AWR TOP SQL實現方法是什么

本篇內容介紹了“AWR TOP SQL實現方法是什么”的有關知識，在實際案例的操作過程中，不少人都會遇到這樣的困境，接下來就讓小編帶領大家學習一下如何處理這些情況吧！希望大家仔細閱讀，能夠學有所成！

1 按解析次數排序

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 按執行時間排序

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時間排序

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 按邏輯讀(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 按物理讀(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 按執行次數排序

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實現方法是什么”的內容就介紹到這里了，感謝大家的閱讀。如果想了解更多行業相關的知識可以關注億速云網站，小編將為大家輸出更多高質量的實用文章！