oracle 中如何定位重要(消耗资源多)的SQL【转】

select substr(to_char(s.pct,‘99.00‘),2)||‘%‘load,
       s.executions executes,
       p.sql_text
from(select address,
               disk_reads,
               executions,
               pct,
               rank()over(order by disk_reads desc) ranking
         from(select address,
                       disk_reads,
                       executions,
                      100*ratio_to_report(disk_reads)over() pct
                 from sys.v_$sql
                where command_type!=47)
        where disk_reads>50*executions) s,
       sys.v_$sqltext p
where s.ranking<=5
  and p.address=s.address
order by 1, s.address, p.piece;

select b.username,
       a. buffer_gets,
       a.executions,
       a.disk_reads / decode(a.executions, 0, 1, a.executions),
       a.sql_text SQL
  from v$sqlarea a, dba_users b
 where a.parsing_user_id = b.user_id
   and a.disk_reads > 10000
 order by disk_reads desc;

select*
from(select buffer_gets, sql_text
         from v$sqlarea
        where buffer_gets>500000
        order by buffer_gets desc)
where rownum<=30;

select sql_text, executions
  from (select sql_text, executions from v$sqlarea order by executions desc)
 where rownum < 81;

select sql_text, disk_reads
from(select sql_text, disk_reads from v$sqlarea order by disk_reads desc)
where rownum<21;

select sql_text, sorts
from(select sql_text, sorts from v$sqlarea order by sorts desc)
where rownum<21;

select substr(sql_text, 1, 80) "sql", count(*), sum(executions) "totexecs"
  from v$sqlarea
 where executions < 5
 group by substr(sql_text, 1, 80)
having count(*) > 30
 order by 2;