[20180607]函数与标量子查询8.txt

--//前面看http://www.cnblogs.com/kerrycode/p/9099507.html链接,里面提到:

通俗来将，当使用标量子查询的时候，ORACLE会将子查询结果缓存在哈希表中， 如果后续的记录出现同样的值，优化器通过缓存在哈希

表中的值，判断重复值不用重复调用函数，直接使用上次计算结果即可。从而减少调用函数次数，从而达到优化性能的效果。另外在 ORACLE 10和11中， 哈希表只包含了255个Buckets，也就是说它能存储255个不同值，如果超过这个范围，就会出现散列冲突，那些出现 散列冲突的值就会重复调用函数，即便如此，依然能达到大幅改善性能的效果。

--//昨天我测试11.2.0.4 for linux下,哈希表不止255个Buckets.

--//今天测试看看10g下到底有多少个Buckets.因为我感觉10g可能哈希表的buckets可能不大.

1.环境:

SYS@test> @ &r/ver1

PORT_STRING                    VERSION        BANNER

------------------------------ -------------- ---------------------------------------------------------------- x86_64/Linux 2.4.xx            10.2.0.4.0     Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - 64bi

grant execute on sys.dbms_lock to scott;

CREATE OR REPLACE FUNCTION sleep1 (seconds IN NUMBER)

RETURN NUMBER is d_date date; BEGIN   select sysdate into d_date from dual;   sys.dbms_lock.sleep(seconds/10);   RETURN seconds; END; /

CREATE OR REPLACE FUNCTION sleep (seconds IN NUMBER)

RETURN NUMBER is d_date date; BEGIN   select sysdate into d_date from dual; --//sys.dbms_lock.sleep(0.01);   RETURN seconds; END; /

create table t as select rownum id1,mod(rownum-1,4000)+1 id2 from dual connect by level<=8000;

--//ALTER TABLE t MINIMIZE RECORDS_PER_BLOCK ; --//注意插入数据的顺序,我以前的插入有1点问题,导致id2显示不按照1-4000,1-4000显示(执行select * from t). --//导致测试出现一些奇怪情况.

2.测试:

SELECT    'exec :x := '            || LEVEL            || ';'            || CHR (10)            || 'select rowid,t.*,(select sleep(id2) from dual) s from t where id2<=:x;'            || CHR (10)            || '@ &r/dpc '''' '''''       FROM DUAL CONNECT BY LEVEL <= 4000;

--//建立脚本ay.txt:

alter session set statistics_level=all; set feed on variable x number; exec :x := 1; select rowid,t.*,(select sleep(id2) from dual) s from t where id2<=:x; @ '' '' ... exec :x := 4000; select rowid,t.*,(select sleep(id2) from dual) s from t where id2<=:x; @ '' ''

--//多执行几次,避免其它递归影响.

spool cz.txt @ ay.txt spool off

--//取出数字

$ egrep 'FAST DUAL|rows selected' cz.txt | sed '/^29 rows selected./d' > c1.txt $ grep  "rows selected." c1.txt  | cut -f1 -d' '> c2.txt $ grep "FAST DUAL"  c1.txt | cut -f5 -d"|" > c3.txt $ paste c2.txt c3.txt -d"," > c4.txt

SCOTT@book> create table t1 ( b number ,a number);

Table created. --//注b在前,表示查询记录数量,a表示执行fast dual次数,也就是递归次数. --//修改b4.txt ,改写成inert插入表t1.执行如下: :%s/^/insert into t1 values(/g :%s/$/);/g

select max(id2) from (

SELECT id2, r, rp   FROM (  SELECT b / 2 id2, a r, LAG (a) OVER (ORDER BY b) rp             FROM t1         ORDER BY b/2) WHERE r - rp = 1 order by id2);   MAX(ID2) ----------       3152

--//3152值还会进入backupset,后面的数字带入都是出现hash 冲突的情况.

SELECT id2, r, rp

  FROM (  SELECT b / 2 id2, a r, LAG (a) OVER (ORDER BY b) rp             FROM t1         ORDER BY b/2) WHERE r - rp >= 2 and id2<=3152 ;

--//输出太长,一共2640个值,略,这个结果就是在1-3152之间,出现hash冲突的值.

select  rowid,t.*,(select sleep(id2) from dual) s from t where id2<=3152

and id2 not in ( SELECT /*+ NL_AJ */ id2   FROM (  SELECT b / 2 id2, a r, LAG (a) OVER (ORDER BY b) rp             FROM t1         ORDER BY b/2) WHERE r - rp >= 2 and id2<=3152 );

SCOTT@test> @ &r/dpc '' ''

PLAN_TABLE_OUTPUT ------------------------------------- SQL_ID  7r8dyxjmdwucp, child number 0 ------------------------------------- select  rowid,t.*,(select sleep(id2) from dual) s from t where id2<=3152 and id2 not in ( SELECT /*+ NL_AJ */ id2   FROM (  SELECT b / 2 id2, a r, LAG (a) OVER (ORDER BY b) rp             FROM t1         ORDER BY b/2)  WHERE r - rp >= 2 and id2<=3152 ) Plan hash value: 4130365942 --------------------------------------------------------------------------------------------------------------------------------------------------- | Id  | Operation             | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time   | A-Rows |   A-Time   | Buffers |  OMem |  1Mem | Used-Mem | --------------------------------------------------------------------------------------------------------------------------------------------------- |   1 |  FAST DUAL            |      |    512 |      1 |       |     2   (0)| 00:00:01 |    512 |00:00:00.01 |       0 |       |       |          | |   2 |  NESTED LOOPS ANTI    |      |      1 |   6304 |   283K| 46284  (32)| 00:09:16 |   1024 |00:00:07.65 |      40 |       |       |          | |*  3 |   TABLE ACCESS FULL   | T    |      1 |   6306 | 44142 |     6   (0)| 00:00:01 |   6304 |00:00:00.01 |      24 |       |       |          | |*  4 |   VIEW                |      |   6304 |      1 |    39 |     7  (29)| 00:00:01 |   5280 |00:00:07.65 |      16 |       |       |          | |   5 |    SORT ORDER BY      |      |   6304 |   4000 | 28000 |     7  (29)| 00:00:01 |   4096K|00:00:06.14 |      16 |   160K|   160K|  142K (0)| |   6 |     WINDOW SORT       |      |   1024 |   4000 | 28000 |     7  (29)| 00:00:01 |   4096K|00:00:04.10 |      16 |   196K|   196K|  174K (0)| |   7 |      TABLE ACCESS FULL| T1   |      1 |   4000 | 28000 |     5   (0)| 00:00:01 |   4000 |00:00:00.01 |      16 |       |       |          | ---------------------------------------------------------------------------------------------------------------------------------------------------

Query Block Name / Object Alias (identified by operation id):

-------------------------------------------------------------

   1 - SEL$2        / DUAL@SEL$2

   2 - SEL$8771BF6C    3 - SEL$8771BF6C / T@SEL$1    4 - SEL$4        / from$_subquery$_003@SEL$3    5 - SEL$4    7 - SEL$4        / T1@SEL$4

Predicate Information (identified by operation id):

---------------------------------------------------

   3 - filter("ID2"<=3152)

   4 - filter(("ID2"="ID2" AND "R"-"RP">=2 AND "ID2"<=3152)) 35 rows selected.

--//正好512,说明10.2.0.4,哈希表只包含了512个Buckets，也就是说它能存储512个不同值，

--//删除冲突的记录看看.

delete from t where id2 in (SELECT id2   FROM (  SELECT b / 2 id2, a r, LAG (a) OVER (ORDER BY b) rp             FROM t1         ORDER BY b/2) WHERE r - rp >= 2 and id2<=3152 ); commit;

select rowid,t.*,(select sleep(id2) from dual) s from t where id2<=3152;

SCOTT@test> @ &r/dpc '' ''

PLAN_TABLE_OUTPUT ------------------------------------- SQL_ID  9q5bnk36nnq68, child number 0 ------------------------------------- select rowid,t.*,(select sleep(id2) from dual) s from t where id2<=3152

Plan hash value: 1032660217

--------------------------------------------------------------------------------------------------------------------

| Id  | Operation         | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time   | A-Rows |   A-Time   | Buffers | -------------------------------------------------------------------------------------------------------------------- |   1 |  FAST DUAL        |      |    512 |      1 |       |     2   (0)| 00:00:01 |    512 |00:00:00.01 |       0 | |*  2 |  TABLE ACCESS FULL| T    |      1 |   6306 | 44142 |     6   (0)| 00:00:01 |   1024 |00:00:00.01 |      24 | -------------------------------------------------------------------------------------------------------------------- Query Block Name / Object Alias (identified by operation id): -------------------------------------------------------------    1 - SEL$2 / DUAL@SEL$2    2 - SEL$1 / T@SEL$1 Predicate Information (identified by operation id): ---------------------------------------------------    2 - filter("ID2"<=3152) 24 rows selected.

--//这样如果

select rowid,t.*,(select sleep(id2) from dual) s from t where (id2<=3152 and id2<>1693) or id2=:x; --//:x 选择 3153-4000 任何一个 ,fast dual 的starts都是514,也就是存在冲突.大家可以自行验证.

--//dpc脚本如下:

set verify off select * from table(dbms_xplan.display_cursor(NVL('&1',NULL),NULL,'ALL ALLSTATS LAST PEEKED_BINDS cost partition -projection -outline &2'));

prompt

prompt argment : typical all advanced partition predicate remote note parallel projection alias peeked_binds outline adaptive prompt