Parse Date in Bash

Question

How would you parse a date in bash, with separate fields (years, months, days, hours, minutes, seconds) into different variables?

The date format is: YYYY-MM-D

Answer 1

Another solution to the OP's problem:

IFS=' -:' read y m d h m s<<<'2014-03-26 16:36:41'

Converting a date to another format with BSD date and GNU date:

$ LC_ALL=C date -jf '%a %b %e %H:%M:%S %Z %Y' 'Wed Mar 26 16:36:41 EET 2014' +%F\ %T
2014-03-26 16:36:41
$ gdate -d 'Wed Mar 26 16:36:41 EET 2014' +%F\ %T
2014-03-26 16:36:41

GNU date recognizes Wed and Mar even in non-English locales but BSD date doesn't.

Converting seconds since epoch to a date and time with GNU date and BSD date:

$ gdate -d @1234567890 '+%F %T'
2009-02-14 01:31:30
$ date -r 1234567890 '+%F %T'
2009-02-14 01:31:30

Converting seconds to hours, minutes, and seconds with a POSIX shell, POSIX awk, GNU date, and BSD date:

$ s=12345;printf '%02d:%02d:%02d\n' $((s/3600)) $((s%3600/60)) $((s%60))
05:25:45
$ echo 12345|awk '{printf "%02d:%02d:%02d\n",$0/3600,$0%3600/60,$0%60}'
05:25:45
$ gdate -d @12345 +%T
05:25:45
$ date -r 12345 +%T
05:25:45

Converting seconds to days, hours, minutes, and seconds:

$ t=12345678
$ printf '%d:%02d:%02d:%02d\n' $((t/86400)) $((t/3600%24)) $((t/60%60)) $((t%60))
142:21:21:18