oracle构建一致性读
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。
- hash表，分散热点
- 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。。。