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网站，小编将为大家输出更多高质量的实用文章！
