Shell Programming and Scripting

Hello Forum,

I want to retrieve the old date from "date" command.
For example: It's 02 Nov today and i want 7 days old date say 25 oct
then how can i do that???
As i do
date | awk '{print $3-7}'
It returns -5 (As 2-7)
So can anyone help me to solve this out..... I think by making function, it would be easier but i want to do this by making One liner command as i mentioned.....
Any kind of help and suggession would be appreciated.....

Many Thanks
Lokesh

If you have GNU date, then

Code:

date --date "7 days ago"

I think the Perderabo's datecalc can also handles these requests.

Hi,

I couldnot think of a one-liner command as u said.
Please find below a script, which calculates previous date..i.e current date-1

Can you please check, if modifying this script a bit will work in your case

Script below
---------------
day=`date +%d%`
month=`date +%m%`
year=`date +%Y`
daynm=`date +%a`

case "$daynm" in
"Mon") day=`expr "$day" - 3`
k=3;;
"Sun") day=`expr "$day" - 2`
k=2;;
*) day=`expr "$day" - 1`
k=1;;
esac
if test $day -le 0
then
month=`expr "$month" - 1`
case "$month" in
0)
month=12
year=`expr "$year" - 1`
;;
esac
m=`expr $day - 1`
day=`cal $month $year | grep . | fmt -1 | tail $m | head -1`
fi

if [[ "${day}" = [0-9] ]]
then
day="0"$day
fi

if [[ "${month}" = [0-9] ]]
then
month="0"$month
fi
Final_Date=$year$month$day
echo $Final_Date


Thanks ,
Suresh Sampat

You can print the date if you're looking for "hours earlier/later than right now" by manipulating the TZ variable:

Code:

# date
Mon Nov  5 09:47:32 CST 2007

# TZ=$(((6+24*7))) date  
Mon Oct 29 09:47:37  2007

It's a bit cheap but it works. Note that I needed to start with 6 for my local timezone, and then add hours to go backward (7 days of 24 hours each).