截取字符串中前8个数字

先替换非数字的为空，再截取

with tmp_a as (
  select ‘2018-10-15 12:30:59‘ update_time from dual union all --结果是 20181015
  select ‘2017/10/16 12:30:59‘ update_time from dual union all --结果是 20171016
  select ‘20151015 12:30:59‘   update_time from dual union all --结果是 20151015
  select null                  update_time from dual union all --结果是 空
  select ‘201510‘              update_time from dual union all --结果是 201510
  select ‘20131015123059‘      update_time from dual           --结果是 20131015
)
select substr(regexp_replace(update_time,‘[^0-9]‘),1,8) 截取前8个数字 from tmp_a

截取字符串中前8个数字

原文：https://www.cnblogs.com/jiangqingfeng/p/10254952.html