Reg Exp replace以匹配oracle中的hh:mm格式
Reg Exp replace以匹配oracle中的hh:mm格式
提问于 2021-09-08 16:05:44
使用regexp replace编写oracle查询，以修复空间并返回hh:mm格式。
with
test_data (srt_tm) as (
select '1:00' from dual union all
select '01:00' from dual union all
select ' 01:00' from dual union all
select '4:00' from dual union all
select '04:00' from dual union all
select ' 04:00' from dual
)预期输出：
已尝试查询：
select (strt_tm,regexp_replace(strt_tm,'^([0-1]?[0-9]|2[0-3]):[0-5][0-9]$','hh:mm') as "REPLACE"
from test_dataStack Overflow用户
发布于 2021-09-08 16:49:56
只需在小时数中添加额外的0，并删除除最后两位以外的所有数字：
select
strt_tm,
regexp_replace(
regexp_replace(
strt_tm
,'(\d+):'
,'0\1:'
)
,'.*(\d{2}:)'
,'\1'
) as "REPLACE"
from test_data;完整的test_data示例：
with
test_data (strt_tm) as (
select '1:00' from dual union all
select '01:00' from dual union all
select ' 01:00' from dual union all
select '4:00' from dual union all
select '04:00' from dual union all
select ' 04:00' from dual
)
select
strt_tm,
regexp_replace(
regexp_replace(
strt_tm
,'(\d+):'
,'0\1:'
)
,'.*(\d{2}:)'
,'\1'
) as "REPLACE"
from test_data页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/69106385
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