[20200824]12c sqlplus rowprefetch arraysize 显示行数量的关系.txt
--//以前写的:
[20181108]12c sqlplus rowprefetch参数4.txt => />[20181109]12c sqlplus rowprefetch参数5.txt => />
--//别人问测试一些细节问题,说真的当时测试完成就再没关注这个问题,这种问题对于实际应用根本不重要.前台应用不会使用
--//sqlplus.
1.fetch规律:
--//我当时测试总结一些规律:
--//1.fetch 第1次数量等于rowprefetch.当然必须小于返回记录的数量.
--//2.fetch 第X次数量(X>=2)与参数arraysize的倍数N有关. N=floor(rowprefetch/arraysize+1), 等于N*arraysize.
--//3.fetch 最后一次应该等于剩余记录.不会大于floor(rowprefetch/arraysize+1)*arraysize.
--//4.fetch 最后一次有可能是0.
--//也就是fetch的顺序:
rowprefetch,(floor(rowprefetch/arraysize)+1)*arraysize,(floor(rowprefetch/arraysize)+1)*arraysize,...,剩下的记录.
--//注:我当时测试时忽略了rowprefetch=arraysize的情况.使用floor代替ceil才是正确的.
2.显示行数量规律:
--//而显示记录时看到的情况并不对应fetch的记录数量.以前的分析有误.仅仅需要记住几点点,我自己的总结:
--//1.显示输出行数 第1次 floor(rowprefetch/arraysize)*arraysize.
--//注:.如果rowprefetch < arraysize,第1次fetch后,不足arraysize数量.不会马上输出,而是等待下一个fetch完成,再输出.
--//2.显示输出行数 第Y次(Y>=2) floor((前次剩下的记录+本次fetch的记录)/arraysize)*arraysize.
--//注:因为前次剩下的记录小于arraysize,这样显示输出行数=floor(rowprefetch/arraysize+1)*arraysize.
--//3.显示输出行数 最后1次比较特殊,是全部输出.判断这个依据是最后fetch的数量<(floor(rowprefetch/arraysize)+1)*arraysize,
--// 表示已没有记录需要fetch.
--//通过下面例子说明:
3.再做一个测试说明问题:
SYS@test> @ ver1
PORT_STRING VERSION BANNER CON_ID
------------------------------ -------------- -------------------------------------------------------------------------------- ----------
IBMPC/WIN_NT64-9.1.0 12.2.0.1.0 Oracle Database 12c Enterprise Edition Release 12.2.0.1.0 - 64bit Production 0
create table t as select rownum id1,1 id2 from dual connect by level<=23;
grant EXECUTE ON dbms_lock to scott;
CREATE OR REPLACE FUNCTION SCOTT.sleep (seconds IN NUMBER)
RETURN NUMBER
AS
BEGIN
sys.DBMS_LOCK.sleep (seconds);
RETURN seconds;
END;
/
$ cat aa.txt
set timing on
set arraysize &1
set rowprefetch &2
alter session set events '10046 trace name context forever, level 12';
select rownum,t.*,sleep(id2) n10,&&1 arraysize ,&&2 rowprefetch from t;
alter session set events '10046 trace name context off';
set timing off
quit
$ cat ~/bin/ts.awk
#! /bin/bash
awk '{ print strftime("[%Y-%m-%d %H:%M:%S]"), $0 }'
$ sqlplus -s -l scott/btbtms@test01p @ aa.txt 2 7 | ~/bin/ts.awk
[2020-08-24 20:31:25]
[2020-08-24 20:31:25] Session altered.
[2020-08-24 20:31:25]
[2020-08-24 20:31:25] Elapsed: 00:00:00.01
[2020-08-24 20:31:25] old 1: select rownum,t.*,sleep(id2) n10,&&1 arraysize ,&&2 rowprefetch from t
[2020-08-24 20:31:25] new 1: select rownum,t.*,sleep(id2) n10,2 arraysize ,7 rowprefetch from t
[2020-08-24 20:31:32]
[2020-08-24 20:31:32] ROWNUM ID1 ID2 N10 ARRAYSIZE ROWPREFETCH
[2020-08-24 20:31:32] ------ ---- ---- ---- ---------- -----------
[2020-08-24 20:31:32] 1 1 1 1 2 7
[2020-08-24 20:31:32] 2 2 1 1 2 7
[2020-08-24 20:31:32] 3 3 1 1 2 7
[2020-08-24 20:31:32] 4 4 1 1 2 7
[2020-08-24 20:31:32] 5 5 1 1 2 7
[2020-08-24 20:31:32] 6 6 1 1 2 7
[2020-08-24 20:31:40] 7 7 1 1 2 7
[2020-08-24 20:31:40] 8 8 1 1 2 7
[2020-08-24 20:31:40] 9 9 1 1 2 7
[2020-08-24 20:31:40] 10 10 1 1 2 7
[2020-08-24 20:31:40] 11 11 1 1 2 7
[2020-08-24 20:31:40] 12 12 1 1 2 7
[2020-08-24 20:31:40] 13 13 1 1 2 7
[2020-08-24 20:31:40] 14 14 1 1 2 7
[2020-08-24 20:31:48] 15 15 1 1 2 7
[2020-08-24 20:31:48] 16 16 1 1 2 7
[2020-08-24 20:31:48] 17 17 1 1 2 7
[2020-08-24 20:31:48] 18 18 1 1 2 7
[2020-08-24 20:31:48] 19 19 1 1 2 7
[2020-08-24 20:31:48] 20 20 1 1 2 7
[2020-08-24 20:31:48] 21 21 1 1 2 7
[2020-08-24 20:31:48] 22 22 1 1 2 7
[2020-08-24 20:31:48] 23 23 1 1 2 7
[2020-08-24 20:31:48]
[2020-08-24 20:31:48] 23 rows selected.
[2020-08-24 20:31:48]
[2020-08-24 20:31:48] Elapsed: 00:00:23.03
[2020-08-24 20:31:48]
[2020-08-24 20:31:48] Session altered.
[2020-08-24 20:31:48]
[2020-08-24 20:31:48] Elapsed: 00:00:00.00
--//注意看前面的时间戳,可以发现输出时间间隔7,8,8秒.显示输出行数6,8,9.
$ grep "FETCH" test_ora_7992.trc | grep "#451109600"
FETCH #451109600:c=0,e=6994268,p=0,cr=7,cu=2,mis=0,r=7,dep=0,og=1,plh=2402761124,tim=1748148741
FETCH #451109600:c=0,e=7993084,p=0,cr=1,cu=0,mis=0,r=8,dep=0,og=1,plh=2402761124,tim=1756146829
FETCH #451109600:c=0,e=7992326,p=0,cr=1,cu=0,mis=0,r=8,dep=0,og=1,plh=2402761124,tim=1764140167
FETCH #451109600:c=0,e=17,p=0,cr=0,cu=0,mis=0,r=0,dep=0,og=1,plh=2402761124,tim=1764141275
--//注意看r=??,fetch 7,8,8,0. 而显示输出行数6,8,9.
--//你可以发现第2次fetch不再等于arraysize,而是4*arraysize=8.
--//最后出现最后一次fetch r=0的情况,实际上显示行数是6,8,8,1,为什么看上去是显示9,因为最后1次fetch是0,消耗时间很小.
--//这样感觉上输出9,这也是对方感觉困惑的主要原因.
--//可以这样理解: arraysize = 2,rowprefetch=7
--//第1次fetch 7,而arraysize=2,这样显示输出floor(rowprefetch/arraysize)*arraysize=floor(7/2)*2=6.剩下1条.
--//第2次fetch 8,显示输出floor((8+1)/2)*2=8,还是剩下1条.
--//第3次fetch 8,显示输出floor((8+1)/2)*2=8,还是剩下1条.
--//第4次fetch 0,全部输出,剩下的1行.由于fetch等于0,这次会很快.
--//实际显示输出的是:6,8,8,1.
--//如果最后1次不是fetch=0,看到的情况如下:
$ sqlplus -s -l scott/btbtms@test01p @ aa.txt 5 4 | ~/bin/ts.awk
[2020-08-24 21:04:12]
[2020-08-24 21:04:12] Session altered.
[2020-08-24 21:04:12]
[2020-08-24 21:04:12] Elapsed: 00:00:00.00
[2020-08-24 21:04:12] old 1: select rownum,t.*,sleep(id2) n10,&&1 arraysize ,&&2 rowprefetch from t
[2020-08-24 21:04:12] new 1: select rownum,t.*,sleep(id2) n10,5 arraysize ,4 rowprefetch from t
[2020-08-24 21:04:21]
[2020-08-24 21:04:21] ROWNUM ID1 ID2 N10 ARRAYSIZE ROWPREFETCH
[2020-08-24 21:04:21] ------ ---- ---- ---- ---------- -----------
[2020-08-24 21:04:21] 1 1 1 1 5 4
[2020-08-24 21:04:21] 2 2 1 1 5 4
[2020-08-24 21:04:21] 3 3 1 1 5 4
[2020-08-24 21:04:21] 4 4 1 1 5 4
[2020-08-24 21:04:21] 5 5 1 1 5 4
[2020-08-24 21:04:26] 6 6 1 1 5 4
[2020-08-24 21:04:26] 7 7 1 1 5 4
[2020-08-24 21:04:26] 8 8 1 1 5 4
[2020-08-24 21:04:26] 9 9 1 1 5 4
[2020-08-24 21:04:26] 10 10 1 1 5 4
[2020-08-24 21:04:31] 11 11 1 1 5 4
[2020-08-24 21:04:31] 12 12 1 1 5 4
[2020-08-24 21:04:31] 13 13 1 1 5 4
[2020-08-24 21:04:31] 14 14 1 1 5 4
[2020-08-24 21:04:31] 15 15 1 1 5 4
[2020-08-24 21:04:35] 16 16 1 1 5 4
[2020-08-24 21:04:35] 17 17 1 1 5 4
[2020-08-24 21:04:35] 18 18 1 1 5 4
[2020-08-24 21:04:35] 19 19 1 1 5 4
[2020-08-24 21:04:35] 20 20 1 1 5 4
[2020-08-24 21:04:35] 21 21 1 1 5 4
[2020-08-24 21:04:35] 22 22 1 1 5 4
[2020-08-24 21:04:35] 23 23 1 1 5 4
[2020-08-24 21:04:35]
[2020-08-24 21:04:35] 23 rows selected.
[2020-08-24 21:04:35]
[2020-08-24 21:04:35] Elapsed: 00:00:23.20
[2020-08-24 21:04:35]
[2020-08-24 21:04:35] Session altered.
[2020-08-24 21:04:35]
[2020-08-24 21:04:35] Elapsed: 00:00:00.00
--//注意看前面的时间戳,可以发现输出时间间隔9,5,5,4秒.
--//显示行数:5,5,5,8.
$ grep "FETCH" test_ora_7712.trc | grep "#201226544"
FETCH #201226544:c=78000,e=4070787,p=0,cr=467,cu=2,mis=0,r=4,dep=0,og=1,plh=2402761124,tim=3712159663
FETCH #201226544:c=0,e=4995743,p=0,cr=1,cu=0,mis=0,r=5,dep=0,og=1,plh=2402761124,tim=3717161694
FETCH #201226544:c=0,e=4995417,p=0,cr=1,cu=0,mis=0,r=5,dep=0,og=1,plh=2402761124,tim=3722158913
FETCH #201226544:c=0,e=4995747,p=0,cr=1,cu=0,mis=0,r=5,dep=0,og=1,plh=2402761124,tim=3727156131
FETCH #201226544:c=0,e=3996706,p=0,cr=1,cu=0,mis=0,r=4,dep=0,og=1,plh=2402761124,tim=3731154321
--//可以这样理解: arraysize = 5,rowprefetch=4
--//第1次fetch 4,而arraysize=5,不足arraisize数量不会输出.
--//第2次fetch 5,floor((5+4)/5)*5=5,输出5行,还剩下4行.这也是为什么第1个时间间隔是9秒的原因.
--//第3次fetch 5,floor((5+4)/5)*5=5,输出5行,还剩下4行.
--//第4次fetch 4,需要4秒完成fetch对应前面最后的时间间隔是4秒,全部输出 4+4=8.
--//再做1个特殊情况,ROWPREFETCH正好整除ARRAYSIZE的情况:
$ sqlplus -s -l scott/btbtms@test01p @ aa.txt 2 6 | ~/bin/ts.awk
[2020-08-24 21:11:58]
[2020-08-24 21:11:58] Session altered.
[2020-08-24 21:11:58]
[2020-08-24 21:11:58] Elapsed: 00:00:00.00
[2020-08-24 21:11:58] old 1: select rownum,t.*,sleep(id2) n10,&&1 arraysize ,&&2 rowprefetch from t
[2020-08-24 21:11:58] new 1: select rownum,t.*,sleep(id2) n10,2 arraysize ,6 rowprefetch from t
[2020-08-24 21:12:04]
[2020-08-24 21:12:04] ROWNUM ID1 ID2 N10 ARRAYSIZE ROWPREFETCH
[2020-08-24 21:12:04] ------ ---- ---- ---- ---------- -----------
[2020-08-24 21:12:04] 1 1 1 1 2 6
[2020-08-24 21:12:04] 2 2 1 1 2 6
[2020-08-24 21:12:04] 3 3 1 1 2 6
[2020-08-24 21:12:04] 4 4 1 1 2 6
[2020-08-24 21:12:04] 5 5 1 1 2 6
[2020-08-24 21:12:04] 6 6 1 1 2 6
[2020-08-24 21:12:12] 7 7 1 1 2 6
[2020-08-24 21:12:12] 8 8 1 1 2 6
[2020-08-24 21:12:12] 9 9 1 1 2 6
[2020-08-24 21:12:12] 10 10 1 1 2 6
[2020-08-24 21:12:12] 11 11 1 1 2 6
[2020-08-24 21:12:12] 12 12 1 1 2 6
[2020-08-24 21:12:12] 13 13 1 1 2 6
[2020-08-24 21:12:12] 14 14 1 1 2 6
[2020-08-24 21:12:20] 15 15 1 1 2 6
[2020-08-24 21:12:20] 16 16 1 1 2 6
[2020-08-24 21:12:20] 17 17 1 1 2 6
[2020-08-24 21:12:20] 18 18 1 1 2 6
[2020-08-24 21:12:20] 19 19 1 1 2 6
[2020-08-24 21:12:20] 20 20 1 1 2 6
[2020-08-24 21:12:20] 21 21 1 1 2 6
[2020-08-24 21:12:20] 22 22 1 1 2 6
[2020-08-24 21:12:21] 23 23 1 1 2 6
[2020-08-24 21:12:21]
[2020-08-24 21:12:21] 23 rows selected.
[2020-08-24 21:12:21]
[2020-08-24 21:12:21] Elapsed: 00:00:23.03
[2020-08-24 21:12:21]
[2020-08-24 21:12:21] Session altered.
[2020-08-24 21:12:21]
[2020-08-24 21:12:21] Elapsed: 00:00:00.00
--//时间间隔:6,8,8,1
--//显示行数:6,8,8,1.
$ grep "FETCH" test_ora_4344.trc | grep "#201597384"
FETCH #201597384:c=0,e=6007379,p=0,cr=56,cu=2,mis=0,r=6,dep=0,og=1,plh=2402761124,tim=4180072163
FETCH #201597384:c=0,e=7998394,p=0,cr=1,cu=0,mis=0,r=8,dep=0,og=1,plh=2402761124,tim=4188072679
FETCH #201597384:c=0,e=8000802,p=0,cr=1,cu=0,mis=0,r=8,dep=0,og=1,plh=2402761124,tim=4196074146
FETCH #201597384:c=0,e=999593,p=0,cr=1,cu=0,mis=0,r=1,dep=0,og=1,plh=2402761124,tim=4197074434
--//fetch 6,8,8,1.
--//可以这样理解: arraysize = 2,rowprefetch=6
--//第1次fetch 6,而arraysize=2, floor(rowprefetch/arraysize)*arraysize=floor(6/2)*2=6.显示输出6行.剩下0条.
--//第2次fetch 8,(floor(8+0)/2)*2=8,显示输出8行.剩下0条.
--//第3次fetch 8,(floor(8+0)/2)*2=8,显示输出8行.剩下0条.
--//第4次fetch 1 ,需要1秒完成fetch对应前面最后的时间间隔是1秒,全部输出1.
--//你可以使用下面的sleept函数替换脚本aa.txt里面的sleep函数.
CREATE OR REPLACE FUNCTION SCOTT.sleepT (seconds IN NUMBER)
RETURN timestamp
AS
BEGIN
sys.DBMS_LOCK.sleep (seconds);
RETURN SYSTIMESTAMP-1/86400;
END;
/
$ cat ba.txt
set timing on
set arraysize &1
set rowprefetch &2
alter session set events '10046 trace name context forever, level 12';
select rownum,t.*,sleept(id2) n10,&&1 arraysize ,&&2 rowprefetch from t;
alter session set events '10046 trace name context off';
set timing off
quit
$ sqlplus -s -l scott/btbtms@test01p @ ba.txt 7 7 | ~/bin/ts.awk
[2020-08-25 20:22:00]
[2020-08-25 20:22:00] Session altered.
[2020-08-25 20:22:00]
[2020-08-25 20:22:00] Elapsed: 00:00:00.00
[2020-08-25 20:22:00] old 1: select rownum,t.*,sleept(id2) n10,&&1 arraysize ,&&2 rowprefetch from t
[2020-08-25 20:22:00] new 1: select rownum,t.*,sleept(id2) n10,7 arraysize ,7 rowprefetch from t
[2020-08-25 20:22:07]
[2020-08-25 20:22:07] ROWNUM ID1 ID2 N10 ARRAYSIZE ROWPREFETCH
[2020-08-25 20:22:07] ------ ---- ---- ----------------------------- --------- -----------
[2020-08-25 20:22:07] 1 1 1 2020-08-25 20:22:00.000000000 7 7
[2020-08-25 20:22:07] 2 2 1 2020-08-25 20:22:01.000000000 7 7
[2020-08-25 20:22:07] 3 3 1 2020-08-25 20:22:02.000000000 7 7
[2020-08-25 20:22:07] 4 4 1 2020-08-25 20:22:03.000000000 7 7
[2020-08-25 20:22:07] 5 5 1 2020-08-25 20:22:04.000000000 7 7
[2020-08-25 20:22:07] 6 6 1 2020-08-25 20:22:05.000000000 7 7
[2020-08-25 20:22:07] 7 7 1 2020-08-25 20:22:06.000000000 7 7
[2020-08-25 20:22:21] 8 8 1 2020-08-25 20:22:07.000000000 7 7
[2020-08-25 20:22:21] 9 9 1 2020-08-25 20:22:08.000000000 7 7
[2020-08-25 20:22:21] 10 10 1 2020-08-25 20:22:09.000000000 7 7
[2020-08-25 20:22:21] 11 11 1 2020-08-25 20:22:10.000000000 7 7
[2020-08-25 20:22:21] 12 12 1 2020-08-25 20:22:11.000000000 7 7
[2020-08-25 20:22:21] 13 13 1 2020-08-25 20:22:12.000000000 7 7
[2020-08-25 20:22:21] 14 14 1 2020-08-25 20:22:13.000000000 7 7
[2020-08-25 20:22:21] 15 15 1 2020-08-25 20:22:14.000000000 7 7
[2020-08-25 20:22:21] 16 16 1 2020-08-25 20:22:15.000000000 7 7
[2020-08-25 20:22:21] 17 17 1 2020-08-25 20:22:16.000000000 7 7
[2020-08-25 20:22:21] 18 18 1 2020-08-25 20:22:17.000000000 7 7
[2020-08-25 20:22:21] 19 19 1 2020-08-25 20:22:18.000000000 7 7
[2020-08-25 20:22:21] 20 20 1 2020-08-25 20:22:19.000000000 7 7
[2020-08-25 20:22:21] 21 21 1 2020-08-25 20:22:20.000000000 7 7
[2020-08-25 20:22:23] 22 22 1 2020-08-25 20:22:21.000000000 7 7
[2020-08-25 20:22:23] 23 23 1 2020-08-25 20:22:22.000000000 7 7
[2020-08-25 20:22:23]
[2020-08-25 20:22:23] 23 rows selected.
[2020-08-25 20:22:23]
[2020-08-25 20:22:23] Elapsed: 00:00:23.00
[2020-08-25 20:22:23]
[2020-08-25 20:22:23] Session altered.
[2020-08-25 20:22:23]
[2020-08-25 20:22:23] Elapsed: 00:00:00.00
--//时间间隔:7,14,2
--//显示行数:7,14,2.
$ grep "FETCH" test_ora_4444.trc | grep 45247739
FETCH #452477392:c=15600,e=7003344,p=0,cr=30,cu=2,mis=0,r=7,dep=0,og=1,plh=2402761124,tim=2850476765
FETCH #452477392:c=0,e=13988663,p=0,cr=1,cu=0,mis=0,r=14,dep=0,og=1,plh=2402761124,tim=2864469452
FETCH #452477392:c=0,e=1999838,p=0,cr=1,cu=0,mis=0,r=2,dep=0,og=1,plh=2402761124,tim=2866471508
--//不再展开分析.
4.总结:
--//实际上这其中细节不重要,你可以理解改变12c下改变sqlplus的rowprefetch参数,有可能隐含改变了fetch方式与数量.
--//第1次fetch = rowpefetch, 第2次 fetch 等于 (floor(rowprefetch/arraysize)+1)*arraysize.
--//个人建议还是不要设置rowprefetch >= arraysize的情况.因为这样改变fetch的模式.
--//显示输出 : floor(rowprefetch/arraysize)*arraysize,floor(rowprefetch/arraysize+1)*arraysize, ...,剩下的记录.
--//注:fetch在前,输出在后.
--// 如果rowprefetch < arraysize,第1次fetch后,不足arraysize数量.不会马上输出,而是等待下一个fetch完成,再输出.
5.补充:
--//另外我还找到一个链接,结论跟我的测试一样,表述的方式不同罢了.
--//https://blog.dbi-services.com/arraysize-or-rowprefetch-in-sqlplus/的测试：
We can see 3 things here:
- The first FETCH (from the internal OCI execute) contains always the number of rows as defined in the ROWPREFETCH
setting
- The second FETCH (and all subsequent fetches) contains a multiple of the ARRAYSIZE setting rows. The following code
fragment should show the logic:
2nd_Fetch_Rows = if ROWPREFETCH <ARRAYSIZE
then ARRAYSIZE
else (TRUNC(ROWPREFETCH/ARRAYSIZE)+1)*ARRAYSIZE
- If a fetch does not detect the end of the data in the cursor then an additional fetch is necessary. In 3 cases above a
last fetch fetched 0 rows.
2020-08-25
[20200824]12c sqlplus rowprefetch arraysize 显示行数量的关系.txt
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