您的位置:首页 > 数据库 > Oracle

oracle按时间段查询

2015-10-10 13:35 441 查看
with a as (

select

to_date('20150423','yyyymmdd')+1-(3/24)*(level-1) dt

from

dual connect by level <= 8

)

select

nvl(count,0) count,a.dt

from a

left join (

select

count(*) as count , a.dt

from

alarm_record b ,

a

where

b.alarm_time <= a.dt

and b.alarm_time >= a.dt-(3/24)

group by dt

) c

on c.dt = a.dt

order by a.dt

效果如下:

内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: