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问题:

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]

回答1:

Try this below thing. It should work

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

回答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

回答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

回答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

回答5:

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

回答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)

回答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