I've got a shell script which does the following to store the current day's date in a variable 'dt':
date "+%a %d/%m/%Y" | read dt
echo ${dt}
How would i go about getting yesterdays date into a variable?
Basically what i'm trying to achieve is to use grep to pull all of yesterday's lines from a log file, since each line in the log contains the date in "Mon 01/02/2010" format.
Thanks a lot
dt=$(date --date yesterday "+%a %d/%m/%Y")
echo $dt
On Linux, you can use
date -d "-1 days" +"%a %d/%m/%Y"
You can use GNU date command as shown below
Getting Date In the Past
To get yesterday and earlier day in the past use string day ago:
date --date='yesterday'
date --date='1 day ago'
date --date='10 day ago'
date --date='10 week ago'
date --date='10 month ago'
date --date='10 year ago'
Getting Date In the Future
To get tomorrow and day after tomorrow (tomorrow+N) use day word to get date in the future as follows:
date --date='tomorrow'
date --date='1 day'
date --date='10 day'
date --date='10 week'
date --date='10 month'
date --date='10 year'
If you have Perl available (and your date doesn't have nice features like yesterday), you can use:
pax> date
Thu Aug 18 19:29:49 XYZ 2010
pax> dt=$(perl -e 'use POSIX;print strftime "%d/%m/%Y%",localtime time-86400;')
pax> echo $dt
17/08/2010
If you are on a Mac or BSD or something else without the --date option, you can use:
date -r `expr \`date +%s\` - 86400` '+%a %d/%m/%Y'
Update: or perhaps...
date -r $((`date +%s` - 86400)) '+%a %d/%m/%Y'
I have shell script in Linux and following code worked for me:
#!/bin/bash
yesterday=`TZ=EST+24 date +%Y%m%d` # Yesterday is a variable
mkdir $yesterday # creates a directory with YYYYMMDD format
Try the following method:
dt=`case "$OSTYPE" in darwin*) date -v-1d "+%s"; ;; *) date -d "1 days ago" "+%s"; esac`
echo $dt
It works on both Linux and OSX.
You have atleast 2 options
Use perl:
perl -e '@T=localtime(time-86400);printf("%02d/%02d/%02d",$T[4]+1,$T[3],$T[5]+1900)'
Install GNU date (it's in the sh_utils package if I remember correctly)
date --date yesterday "+%a %d/%m/%Y" | read dt
echo ${dt}
Not sure if this works, but you might be able to use a negative timezone. If you use a timezone that's 24 hours before your current timezone than you can simply use date.
Here is a ksh script to calculate the previous date of the first argument, tested on Solaris 10.
#!/bin/ksh
sep=""
today=$(date '+%Y%m%d')
today=${1:-today}
ty=`echo $today|cut -b1-4` # today year
tm=`echo $today|cut -b5-6` # today month
td=`echo $today|cut -b7-8` # today day
yy=0 # yesterday year
ym=0 # yesterday month
yd=0 # yesterday day
if [ td -gt 1 ];
then
# today is not first of month
let yy=ty # same year
let ym=tm # same month
let yd=td-1 # previous day
else
# today is first of month
if [ tm -gt 1 ];
then
# today is not first of year
let yy=ty # same year
let ym=tm-1 # previous month
if [ ym -eq 1 -o ym -eq 3 -o ym -eq 5 -o ym -eq 7 -o ym -eq 8 -o ym - eq 10 -o ym -eq 12 ];
then
let yd=31
fi
if [ ym -eq 4 -o ym -eq 6 -o ym -eq 9 -o ym -eq 11 ];
then
let yd=30
fi
if [ ym -eq 2 ];
then
# shit... :)
if [ ty%4 -eq 0 ];
then
if [ ty%100 -eq 0 ];
then
if [ ty%400 -eq 0 ];
then
#echo divisible by 4, by 100, by 400
leap=1
else
#echo divisible by 4, by 100, not by 400
leap=0
fi
else
#echo divisible by 4, not by 100
leap=1
fi
else
#echo not divisible by 4
leap=0 # not divisible by four
fi
let yd=28+leap
fi
else
# today is first of year
# yesterday was 31-12-yy
let yy=ty-1 # previous year
let ym=12
let yd=31
fi
fi
printf "%4d${sep}%02d${sep}%02d\n" $yy $ym $yd
bin$ for date in 20110902 20110901 20110812 20110801 20110301 20100301 20080301 21000301 20000301 20000101 ; do yesterday $date; done
20110901
20110831
20110811
20110731
20110228
20100228
20080229
21000228
20000229
19991231
ksh93:
dt=${ printf "%(%a %d/%m/%Y)T" yesterday; }
or:
dt=$(printf "%(%a %d/%m/%Y)T" yesterday)
The first one runs in the same process, the second one in a subshell.
For Hp-UX only below command worked for me:
TZ=aaa24 date +%Y%m%d
you can use it as :
ydate=`TZ=aaa24 date +%Y%m%d`
echo $ydate
If you have access to python, this is a helper that will get the yyyy-mm-dd date value for any arbitrary n days ago:
function get_n_days_ago {
local days=$1
python -c "import datetime; print (datetime.date.today() - datetime.timedelta(${days})).isoformat()"
}
# today is 2014-08-24
$ get_n_days_ago 1
2014-08-23
$ get_n_days_ago 2
2014-08-22
$var=$TZ;
TZ=$TZ+24;
date;
TZ=$var;
Will get you yesterday in AIX and set back the TZ variable back to original
Thanks for the help everyone, but since i'm on HP-UX (after all: the more you pay, the less features you get...) i've had to resort to perl:
perl -e '@T=localtime(time-86400);printf("%02d/%02d/%04d",$T[3],$T[4]+1,$T[5]+1900)' | read dt
If your HP-UX installation has Tcl installed, you might find it's date arithmetic very readable (unfortunately the Tcl shell does not have a nice "-e" option like perl):
dt=$(echo 'puts [clock format [clock scan yesterday] -format "%a %d/%m/%Y"]' | tclsh)
echo "yesterday was $dt"
This will handle all the daylight savings bother.
If you don't have a version of date that supports --yesterday and you don't want to use perl, you can use this handy ksh script of mine. By default, it returns yesterday's date, but you can feed it a number and it tells you the date that many days in the past. It starts to slow down a bit if you're looking far in the past. 100,000 days ago it was 1/30/1738, though my system took 28 seconds to figure that out.
#! /bin/ksh -p
t=`date +%j`
ago=$1
ago=${ago:=1} # in days
y=`date +%Y`
function build_year {
set -A j X $( for m in 01 02 03 04 05 06 07 08 09 10 11 12
{
cal $m $y | sed -e '1,2d' -e 's/^/ /' -e "s/ \([0-9]\)/ $m\/\1/g"
} )
yeardays=$(( ${#j[*]} - 1 ))
}
build_year
until [ $ago -lt $t ]
do
(( y=y-1 ))
build_year
(( ago = ago - t ))
t=$yeardays
done
print ${j[$(( t - ago ))]}/$y
Though all good answers, unfortunately none of them worked for me. So I had to write something old school. ( I was on a bare minimal Linux OS )
$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) )
You can combine this with date's usual formatting. Eg.
$ date -d @$( echo $(( $(date +%s)-$((60*60*24)) )) ) +%Y-%m-%d
Explanation : Take date input in terms of epoc seconds ( the -d option ), from which you would have subtracted one day equivalent seconds. This will give the date precisely one day back.
