oracle资源检查 | 易学教程

一、查询执行最慢的sql
select *
from(select sa.SQL_TEXT,
sa.SQL_FULLTEXT,
sa.EXECUTIONS"执行次数",
round(sa.ELAPSED_TIME / 1000000, 2)“总执行时间”,
round(sa.ELAPSED_TIME / 1000000 / sa.EXECUTIONS, 2)“平均执行时间”,
sa.COMMAND_TYPE,
sa.PARSING_USER_ID"用户ID",
u.username"用户名",
sa.HASH_VALUE
from v$sqlarea sa
left join all_users u
on sa.PARSING_USER_ID = u.user_id
where sa.EXECUTIONS > 0
order by(sa.ELAPSED_TIME / sa.EXECUTIONS)desc)
where rownum <= 50;

二、查询次数最多的 sql
select *
from(select s.SQL_TEXT,
s.EXECUTIONS"执行次数",
s.PARSING_USER_ID"用户名",
rank() over(order by EXECUTIONS desc) EXEC_RANK
from v$sql s
left join all_users u
on u.USER_ID = s.PARSING_USER_ID) t
where exec_rank <= 100;

三、查看总消耗时间最多的前10条SQL语句
select *
from (select v.sql_id,
v.child_number,
v.sql_text,
last_load_time,
v.PARSING_USER_ID,
ROUND(v.ELAPSED_TIME / 1000000 / (CASE
WHEN (EXECUTIONS = 0 OR NVL(EXECUTIONS, 1 ) = 1) THEN
1
ELSE
EXECUTIONS
END),
2) “执行时间’S’”,
v.SQL_FULLTEXT,
v.cpu_time,
v.disk_reads,
rank() over(order by v.elapsed_time desc) elapsed_rank
from v$sql v ) a
where elapsed_rank <= 100 and last_load_time > to_char(sysdate - 1/24, ‘YYYY-MM-DD/HH:MI:SS’) order by “执行时间’S’” desc

四、查询最近一小时内最慢的SQL:
select *
from (select v.sql_id,
v.child_number,
v.sql_text,
v.elapsed_time,
v.cpu_time,
v.disk_reads,
rank() over(order by v.cpu_time desc) elapsed_rank
from v$sql v) a
where elapsed_rank <= 10;

五、查看CPU消耗时间最多的前10条SQL语句
select *
from (select v.sql_id,
v.child_number,
v.sql_text,
v.elapsed_time,
v.cpu_time,
v.disk_reads,
rank() over(order by v.cpu_time desc) elapsed_rank
from v$sql v) a
where elapsed_rank <= 10;

六、查看消耗磁盘读取最多的前10条SQL语句
select *
from (select v.sql_id,
v.child_number,
v.sql_text,
v.elapsed_time,
v.cpu_time,
v.disk_reads,
rank() over(order by v.disk_reads desc) elapsed_rank
from v$sql v) a
where elapsed_rank <= 10;

来源：CSDN

作者：计划式

链接：https://blog.csdn.net/qq_28429443/article/details/103984549

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