# explain analyze verbose select * from t_fts;
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------------
--
Seq Scan on public.t_fts (cost=0.00..262223.14 rows=4000014 width=412) (actual time=3168.414..6117.780 rows=2000000 loops=1
)
Output: id, c1, c2
Planning Time: 5.493 ms
Execution Time: 6205.705 ms
(4 rows)
testdb=# explain analyze verbose select * from t_fts;
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------------
-
Seq Scan on public.t_fts (cost=0.00..262223.14 rows=4000014 width=412) (actual time=776.660..2311.270 rows=2000000 loops=1)
Output: id, c1, c2
Planning Time: 0.426 ms
Execution Time: 2391.895 ms
(4 rows)
testdb=# explain analyze verbose select * from t_fts;
QUERY PLAN
-----------------------------------------------------------------------------------------------------------------------------
-
Seq Scan on public.t_fts (cost=0.00..262223.14 rows=4000014 width=412) (actual time=728.758..2293.157 rows=2000000 loops=1)
Output: id, c1, c2
Planning Time: 0.481 ms
Execution Time: 2373.241 ms
(4 rows)感谢各位的阅读，以上就是“怎么理解PostgreSQL全表扫描问题”的内容了，经过本文的学习后，相信大家对怎么理解PostgreSQL全表扫描问题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。这里是创新互联，小编将为大家推送更多相关知识点的文章，欢迎关注！
