Get the date (a day before current time) in Bash

Question

How can I print the date which is a day before current time in Bash?

Answer 1

yesterday=`date -d "-1 day" %F`

Puts yesterday's date in YYYY-MM-DD format into variable $yesterday.

Answer 2

#!/bin/bash
OFFSET=1;
eval `date "+day=%d; month=%m; year=%Y"`
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=`expr $day - $OFFSET`
if [ $day -le 0 ] ;then
month=`expr $month - 1`
if [ $month -eq 0 ] ;then
year=`expr $year - 1`
month=12
fi
set `cal $month $year`
xday=${$#}
day=`expr $xday + $day`
fi
echo $year-$month-$day

Answer 3

date --date='-1 day'

Answer 4

Not very sexy but might do the job:

perl -e 'my ($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = localtime(time - 86400);$year += 1900; $mon+= 1; printf ("YESTERDAY: %04d%02d%02d \n", $year, $mon, $mday)'

Formated from "martin clayton" answer.

Answer 5

Well this is a late answer,but this seems to work!!

     YESTERDAY=`TZ=GMT+24 date +%d-%m-%Y`;
     echo $YESTERDAY;