Shell – Subtract 7 days from today’s date

dateosxshell-script

Here is my script. I need to subtract 7 days from today's date and use it in a file name. I am using a Mac.

#/bin/bash
DATE=$(date -d "-7 days")
echo $DATE

When I run this .sh script, I get this:

$ /Users/xxxxxxx/xxxxxxxx/dateTest.sh 
usage: date [-jnRu] [-d dst] [-r seconds] [-t west] [-v[+|-]val[ymwdHMS]] ... 
            [-f fmt date | [[[mm]dd]HH]MM[[cc]yy][.ss]] [+format]

Best Answer

With BSD date you need a different syntax:

DATE="$(date -v-7d)"

On my FreeBSD man date includes:

 -v      Adjust (i.e., take the current date and display the result of the  
         adjustment; not actually set the date) the second, minute, hour,  
         month day, week day, month or year according to val.  If val is  
         preceded with a plus or minus sign, the date is adjusted forwards  
         or backwards according to the remaining string, otherwise the  
         relevant part of the date is set.

Related Solutions

How to convert a local date-time into UTC date-time

To pass a date to use into date, use the -d option. So your command would look something like echo "20150903154607" | xargs date +"%Y%m%d%H%M%S" -u -d.

It doesn't take exactly the date format you're supplying, though:

$ date -d 20150903154607
date: invalid date ‘20150903154607’
$ date -d 20150903\ 15:46:07
Thu Sep  3 15:46:07 BST 2015

So massage it a little first (GNU sed):

$ echo "20150903154607" \
   | sed -re 's/^([0-9]{8})([0-9]{2})([0-9]{2})([0-9]{2})$/\1\\ \2:\3:\4/' \
   | xargs date -u -d
Thu Sep  3 15:46:07 UTC 2015

To convert from local to UTC, you need to do a bit more, because -u affects the interpretation of both input and output dates:

$ echo "20150903171734" \
   | sed -re 's/^([0-9]{8})([0-9]{2})([0-9]{2})([0-9]{2})$/\1\\ \2:\3:\4/' \
   | xargs date +@%s -d \
   | xargs date -u +%Y%m%d%H%M%S -d
20150903161734

The first invocation of date above converts local time (according to $TZ) into seconds since the epoch; the second then converts epoch-seconds into UTC date and time.

Bash – determine if date is beyond 90 days in bash

You can use GNU date to convert a date-time string into a number of seconds (since "the epoch", 1st January 1970). From there it's a simple arithmetic comparison

datetime='2016-08-31T15:38:18Z'
timeago='90 days ago'

dtSec=$(date --date "$datetime" +'%s')    # For "now", use $(date +'%s')
taSec=$(date --date "$timeago" +'%s')

echo "INFO: dtSec=$dtSec, taSec=$taSec" >&2

[ $dtSec -lt $taSec ] && echo too old

Best Answer

Related Solutions

How to convert a local date-time into UTC date-time

Bash – determine if date is beyond 90 days in bash

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