oracle构建一致性读

时间:2020-11-30 06:18:11

对于实际的业务系统,通常有一些热点的表,insert和delete的量非常大,这个时候就会发现一些查询语句的逻辑读比较偏高,
这时可能就是oracle在构建一致性块的进行的consistent read。
下面做一个测试看下:
第一步准备数据:

create table test(
col1 varchar2(12)
col2 number
ext varchar2(4000)
);
create index test_ind on test(user_id, col2);
create sequence seq_test cache 200;

这样这样的表,我们假设有频繁的插入和删除操作,那么下面来测试一下select的逻辑读的情况。

开启两个session:
1,创建表保存snapshot

在session1:
create table prefix_stats tablespace IW_ACCOUNT_LOG_01 as select * from v$sesstat where sid=&1;

2,在session2查询

select  * from (select  * from test t where col1 = 'xpchild001' order by trans_log_id) where rownum <= 200;

3,在session1监控session2的统计信息

  

select *
from (select t.name,
pre.value as pre,
suf.value as suf,
(suf.value - pre.value) as diff
from prefix_stats pre, v$sesstat suf, v$statname t
where pre.sid = suf.sid
and pre.STATISTIC# = suf.STATISTIC#
and pre.STATISTIC# = t.STATISTIC#) tmp
where tmp.diff > 0
order by tmp.diff desc Name PRE SUF DIFF
---------------------------------------------------------------------- ---------- ---------- ----------
session pga memory max 957208 1153816 196608
session pga memory 957208 1153816 196608
bytes sent via SQL*Net to client 6692 37013 30321
redo size 0 8256 8256
session logical reads 52 1508 1456
consistent gets from cache 52 1508 1456
consistent gets 52 1508 1456
bytes received via SQL*Net from client 4385 5639 1254
consistent gets - examination 21 1253 1232
data blocks consistent reads - undo records applied 0 920 920
consistent changes 0 920 920
buffer is not pinned count 17 222 205
table fetch by rowid 6 206 200
buffer is pinned count 0 197 197
CR blocks created 0 160 160
calls to kcmgas 0 160 160
db block changes 0 120 120
redo entries 0 120 120
cleanout - number of ktugct calls 0 120 120
cleanouts and rollbacks - consistent read gets 0 120 120
immediate (CR) block cleanout applications 0 120 120
no work - consistent read gets 19 83 64
heap block compress 0 51 51
rollbacks only - consistent read gets 0 40 40
shared hash latch upgrades - no wait 0 5 5
user calls 28 33 5
execute count 21 23 2
DB time 0 2 2
parse count (total) 22 24 2
session cursor cache count 16 17 1
CPU used when call started 0 1 1
recursive calls 92 93 1
parse count (hard) 0 1 1
session cursor cache hits 4 5 1
CPU used by this session 0 1 1

这一次的查询,返回记录200条。table fetch by rowid=200;

1,逻辑读session logical reads=consistent gets(一致读)+db block gets(当前读);
这个sql只有一致性读session logical reads=consistent gets=1456

2,构建一致性读应用回滚记录统计:data blocks consistent reads(undo records applied):920
等价于consistent changes。

3,需要回滚或者块清除产生的一致性读(这里边没有回滚,只可能有块清除)cleanouts and rollbacks - consistent read gets:120
跟db block changes=120一致,这里进行了块清楚,从而改变了db block。

4,构建一致性读clone的块数:CR blocks created=160
5,块清除产生的redo:redo size 8256

验证了开始的猜测:大量的构建一致性读。

对于这样的热点表,通常有几种手动去调整,但核心都是要分散热点,减少争用。

  1. hash表,分散热点
  2. 调整pctfree,增加pctfred的大小。使用块中的记录数变少,减少构建一致性读的问题。

未完待续。。。