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

The array method is perhaps better, but this is what you were specifically asking for:

IFS=" :-"
read year month day hour minute second < <(echo "YYYY-MM-DD hh:mm:ss")

Answer 2

have you tried using cut? something like this: dayofweek=date|cut -d" " -f1

Answer 3

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

Answer 4

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