[20200824]12c sqlplus rowprefetch arraysize 显示行数量的关系.txt

来源：这里教程网时间：2026-03-03 16:08:45 作者：

[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 => http://blog.itpub.net/267265/viewspace-2219260/ [20181109]12c sqlplus rowprefetch参数5.txt => http://blog.itpub.net/267265/viewspace-2219334/ --//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.

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[20200824]12c sqlplus rowprefetch arraysize 显示行数量的关系.txt

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