1 2017/7/17 0
2 2017/7/18 0
3 2017/7/19 0
4 2017/7/20 0
5 2017/7/21 0
6 2017/7/22 1
7 2017/7/23 1
8 2017/7/24 0
9 2017/7/25 0
select t1.date_d,max(t2.date_d)
from table_a t1 left join table_a t2 on (t1.date_d >= t2.date_d and t2.is_holiday=‘0‘)
group by t1.date_d
order by t1.date_d
结果:
DATE_D MAX(T2.DATE_D)
1 2017/7/17 2017/7/17
2 2017/7/18 2017/7/18
3 2017/7/19 2017/7/19
4 2017/7/20 2017/7/20
5 2017/7/21 2017/7/21
6 2017/7/22 2017/7/21
7 2017/7/23 2017/7/21
Oracle取当前日期的最近工作日
标签:order by tab 当前日期 acl 需求 描述 rom ble order