Hi Krishnan, Try writing your own UDF for it, it should be simple.
For the internals of the UDF, I'd recommend Joda Time library: http://www.joda.org/joda-time/ For a good resource on how to write UDFs, here's something: http://blog.matthewrathbone.com/2013/08/10/guide-to-writing-hive-udfs.html For more complex GenericUDFs (which you won't need here, probably), here's something I wrote: http://blog.spryinc.com/2013/10/writing-hive-genericudfs.html#more Internal to the UDF, your input format should be 'dd-MON-yyyy' and output format as 'dd-MM-yyyy'. Good luck! Best, Nishant Kelkar On Thu, Jun 12, 2014 at 8:33 PM, Andre Araujo <ara...@pythian.com> wrote: > 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 > > Office (calls from within Australia): 1300 366 021 x1270 > Office (international): +61 2 8016 7000 x270 *OR* +1 613 565 8696 x1270 > Mobile: +61 410 323 559 > Fax: +61 2 9805 0544 > IM: pythianaraujo @ AIM/MSN/Y! or ara...@pythian.com @ GTalk > > “Success is not about standing at the top, it's the steps you leave behind.” > — Iker Pou (rock climber) > > -- > > > >
