AWRTOPSQL实现方法是什么-创新互联

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！

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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！

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