Bash – How to convert date to timestamp

bashdateshell

I saw this date command:

date -d @1332468005 '+%Y-%m-%d %H:%M:%S'

Resulting in :

2012-03-22 22:00:05

How can I convert it back from 2012-03-22 22:00:05 to 1332468005 with bash shell?

man date will give you the details of how to use the date command.

To convert long date in standard format into Unix epoc time (%s): date -d '2012-03-22 22:00:05 EDT' +%s

You could do something like:

awk -F- 'BEGIN{split("Jan,Feb,Mar,Apr,May,Jun,Jul,Aug,Sep,Oct,Nov,Dec",m,",")}
         {print m[$2], substr($0,9)}'

Or:

awk -F- '{print substr("JanFebMarAprMayJunJulAugSepOctNovDec", \
         $2*3-2,3), substr($0,9)}'

Or:

sed 's/$/:01Jan02Feb03Mar04Apr05May06Jun07Jul08Aug09Sep10Oct11Nov12Dec/
     s/.....\(..\).\(.*\):.*\1\(...\).*/\3 \2/'

Try

date @`date +%s`

I don't think it's got anything to do with bash.

Busybox's date command is a lightweight version of the more classic GNU/FSF date