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

Pure Bash:

date="2009-12-03 15:35:11"
saveIFS="$IFS"
IFS="- :"
date=($date)
IFS="$saveIFS"
for field in "${date[@]}"
do
    echo $field
done

2009
12
03
15
35
11