Timestamp class is there to do this. On Oct 27, 2014 12:26 AM, "Vidya Sujeet" <sjayatheer...@gmail.com> wrote:
> Hi, > > The date time format coming from the source is* "25-FEB-01' .* I want to > convert it to the following format. *'MM/DD/YYYY' *. How can we do this > in Hive? > > > I see that as per the documentation > > > https://cwiki.apache.org/confluence/display/Hive/LanguageManual+UDF#LanguageManualUDF-DateFunctions > > I could possibly convert the string date to a Unix time stamp in seconds > using the below UDF. However, what is the string pattern if the date at > source is coming this way* "25-FEB-01'* ? The link provided to look up > for the patterns does not work. > > Please help. > > *Name:* unix_timestamp (string date, string pattern) function. > *Description: *Convert time string with given pattern (see [ > http://java.sun.com/j2se/1.4.2/docs/api/java/text/SimpleDateFormat.html]) > to Unix time stamp (in seconds), return 0 if fail: > unix_timestamp('2009-03-20', 'yyyy-MM-dd') = 1237532400. > > > > Vidya > > > >
