You can find this is a cheeky trick, but it works as a treat :)
select
printf('%s-%02.0f-%s',
substr(started_dt,1,2),
(2+instr('JanFebMarAprMayJunJulAugSepOctNovDec',
substr(started_dt,4,3)))/3,
substr(started_dt,8,4)
)
On 13 June 2014 10:01, Krishnan Narayanan <krishnan.sm...@gmail.com> wrote:
> Hi All,
>
> I have my date format as 08-Mar-2014 how to I change it to 08-03-2014?
> Can I use regexp_replace.
>
> I tried below but not getting the desired output.
>
>
> regexp_replace(started_dt,"\Jan|\Feb|\Mar|\Apr|\May|\Jun|\Jul|\Aug|\Sep|\Oct|\Nov|\Dec","\01|\02|\03|\04|\05|\06|\07|\08|\09|\10|\11|\12")
>
> Output: 09-1|2|3|4|5|6|7|8|9|10|11|12-2014 16:19:56.
> Help is much appreciated.
>
>
> Thanks
> Krishnan
>
>
--
André Araújo
Big Data Consultant/Solutions Architect
The Pythian Group - Australia - www.pythian.com
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