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Reformating unix data

 
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# 1  
Old 07-08-2011
Reformating unix date
Hi i have a unix date in file a file like this '1313675999'
in oracle i would do it like this

Code:
select TO_CHAR (  TO_DATE ('01011970', 'DDMMYYYY')+ 1 / 24 / 60 / 60 * 1313675999,'YYYYMMDD') from dual

how to achive the same in unix ?
Last edited by phpsnook; 07-08-2011 at 06:54 AM..
# 2  
Old 07-08-2011
if your's is gnu date then try as below

Code:
 
date -d @1313675999

# 3  
Old 07-08-2011
No luck with that Smilie
# 4  
Old 07-08-2011
refer here

https://www.unix.com/shell-programmin...timestamp.html
# 5  
Old 07-08-2011
can someone tell me in what format this timestamp is '1313675999'
# 6  
Old 07-08-2011
It's POSIX or UNIX time
Unix time - Wikipedia, the free encyclopedia

Regards
Peasant.
# 7  
Old 07-08-2011
Code:
perl -e 'print scalar(localtime(1313675999)),"\n"'
Thu Aug 18 13:59:59 2011

So, epoch time, the most common way to store time on a Unix system it is a measure of elapsed seconds since midnight on Jan 1 1970
 
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