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 pure bash

$ d="2009-12-03 15:35:11"
$ d=${d//[- :]/|}
$ IFS="|"
$ set -- $d
$ echo $1
2009
$ echo $2
12
$ echo $@
2009 12 03 15 35 11