[20190322]测试相同语句遇到导致cursor pin s的疑问.txt
–//昨天测试遇到的情况,链接:http://blog.itpub.net/267265/viewspace-2638857/
–//我一直认为打散sql语句,避开cursor: pin s等待事件,能够提高执行效率.而测试结果有点出乎意料.
–//反而是测试2快于测试1,很难理解为什么会出现这样的情况,今天继续探究看看.
1.环境:
scott@book> @ 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
2.建立测试脚本:
create table job_times (sid number, time_ela number,method varchar2(20));
$ cat m1.txt
set verify off
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
select /*+ &&3 */ 1 into v_id from dual ;
–select /*+ &&3 */ sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
$ cat m2.txt
set verify off
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
select 1 into v_id from dual ;
–//select sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
–//通过加入注解&&3,产生每个会话执行语句不同,对比看看.昨天的测试第2种情况快于第1种情况,不理解,我再次重复测试:
3.测试:
–//测试最好避开整点的awr生成以及其它对数据库的操作,
seq 1 40 | xargs -i {} echo ‘seq {} | xargs -i %# -p {} bash -c “sqlplus -s -l scott/book @m1.txt 1e6 m1_{} %# >/dev/null”‘ | bash
seq 1 40 | xargs -i {} echo ‘seq {} | xargs -i %# -p {} bash -c “sqlplus -s -l scott/book @m2.txt 1e6 m2_{} %# >/dev/null”‘ | bash
$ sqlplus -s -l scott/book <<< “select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times where method like ‘m%’ group by method order by to_number(substr(method,4)),4;” | egrep “^|m1_.*$”
method count(*) round(avg(time_ela),0) sum(time_ela)
—— ——– ———————- ————-
m2_1 1 1802 1802
m1_1 1 1841 1841
m1_2 2 1677 3354
m2_2 2 1697 3393
m2_3 3 1660 4981
m1_3 3 1691 5074
m1_4 4 1672 6688
m2_4 4 1704 6816
m1_5 5 1588 7941
m2_5 5 1717 8586
m1_6 6 1584 9505
m2_6 6 1655 9931
m1_7 7 1595 11162
m2_7 7 1628 11395
m1_8 8 1599 12790
m2_8 8 1758 14067
m1_9 9 1600 14400
m2_9 9 1659 14935
m1_10 10 1595 15953
m2_10 10 1670 16701
m1_11 11 1584 17429
m2_11 11 1696 18653
m1_12 12 1586 19029
m2_12 12 1710 20519
m1_13 13 1683 21877
m2_13 13 1834 23844
m1_14 14 1757 24596
m2_14 14 1913 26777
m1_15 15 1832 27473
m2_15 15 2061 30919
m1_16 16 1913 30600
m2_16 16 2078 33247
m1_17 17 1975 33568
m2_17 17 2130 36203
m1_18 18 2057 37023
m2_18 18 2245 40418
m1_19 19 2131 40485
m2_19 19 2332 44301
m1_20 20 2246 44913
m2_20 20 2397 47932
m1_21 21 2318 48674
m2_21 21 2537 53285
m1_22 22 2430 53456
m2_22 22 2646 58218
m1_23 23 2538 58365
m2_23 23 2735 62906
m1_24 24 2664 63927
m2_24 24 2866 68789
m1_25 25 2795 69875
m2_25 25 2998 74952
m1_26 26 2835 73716
m2_26 26 3134 81489
m1_27 27 2967 80114
m2_27 27 3239 87444
m1_28 28 3088 86477
m2_28 28 3371 94391
m1_29 29 3172 91990
m2_29 29 3536 102550
m1_30 30 3303 99083
m2_30 30 3660 109802
m1_31 31 3377 104679
m2_31 31 3979 123359
m1_32 32 3560 113920
m2_32 32 4179 133740
m1_33 33 3608 119055
m2_33 33 4335 143050
m1_34 34 3732 126880
m2_34 34 4404 149722
m1_35 35 3760 131615
m2_35 35 4277 149679
m1_36 36 3948 142134
m2_36 36 4714 169701
m1_37 37 4098 151626
m2_37 37 4889 180908
m1_38 38 4066 154523
m2_38 38 5033 191253
m1_39 39 4153 161969
m2_39 39 5149 200827
m1_40 40 4394 175767
m2_40 40 5182 207291
80 rows selected.
–//大部分情况下都是测试1快于测试2.设置grep支持彩色.比较好观察.还有一种方法在vim下:set hls,查询/^m1.*$,这样也可以.
$ sqlplus -s -l scott/book <<< “select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times where method like ‘m%’ group by method order by to_number(substr(method,4)),4;” | egrep “^|m1_.*$”
–//也就是我前面的测试有问题??? 如果执行如下:
–//注:我注解select /*+ &&3 */ 1 into v_id from dual ;改为执行select /*+ &&3 */ sysdate into v_d from dual ;
seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m1.txt 1e6 x1_150 %# >/dev/null”
seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m2.txt 1e6 x2_150 %# >/dev/null”
scott@book> select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times where method like ‘x%’ group by method order by to_number(substr(method,4)),3;
method count(*) round(avg(time_ela),0) sum(time_ela)
——————– ———- ———————- ————-
x2_150 150 19268 2890191
x1_150 150 19768 2965136
–//为什么呢?并发用户多了吗?我仔细想想,问题估计在开始阶段,测试1 150个连接开始执行大量的非绑定变量语句要硬解析,这样开始出现大量
–//latch: shared pool等待事件,而引起的这样的情况,如果开始脚本加入适当的延迟,测试才比较准确.修改脚本如下:
$ cat m1.txt
set verify off
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
host sleep $(echo &&3/50 | bc -l )
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
select /*+ &&3 */ 1 into v_id from dual ;
–select /*+ &&3 */ sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
$ cat m2.txt
set verify off
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
host sleep $(echo &&3/50| bc -l )
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
select 1 into v_id from dual ;
–select sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m1.txt 1e6 a1_150 %# >/dev/null”
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m2.txt 1e6 a2_150 %# >/dev/null”
–//实际上我在测试时还犯了一个严重错误,实际上到一定时间,看latch: shared pool的等待事件最后seconds_in_wait是不变的,
–//实际上v$session记录的最后1个等待事件,非常容易存在疑惑.
scott@book> @ wait
p1raw p2raw p3raw p1 p2 p3 sid serial# seq# event status state wait_time_micro seconds_in_wait wait_class
—————- —————- —– ———- — — — ——- —- ————————— ——– —————– ————— ————— ————
…
0000000062657100 0000000000000001 00 1650815232 1 0 333 95 28 sql*net message from client active waited known time 305362 144 idle
000000006010d860 0000000000000150 00 1611716704 336 0 6 145 30 latch: shared pool active waited known time 34009 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 21 131 30 latch: shared pool active waited known time 30997 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 67 115 31 latch: shared pool active waited known time 31987 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 132 87 30 latch: shared pool active waited short time 17 141 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 170 719 31 latch: shared pool active waited known time 33078 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 246 89 33 latch: shared pool active waited known time 28119 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 320 57 30 latch: shared pool active waited known time 35783 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 255 789 31 latch: shared pool active waited known time 28109 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 257 157 33 latch: shared pool active waited known time 34000 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 262 125 30 latch: shared pool active waited known time 30084 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 284 111 30 latch: shared pool active waited known time 26152 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 290 33 32 latch: shared pool active waited known time 31020 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 298 403 30 latch: shared pool active waited known time 35788 138 concurrency
000000006010d860 0000000000000150 00 1611716704 336 0 247 121 30 latch: shared pool active waited known time 32958 138 concurrency
150 rows selected.
scott@book> select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times where method like ‘a%’ group by method order by to_number(substr(method,4)),3;
method count(*) round(avg(time_ela),0) sum(time_ela)
——————– ———- ———————- ————-
a1_150 150 16718 2507696
a2_150 150 18234 2735046
–//这样测试1就快于测试2.
–//改成执行select sysdate into v_d from dual ;代码看看.
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m1.txt 1e6 b1_150 %# >/dev/null”
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m2.txt 1e6 b2_150 %# >/dev/null”
scott@book> select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times group by method order by method;
method count(*) round(avg(time_ela),0) sum(time_ela)
——————– ———- ———————- ————-
a1_150 150 16718 2507696
a2_150 150 18234 2735046
b1_150 150 17613 2641914
b2_150 150 21185 3177713
–//你可以可以看出sql语句分散后大概提高10%上下.
–//另外我的测试还加入了sleep,
$ seq 150 | xargs | tr ‘ ‘ + | bc -l
11325
–//11325/50 = 226.5秒,排除这个因素.平均每个扣除226.5/150 = 1.51秒.哎!!脚本写的有问题,考虑欠缺,应该sleep在前面,再次重复测试:
–//从这次测试也看出,自己的测试设计不严谨,没有考虑一些细节问题,从另外一个方面也可以看出不使用绑定变量对数据库的危害,特别是oltp系统.
–//补充:
$ cat m1.txt
set verify off
host sleep $(echo &&3/50 | bc -l )
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
–select /*+ &&3 */ 1 into v_id from dual ;
select /*+ &&3 */ sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
$ cat m2.txt
set verify off
host sleep $(echo &&3/50| bc -l )
insert into job_times values ( sys_context (‘userenv’, ‘sid’) ,dbms_utility.get_time ,’&&2′) ;
commit ;
declare
v_id number;
v_d date;
begin
for i in 1 .. &&1 loop
–select 1 into v_id from dual ;
select sysdate into v_d from dual ;
end loop;
end ;
/
update job_times set time_ela = dbms_utility.get_time – time_ela where sid=sys_context (‘userenv’, ‘sid’) and method=’&&2′;
commit;
quit
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m1.txt 1e6 c1_150 %# >/dev/null”
$ seq 150 | xargs -i %# -p 150 bash -c “sqlplus -s -l scott/book @m2.txt 1e6 c2_150 %# >/dev/null”
scott@book> select method,count(*),round(avg(time_ela),0),sum(time_ela) from job_times group by method order by method;
method count(*) round(avg(time_ela),0) sum(time_ela)
——————– ———- ———————- ————-
a1_150 150 16718 2507696
a2_150 150 18234 2735046
b1_150 150 17613 2641914
b2_150 150 21185 3177713
c1_150 150 16024 2403599
c2_150 150 21062 3159275
6 rows selected.