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

if you have GNU date and i understood you correctly

$ date +%Y:%m:%d -d "yesterday"
2009:11:09

or

$ date +%Y:%m:%d -d "1 day ago"
2009:11:09

Answer 2

date --date='-1 day'

Answer 3

You can use the date command:

date --date 'yesterday' +%F

Answer 4

#!/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 5

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

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

Answer 6

Advanced Bash-scripting Guide

date +%Y:%m:%d -d "yesterday"

For details about the date format see the man page for date

date --date='-1 day'

Answer 7

date --date='-1 day'

Answer 8

date +%Y:%m:%d|awk -vFS=":" -vOFS=":" '{$3=$3-1;print}'
2009:11:9

Answer 9

Use Perl instead perhaps?

perl -e 'print scalar localtime( time - 86400 ) . "\n";'

Or, use nawk and (ab)use /usr/bin/adb:

nawk 'BEGIN{printf "0t%d=Y\n", srand()-86400}' | adb

Came across this too ... insane!

/usr/bin/truss /usr/bin/date 2>&1 | nawk -F= '/^time\(\)/ {gsub(/ /,"",$2);printf "0t%d=Y\n", $2-86400}' | adb

Answer 10

Sorry not mentioning I on Solaris system. As such, the -date switch is not available on Solaris bash.

I find out I can get the previous date with little trick on timezone.

DATE=`TZ=MYT+16 date +%Y-%m-%d_%r`
echo $DATE