[20170909]为什么是12秒.txt
[20170909]为什么是12秒.txt
--//在开发程序时我一般会强调开发尽量不要写一些自定义函数，往往可能导致CPU忙。
--//例子很像这样:
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
--//这样定义实际上非常不好,如果执行如下,相当于每行都会调用函数get_dept,形成不必要的递归.
--//本来应该使用表连接解决的,使用函数,有点像标量子查询.
--//当时我建议他好好看一下我写的blog,链接如下:
--// http://blog.itpub.net/267265/viewspace-2138042/ => [20170426]为什么是4秒.txt
--// 结果对方发现在12c for windows下,测试结果与我的测试不符,我重新测试看看.
1.环境:
SCOTT@test01p> @ ver1
PORT_STRING VERSION BANNER CON_ID
------------------------------ -------------- -------------------------------------------------------------------------------- ----------
IBMPC/WIN_NT64-9.1.0 12.1.0.1.0 Oracle Database 12c Enterprise Edition Release 12.1.0.1.0 - 64bit Production 0
2.建立函数:
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
set timing on
set autot traceonly;
SCOTT@test01p> select empno, ename, deptno, get_dept(deptno) c20 from emp;
14 rows selected.
Elapsed: 00:00:14.02
--//执行多次稳定递归的数量.需要14秒完成.
Execution Plan
----------------------------------------------------------
Plan hash value: 3956160932
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 14 | 182 | 3 (0)| 00:00:01 |
| 1 | TABLE ACCESS FULL| EMP | 14 | 182 | 3 (0)| 00:00:01 |
--------------------------------------------------------------------------
Statistics
----------------------------------------------------------
14 recursive calls
0 db block gets
36 consistent gets
0 physical reads
0 redo size
1130 bytes sent via SQL*Net to client
544 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
14 rows processed
--//可以发现递归执行14次(如果不是,执行多次最后会稳定在14次)
--//换成标量子查询:
SCOTT@test01p> select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
14 rows selected.
Elapsed: 00:00:12.04
Execution Plan
----------------------------------------------------------
Plan hash value: 1340320406
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 14 | 182 | 9 (0)| 00:00:01 |
| 1 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS FULL| EMP | 14 | 182 | 3 (0)| 00:00:01 |
--------------------------------------------------------------------------
Statistics
----------------------------------------------------------
12 recursive calls
0 db block gets
32 consistent gets
0 physical reads
0 redo size
1130 bytes sent via SQL*Net to client
544 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
14 rows processed
--//执行时间是12秒,递归12次.与我在11g下测试不同,11g下测试3秒.
set autot off
alter session set statistics_level=all;
SCOTT@test01p> show array
arraysize 200
SCOTT@test01p> select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
...
SCOTT@test01p> @ dpc '' ''
PLAN_TABLE_OUTPUT
-------------------------------------
SQL_ID apagxtf1p2puy, child number 1
-------------------------------------
select empno, ename, deptno, (select get_dept(deptno) from dual )c20
from emp
Plan hash value: 1340320406
--------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time | A-Rows | A-Time | Buffers |
--------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | | 9 (100)| | 14 |00:00:00.01 | 8 |
| 1 | FAST DUAL | | 12 | 1 | | 2 (0)| 00:00:01 | 12 |00:00:00.01 | 0 |
| 2 | TABLE ACCESS FULL| EMP | 1 | 14 | 182 | 3 (0)| 00:00:01 | 14 |00:00:00.01 | 8 |
--------------------------------------------------------------------------------------------------------------------
Query Block Name / Object Alias (identified by operation id):
-------------------------------------------------------------
1 - SEL$2 / DUAL@SEL$2
2 - SEL$1 / EMP@SEL$1
21 rows selected.
--//从执行计划也可以发现fast dual执行了12次.为什么?不好理解?要么3次,要么14次?什么也不应该12次.
3.分析:
--//使用10046跟踪看看
@ 10046on 12
select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
@ 10046off
--//检查跟踪文件.
*** 2017-09-09 21:20:16.366
WAIT #181607424: nam='PL/SQL lock timer' ela= 999265 duration=0 p2=0 p3=0 obj#=665 tim=5412716150
=====================
PARSING IN CURSOR #181597968 len=42 dep=1 uid=109 oct=3 lid=109 tim=5412716418 hv=328320763 ad='7ff13e15ba0' sqlid='56r5sd49t3jrv'
SELECT DNAME FROM DEPT WHERE DEPTNO = :B1
END OF STMT
BINDS #181597968:
Bind#0
oacdty=02 mxl=22(21) mxlc=00 mal=00 scl=00 pre=00
oacflg=03 fl2=1206001 frm=00 csi=00 siz=24 off=0
kxsbbbfp=212a7fb0 bln=22 avl=02 flg=05
value=20
EXEC #181597968:c=0,e=8900,p=0,cr=0,cu=0,mis=0,r=0,dep=1,og=1,plh=2852011669,tim=5412725274
FETCH #181597968:c=0,e=28,p=0,cr=2,cu=0,mis=0,r=1,dep=1,og=1,plh=2852011669,tim=5412725381
STAT #181597968 id=1 cnt=1 pid=0 pos=1 obj=92285 op='TABLE ACCESS BY INDEX ROWID DEPT (cr=2 pr=0 pw=0 time=30 us cost=1 size=13 card=1)'
STAT #181597968 id=2 cnt=1 pid=1 pos=1 obj=92286 op='INDEX UNIQUE SCAN PK_DEPT (cr=1 pr=0 pw=0 time=19 us cost=0 size=0 card=1)'
CLOSE #181597968:c=0,e=3,dep=1,type=3,tim=5412735882
FETCH #181607424:c=0,e=1019217,p=0,cr=7,cu=0,mis=0,r=1,dep=0,og=1,plh=1340320406,tim=5412735943
WAIT #181607424: nam='SQL*Net message from client' ela= 1165 driver id=1413697536 #bytes=1 p3=0 obj#=665 tim=5412737203
...
d:\tools\rlwrap>grep "BINDS #181597968:" D:\app\oracle\diag\rdbms\test\test\trace\test_ora_6436.trc |wc
12 24 216
--//确实12次.
d:\tools\rlwrap> grep -A5 "BINDS #181597968:" D:\app\oracle\diag\rdbms\test\test\trace\test_ora_6436.trc | grep "^ value="
value=20
value=30
value=20
value=30
value=10
value=20
value=10
value=30
value=20
value=30
value=20
value=10
SCOTT@test01p> select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
EMPNO ENAME DEPTNO C20
---------- ---------- ---------- --------------------
7369 SMITH 20 RESEARCH
7499 ALLEN 30 SALES
7521 WARD 30 SALES
7566 JONES 20 RESEARCH
7654 MARTIN 30 SALES
7698 BLAKE 30 SALES
7782 CLARK 10 ACCOUNTING
7788 SCOTT 20 RESEARCH
7839 KING 10 ACCOUNTING
7844 TURNER 30 SALES
7876 ADAMS 20 RESEARCH
7900 JAMES 30 SALES
7902 FORD 20 RESEARCH
7934 MILLER 10 ACCOUNTING
14 rows selected.
--//对比你可以发现,如果邻近的值一样,函数调用仅仅1次.
SCOTT@test01p> select * from (select empno, ename, deptno, lag(deptno,1) over(order by rownum) next from emp) where deptno=next;
EMPNO ENAME DEPTNO NEXT
---------- ---------- ---------- ----------
7521 WARD 30 30
7698 BLAKE 30 30
--//这样减少2次调用,说明12c与11g在标量子查询上处理与原来不同.不知道是什么回事,上班再仔细对比看看(与11g).
--//或者是OS版本的问题....
--//一个简单的验证,建立一个按照deptno排序的表.
create table empx as select * from emp order by deptno;
SCOTT@test01p> select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from empx;
EMPNO ENAME DEPTNO C20
---------- ---------- ---------- --------------------
7782 CLARK 10 ACCOUNTING
...
7654 MARTIN 30 SALES
14 rows selected.
Elapsed: 00:00:03.03
--//现在3秒...如果这样执行:
select t.* ,(select get_dept(t.deptno) from dual ) c20 from ( select empno, ename, deptno from emp order by deptno) t;
...
Elapsed: 00:00:12.00
--//竟然也是12秒.不过看执行计划马上明白:
SCOTT@test01p> @ dpc '' ''
PLAN_TABLE_OUTPUT
-------------------------------------
SQL_ID 248fpg2995yz8, child number 0
-------------------------------------
select t.* ,(select get_dept(t.deptno) from dual ) c20 from ( select
empno, ename, deptno from emp order by deptno) t
Plan hash value: 3996307794
------------------------------------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time | A-Rows | A-Time | Buffers | OMem | 1Mem | Used-Mem |
------------------------------------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | | 9 (100)| | 14 |00:00:12.00 | 31 | | | |
| 1 | FAST DUAL | | 12 | 1 | | 2 (0)| 00:00:01 | 12 |00:00:00.01 | 0 | | | |
| 2 | SORT ORDER BY | | 1 | 14 | 182 | 9 (0)| 00:00:01 | 14 |00:00:12.00 | 31 | 2048 | 2048 | 2048 (0)|
| 3 | TABLE ACCESS FULL| EMP | 1 | 14 | 182 | 3 (0)| 00:00:01 | 14 |00:00:00.01 | 7 | | | |
------------------------------------------------------------------------------------------------------------------------------------------------
--//可以发现 SORT ORDER BY 消耗12秒.也就是在排序前就进行递归调用函数了.
--//对比前面的执行计划,再次发现奇怪问题,看不出前面的执行计划12秒消耗在那里?
--//使用提示MATERIALIZE改写如下:
SCOTT@test01p> with t as ( select /*+MATERIALIZE */ empno, ename, deptno from emp order by deptno) select t.*,(select get_dept(t.deptno) from dual ) c20 from t;
EMPNO ENAME DEPTNO C20
---------- ---------- ---------- --------------------
7782 CLARK 10 ACCOUNTING
7839 KING 10 ACCOUNTING
..
7654 MARTIN 30 SALES
14 rows selected.
Elapsed: 00:00:03.02
SCOTT@test01p> @ dpc '' ''
PLAN_TABLE_OUTPUT
-------------------------------------
SQL_ID fwvgq6403j0nx, child number 0
-------------------------------------
with t as ( select /*+MATERIALIZE */ empno, ename, deptno from emp
order by deptno) select t.*,(select get_dept(t.deptno) from dual ) c20
from t
Plan hash value: 2014637640
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time | A-Rows | A-Time | Buffers | Reads | Writes | OMem | 1Mem | Used-Mem |
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | | 11 (100)| | 14 |00:00:00.01 | 21 | 1 | 1 | | | |
| 1 | FAST DUAL | | 3 | 1 | | 2 (0)| 00:00:01 | 3 |00:00:00.01 | 0 | 0 | 0 | | | |
| 2 | TEMP TABLE TRANSFORMATION | | 1 | | | | | 14 |00:00:00.01 | 21 | 1 | 1 | | | |
| 3 | LOAD AS SELECT | | 1 | | | | | 0 |00:00:00.01 | 11 | 0 | 1 | 1036K| 1036K| |
| 4 | SORT ORDER BY | | 1 | 14 | 182 | 3 (0)| 00:00:01 | 14 |00:00:00.01 | 7 | 0 | 0 | 2048 | 2048 | 2048 (0)|
| 5 | TABLE ACCESS FULL | EMP | 1 | 14 | 182 | 3 (0)| 00:00:01 | 14 |00:00:00.01 | 7 | 0 | 0 | | | |
| 6 | VIEW | | 1 | 14 | 322 | 2 (0)| 00:00:01 | 14 |00:00:00.01 | 7 | 1 | 0 | | | |
| 7 | TABLE ACCESS FULL | SYS_TEMP_0FD9D660E_18A8CC5 | 1 | 14 | 182 | 2 (0)| 00:00:01 | 14 |00:00:00.01 | 7 | 1 | 0 | | | |
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
4.继续测试使用 DETERMINISTIC Functions:
--//一般如果在在某个函数定义索引，需要DETERMINISTIC,表示返回结果固定。其实即使不固定，也可以这样定义。
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
DETERMINISTIC
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
SCOTT@test01p> select empno, ename, deptno, get_dept(deptno) c20 from emp;
...
14 rows selected.
Elapsed: 00:00:14.01
--//奇怪! 又回到14秒,而在11g下这样的测试是4秒(注意设置array=200的情况,如果太小,时间会变化).
5.最后补充测试RESULT CACHE的情况:
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
RESULT_CACHE
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
SCOTT@test01p> select empno, ename, deptno, get_dept(deptno) c20 from emp;
14 rows selected.
Elapsed: 00:00:03.13
SCOTT@test01p> select empno, ename, deptno, get_dept(deptno) c20 from emp;
14 rows selected.
Elapsed: 00:00:00.00
--//第1次执行3秒,第2次执行0秒.修改如下结果一样.
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
RESULT_CACHE
DETERMINISTIC
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
总结:
--//不知道是否是OS版本的问题,有机会安装12c在linux下重复测试看看.
--//看来测试环境该转向12c了....那位有12c for linux的版本测试看看...
--//补充测试11g的情况:
SCOTT@book> @ &r/ver1
PORT_STRING VERSION BANNER
------------------------------ -------------- --------------------------------------------------------------------------------
x86_64/Linux 2.4.xx 11.2.0.4.0 Oracle Database 11g Enterprise Edition Release 11.2.0.4.0 - 64bit Production
SYS@book> GRANT EXECUTE ON SYS.dbms_lock TO SCOTT;
Grant succeeded.
CREATE OR REPLACE FUNCTION get_dept (p_deptno dept.deptno%TYPE)
RETURN dept.dname%TYPE
IS
l_dname dept.dname%TYPE;
BEGIN
DBMS_LOCK.sleep (1);
SELECT dname
INTO l_dname
FROM dept
WHERE deptno = p_deptno;
RETURN l_dname;
END;
/
set timing on
set autot traceonly;
select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
SCOTT@book> select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
14 rows selected.
Elapsed: 00:00:03.01
Execution Plan
----------------------------------------------------------
Plan hash value: 1340320406
--------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
--------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 14 | 182 | 3 (0)| 00:00:01 |
| 1 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
| 2 | TABLE ACCESS FULL| EMP | 14 | 182 | 3 (0)| 00:00:01 |
--------------------------------------------------------------------------
Statistics
----------------------------------------------------------
3 recursive calls
0 db block gets
13 consistent gets
0 physical reads
0 redo size
1114 bytes sent via SQL*Net to client
520 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
14 rows processed
--//执行多次,基本稳定在3个递归调用.
set autot off
alter session set statistics_level=all;
select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp;
SCOTT@book> @ &r/dpc '' ''
PLAN_TABLE_OUTPUT
-------------------------------------
SQL_ID apagxtf1p2puy, child number 1
-------------------------------------
select empno, ename, deptno, (select get_dept(deptno) from dual )c20 from emp
Plan hash value: 1340320406
--------------------------------------------------------------------------------------------------------------------
| Id | Operation | Name | Starts | E-Rows |E-Bytes| Cost (%CPU)| E-Time | A-Rows | A-Time | Buffers |
--------------------------------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | | 3 (100)| | 14 |00:00:00.01 | 7 |
| 1 | FAST DUAL | | 3 | 1 | | 2 (0)| 00:00:01 | 3 |00:00:00.01 | 0 |
| 2 | TABLE ACCESS FULL| EMP | 1 | 14 | 182 | 3 (0)| 00:00:01 | 14 |00:00:00.01 | 7 |
--------------------------------------------------------------------------------------------------------------------
Query Block Name / Object Alias (identified by operation id):
-------------------------------------------------------------
1 - SEL$2 / DUAL@SEL$2
2 - SEL$1 / EMP@SEL$1
21 rows selected.
--//Id=2.starts=3次,也验证函数调用3次.
--//不知道为什么,也许12.2.0.N版本跟11g一样,没有这个oracle版本等以后再测试吧....
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Description 如果对于一个 \(1\sim n\) 的排列满足: 在 \(1\sim n-1\) 这些位置之后将序列断开,使得总可以从右边找一个数,使得该数不会比左边所有数都大,则称该序列是 ...