Parse Date in Bash

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

Solution 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.

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