Hi Stephen, Thanks for your reply.
The problem is that my input date is this : in_co_an_mois (format : YYYYMM, integer), for example, this month, we have 201307 and i have to deal with this date : add one month, compare to over date, etc... The problem is that apparently, there is no way to do this, because Hive can't deal with this type of data because it's not a date format. For hive, this is just a number. Hive can deal with this : 1970-01-01 00:00:00, or this : 2009-03-20, but not with this unusual format : 201307. Thanks. 2013/7/2 Stephen Sprague <sprag...@gmail.com> > not sure i fully understand your dilemma. have you investigated any of > the date functions listed here? > > > https://cwiki.apache.org/confluence/display/Hive/LanguageManual+UDF#LanguageManualUDF-DateFunctions > > seems to me you could pull the year and month from a date. or if you have > an int then do some arithmetic to get the year and month. eg. year = > floor( <your int>/10000) and month = cast( <your int> % 100 as int) [% == > modulus operator] > > or am i not even answering your question? > > > > On Tue, Jul 2, 2013 at 2:42 AM, Jérôme Verdier <verdier.jerom...@gmail.com > > wrote: > >> Hi, >> >> i trying to translate some PL/SQL script in HiveQL, and dealing with >> unusual date format. >> >> i added a variable in my hive script : '${hiveconf:in_co_an_mois}' which >> is a year/month date format, like this : 201307 (INT format). >> >> I would like to transform this in date format, because i have to >> increment this (add one month/one year). >> >> Is there a way to do this in hive ? >> >> Thanks. >> >> >> >> -- >> *Jérôme* >> >> >
