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！
为石峰等地区用户提供了全套网页设计制作服务，及石峰网站建设行业解决方案。主营业务为成都网站设计、
网站制作、石峰网站设计，以传统方式定制建设网站，并提供域名空间备案等一条龙服务，秉承以专业、用心的态度为用户提供真诚的服务。我们深信只要达到每一位用户的要求，就会得到认可，从而选择与我们长期合作。这样，我们也可以走得更远！
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创新互联-成都网站建设公司网站，小编将为大家输出更多高质量的实用文章！
文章题目：AWRTOPSQL实现方法是什么-创新互联
分享URL：
http://cdkjz.cn/article/pgisg.html
