Parse Date in Bash

Question

I need to parse a date (format: YYYY-MM-DD hh:mm:ss) in bash. I'd like to end up with all fields (years, months, days, hours, minutes, seconds) in different variables.

Answer 1

Does it have to be bash? You can use the /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

$ t='2009-12-03 12:38:15'
$ a=(`echo $t | sed -e 's/[:-]/ /g'`)
$ echo ${a[*]}
2009 12 03 12 38 15
$ echo ${a[3]}
12

Answer 3

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

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

Answer 5

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

Answer 6

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 7

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 8

instead of using the shell scripting,incorporate in your scripting itself like below wheever you need:

a=date +%Y 
b=date +%S
c=date +%H

a will be year b will be seconds c will be hours. and so on.