Get Yesterday's date in solaris - linux

Question

I am running SunOS.
bash-3.00$ uname -a
SunOS lvsaishdc3in0001 5.10 Generic_142901-02 i86pc i386 i86pc
I need to find Yesterday's date in linux with the proper formatting passed from command prompt. When I tried like this on my shell prompt-
bash-3.00$ date --date='yesterday' '+%Y%m%d'
date: illegal option -- date=yesterday
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]
I always get date illegal option, why is it so?
Is there anything wrong I am doing?
Update:-
bash-3.00$ date --version
date: illegal option -- version
usage: date [-u] mmddHHMM[[cc]yy][.SS]
date [-u] [+format]
date -a [-]sss[.fff]

Answer 1

Try this below thing. It should work
YESTERDAY=`TZ=GMT+24 date +%Y%m%d`; echo $YESTERDAY

Answer 2

Try this one out:
DATE_STAMP=`TZ=GMT+24 date +%Y%m%d`
where GMT is the time zone and you might need to alter the 24 according to the hours difference you have from GMT. Either that or you can change GMT to a time zone more comfortable to you e.g. CST

Answer 3

As larsks suggested, you can use perl:
perl -e 'use POSIX qw(strftime); print strftime "%a %b %e %H:%M:%S %Y",localtime(time()- 3600*24);'
Slightly modified from
http://blog.rootshell.be/2006/05/04/solaris-yesterday-date/
To get YYYYMMDD format use this
perl -e 'use POSIX qw(strftime); print strftime "%Y%m%d",localtime(time()- 3600*24);'
This link explains how to format date and time with strftime
http://perltraining.com.au/tips/2009-02-26.html

Answer 4

A pure bash solution
#!/bin/bash
# get and split date
today=`date +%Y%m%d`
year=${today:0:4}
month=${today:4:2}
day=${today:6:2}
# avoid octal mismatch
if (( ${day:0:1} == 0 )); then day=${day:1:1}; fi
if (( ${month:0:1} == 0 )); then month=${month:1:1}; fi
# calc
day=$((day-1))
if ((day==0)); then
month=$((month-1))
if ((month==0)); then
year=$((year-1))
month=12
fi
last_day_of_month=$((((62648012>>month*2&3)+28)+(month==2 && y%4==0)))
day=$last_day_of_month
fi
# format result
if ((day<10)); then day="0"$day; fi
if ((month<10)); then month="0"$month; fi
yesterday="$year$month$day"
echo $yesterday

Answer 5

TZ=$TZ+24 date +'%Y/%m/%d' in SunOS.

Answer 6

Playing on Solaris10 with non-GMT environment, I'm getting this:
# date
Fri Jul 26 13:09:38 CEST 2013 (OK)
# (TZ=CEST+24 date)
Thu Jul 25 11:09:38 CEST 2013 (ERR)
# (TZ=GMT+24 date)
Thu Jul 25 11:09:38 GMT 2013 (OK)
# (TZ=CEST+$((24-$((`date "+%H"`-`date -u "+%H"`)))) date)
Thu Jul 25 13:09:38 CEST 2013 (OK)
As You colud see, I have and I want to get CEST , but TZ=CEST+24 giving me wrong CEST data; GMT+24 giving me correct data, but unusable.
To get the proper result, I has to use GMT+22 (wrong command, correct result) or CEST+22 (wrong value, but finnaly correct result for correct TZ)

Answer 7

A pure bash solution given by #olivecoder is very reliable compared to any other solution but there is a mistake to be corrected. when the day fall on 1st of the month the script is failing with date saying "last_day_of_month" in day value. #olivecoder has missed $ in
day=last_day_of_month, that it should be
day=$last_day_of_month;
After this correction it works very good.
Using Timezone -24 is having some issue based on time when use it. in some cases it goes to day before yesterday. So I think its not reliable.