本篇内容介绍了“AWR TOP sql实现方法是什么”的有关知识,在实际案例的操作过程中,不少人都会遇到这样的困境,接下来就让小编带领大家学习一下如何处理这些情况吧!希望大家仔细阅读,能够学有所成!1&nb
本篇内容介绍了“AWR TOP sql实现方法是什么”的有关知识,在实际案例的操作过程中,不少人都会遇到这样的困境,接下来就让小编带领大家学习一下如何处理这些情况吧!希望大家仔细阅读,能够学有所成!
1 按解析次数排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
sum(parse_calls_delta) parse,
sum(executions_delta) exec_nums,
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a.parse desc;
2 按执行时间排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
sum(executions_delta) execs,
round(sum(elapsed_time_delta)/1000/1000/sum(executions_delta),2) elapsed_time_per,
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."elapsed_time(s)" desc;
3 按CPU时间排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(cpu_time_delta)/1000/1000,2) "cpu_time",
sum(executions_delta) execs,
round(sum(cpu_time_delta)/1000/1000/sum(executions_delta),2) cpu_time_per,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."cpu_time" desc;
4 按User I/O wait排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(iowait_delta)/1000/1000,2) "iowait_time(s)",
sum(executions_delta) execs,
round(sum(iowait_delta)/1000/1000/sum(executions_delta),2) iowait_time_per,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."iowait_time(s)" desc;
5 按逻辑读(gets)排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(buffer_gets_delta),2) "buffer_ges",
sum(executions_delta) execs,
round(sum(buffer_gets_delta)/sum(executions_delta),2) iowait_time_per,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."buffer_ges" desc;
7 按物理读(physical read)排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(DISK_READS_DELTA),2) "physical_read",
sum(executions_delta) execs,
round(sum(DISK_READS_DELTA)/sum(executions_delta),2) iowait_time_per,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."physical_read" desc;
8 按执行次数排序
select a.*,
to_char(substr(b.sql_text,1,4000))
from
(select dhs.sql_id,
round(sum(executions_delta),2) "exec_num",
sum(ROWS_PROCESSED_DELTA) row_process,
round(sum(ROWS_PROCESSED_DELTA)/sum(executions_delta),2) rows_per_exec,
round(sum(elapsed_time_delta)/1000/1000,2) "elapsed_time(s)",
dhs.MODULE
from dba_hist_sqlstat dhs
where
snap_id > 22438
and snap_id <= 22440
group by dhs.sql_id,MODULE) a,
dba_hist_sqltext b
where a.sql_id=b.sql_id order by a."exec_num" desc;
“AWR TOP SQL实现方法是什么”的内容就介绍到这里了,感谢大家的阅读。如果想了解更多行业相关的知识可以关注编程网网站,小编将为大家输出更多高质量的实用文章!
--结束END--
本文标题: AWR TOP SQL实现方法是什么
本文链接: https://lsjlt.com/news/63829.html(转载时请注明来源链接)
有问题或投稿请发送至: 邮箱/279061341@qq.com QQ/279061341
2024-10-23
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
2024-10-22
回答
回答
回答
回答
回答
回答
回答
回答
回答
回答
0