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-DD hh:mm:ss

Answer 1

Does it have to be bash? You can use the GNU coreutils /bin/date binary for many transformations:

 $ date --date="2009-01-02 03:04:05" "+%d %B of %Y at %H:%M and %S seconds"  02 January of 2009 at 03:04 and 05 seconds 

This parses the given date and displays it in the chosen format. You can adapt that at will to your needs.

Answer 2

This is simple, just convert your dashes and colons to a space (no need to change IFS) and use 'read' all on one line:

read Y M D h m s <<< ${date//[-:]/ } 

For example:

$ date=$(date +'%Y-%m-%d %H:%M:%S') $ read Y M D h m s <<< ${date//[-: ]/ } $ echo "Y=$Y, m=$m" Y=2009, m=57