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Before you jump down to the comments and post a Gnu date solution (or anything else), please READ the rest of the article AND the comments. Doing so will help prevent you looking like a fool and will save me the time necessary to point out that you didn't bother to read before posting something we already know..
Finding yesterday's date is easy in languages like Perl:
# this example is programmed for clarity, not efficiency $rightnow=time(); $rightnow -= (24 * 60 * 60); print scalar localtime($rightnow);
But in the shell (without Gnu date) it can be harder. You'll see people recommend something like:
#!/bin/ksh yesterday=$(TZ=EST26EDT date +%mm-%d-%yy)
where the timezone is selected for the proper effect, but that isn't actually going to work always.
Tapani Tarvainen posted a better solution:
#! /usr/bin/ksh
# Get yesterday's date in YYYY-MM-DD format.
# With argument N in range 1..28 gets date N days before.
# Tapani Tarvainen January 2002
# This code is in the public domain.
OFFSET=${1:-1}
case $OFFSET in
*[!0-9]* | ???* | 3? | 29) print -u2 "Invalid input" ; exit 1;;
esac
eval `date "+day=%d; month=%m; year=%Y`
typeset -Z2 day month
typeset -Z4 year
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=$((day - OFFSET))
if (( day <= 0 )) ;then
month=$((month - 1))
if (( month == 0 )) ;then
year=$((year - 1))
month=12
fi
set -A days `cal $month $year`
xday=${days[$(( ${#days[*]}-1 ))]}
day=$((xday + day))
fi
print $year-$month-$day
print $month/$day/${year#??}
You may also want to look at http://aplawrence.com/SCOFAQ/FAQ_scotec6datemath.html and http://aplawrence.com/Unix/yesterday.html
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Click here to add your comments
---July 13, 2004
The following command will do the job without any mess.
$ date --date='1 day ago'
- amber
---July 13, 2004
Thanks Amber!
It's only true on Linux, but that does point out the importance of reading "info date" instead of just "man date" - you won't find that option spelled out fully in the man page, but the info doc does explain it.
--
TonyLawrence
---August 4, 2004
The script doesn't work.
When day becomes smaller then -9, then, by the typesetting of -Z2, day == 10. While 10 !<=0, you'll get a date you've never expected.
Norbert Groen
---August 25, 2004
need any formated string? and like perl. so take:
$|=1; # flush output
$rightnow=time() - (24 * 60 * 60);
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday) = localtime
($rightnow);
$mon +=1; # always that
$mon = '0' . $mon if ( length $mon < 2 );
$mday = '0' . $mday if ( length $mday < 2 );
$yesterday = ( $year + 1900 ) . '.' . $mon . '.' . $mday;
print $yesterday . "\n";
exit 0;
---December 8, 2004
echo $(date --date='1 day ago' +%Y%m%d)
worked fine on linux, thankx "Amber"
-merouane
---December 13, 2004
here the script of tapani in pure bash
#!/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-$day
P.s.= i'm sorry but this editing loss indentation
--
stetor
---December 13, 2004
---December 28, 2004
Hi,
My problem is that i want to compute a date (before or after) based on a fixed date.
For example my shell parameter is : D='12/25/2004'
and i want to obtain D-5, D+2, D+4, D+5
Before starting to write a complex shell script, i want to be sure that i cannot use the date command. Any ideas ?
Thanks
Chris
Wed Mar 16 20:44:49 2005: Subject: TonyLawrence
Robert Wolf sent email as follows:
Your script on your web page, has a bug in it, below is the corrected version. For example when you set N to a large value like 25 or 26 or 27 or 28 then depending on the current day of the month, the variable 'day' will be more than 2 digits, i.e. -10 say. Now the value '-10' will not fit in a 'typeset -Z2' variable, it needs to be 'typeset -Z3'.
#!/bin/ksh
# Get yesterday's date in YYYY-MM-DD format.
# With argument N in range 1..28 gets date N days before.
# Tapani Tarvainen January 2002
# This code is in the public domain.
OFFSET=${1:-1}
case $OFFSET in
*[!0-9]* | ???* | 3? | 29) print -u2 "Invalid input" ; exit 1;;
esac
eval `date "+day=%d; month=%m; year=%Y`
typeset -Z3 day month
typeset -Z4 year
# Subtract offset from day, if it goes below one use 'cal'
# to determine the number of days in the previous month.
day=$((day - OFFSET))
if (( day <= 0 )) ;then
month=$((month - 1))
if (( month == 0 )) ;then
year=$((year - 1))
month=12
fi
set -A days `cal $month $year`
xday=${days[$(( ${#days[*]}-1 ))]}
day=$((xday + day))
fi
typeset -Z2 day month
print $year-$month-$day
print $month/$day/${year#??}
Tue Jun 7 15:43:38 2005: Subject: anonymous
"date --date=yesterday" or "date -d yesterday" works with 'gnu date', not only on linux.
"date --version" will tell you, if your system uses 'gnu date':
save-lin:/ # date --version
date (coreutils) 5.2.1
Written by David MacKenzie.
Copyright (C) 2004 Free Software Foundation, Inc.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
-thomas
Mon Aug 14 14:43:00 2006: Subject: anonymous
If you don't have gnu date, the tip at:
http://aplawrence.com/SCOFAQ/FAQ_scotec6datemath.html
is more elegant....
bil
Thu Feb 22 20:22:36 2007: Subject: anonymous
Screw all that calculation crap. Here is a one-liner to provide yesterday's date in a shell variable.
YEST=`TZ="GMT+24" date +'%Y%m%d'`
that will provide yesterday in CCYYMMDD format. Don't like that format?
Change it by consulting strftime on your system.
Fri Feb 23 15:13:54 2007: Subject: BigDumbDinosaur
Screw all that calculation crap. Here is a one-liner to provide yesterday's date in a shell variable.
YEST=`TZ="GMT+24" date +'%Y%m%d'`
that will provide yesterday in CCYYMMDD format. Don't like that format? Change it by consulting strftime on your system.
You can rearrange the date parameters to produce a different format. For example:will produce a MM/DD/YYYY format.
If you need to go back more than a day just increment the GMT parameter by 24 per day. For example:gives you the date of three days ago. To go forward, make the GMT parameter negative, e.g.:gives you the date of the next day. You could also write that as: with KSH or BASH, where ND is the number of days. The maximum range for ND is +/- 365 days.
Fri Feb 23 17:32:33 2007: Subject: TonyLawrence
Sigh..
The TZ method was mentioned in the article body and as pointed out there and in the News articles referenced, IT WON'T ALWAYS WORK.
Fri Feb 23 18:43:19 2007: Subject: BigDumbDinosaur
The TZ method was mentioned in the article body and as pointed out there and in the News articles referenced, IT WON'T ALWAYS WORK.
I wasn't reinventiing the wheel and I didn't say that it was 100 percent reliable. <Smile> I was merely pointing out how that method could be extended with a little twiddling.
For my applications, I use a small C program that reads STDIN for a date in YYYYMMDD format, converts it to a SQL date number, adds X number of days (which if negative, goes back in time), converts the result back to YYYYMMDD format and writes that on STDOUT. It's trustworthy from October 1752 (the first full month after the British empire adopted the Gregorian calendar) to December 31, 9999.
Tue May 1 12:24:04 2007: Subject: www.iphone2die4.com anonymous
errr.... how about:
yday=$(date --date "1 day ago")
Tue May 1 12:26:04 2007: Subject: TonyLawrence
How about you actually READ everything written?
Sun Nov 30 19:00:45 2008: Subject: anonymous
Great solution! Never knew of that option but now that I do it makes things alot easier.
Fri Jun 5 05:58:02 2009: Subject: anonymous
I see why the simpler shell solution doesn't always work. However the improved version is more complicated than I need it.
This is sufficient for my purpose (machine uses UTC):
#!/bin/bash
typeset -i H
H=$(date '+%H')+1
TZ=UTC$H date '+%Y%m%d'
Fri Jun 5 10:31:20 2009: Subject: TonyLawrence
Nothing wrong with that. It's important to realize the potential issues, but if they don't apply to you, go for it.
Mon Jun 8 20:29:16 2009: Subject: I have had a few run-ins with this issue myself. snoopy
Just my two cents, but the TZ really does not work too well. I have used this in my code/scripts for the past two years and found some strange behavior. It used to run on Solaris 8 fine, I think, but now we are on Solaris 10 and anything beyond 6 days causes it to fail. I am not sure if this was failing before or not.
For example:
date
Mon Jun 8 20:22:55 GMT 2009
YEST=`TZ="GMT+72" date +'%m/%d/%Y'`
echo $YEST
06/05/2009
YEST=`TZ="GMT+144" date +'%m/%d/%Y'`
echo $YEST
06/02/2009
YEST=`TZ="GMT+168" date +'%m/%d/%Y'`
echo $YEST
06/08/2009
-- After 144 hours the thing is AFU.
I tried to do this in Perl instead: Name the following as ydate.pl, and run with the argument as the amount to subtract. ./ydate.pl 8 I incorporate it as a subroutine to calculate date. Or you can use it as is and use bash or korn to make an external call to calculate.
#!/usr/bin/perl
$var1 = $ARGV[0];
sub ydate {
my $var=shift;
$subtract = ($var * 24 * 60 * 60); # Calculate the number of seconds in a day, $var is a day multiplier.
($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst) = gmtime( (time() -($subtract))); # Get time
$year += 1900; # Add amount of years to add to current year, necessary for Perl.
$mon += 1; # Add 1 to month of year variable, localtime month is "0" based.
$mon = sprintf("%02d", $mon ); # Add 2 decimal places, necessary if mon is 1-9.
$mday = sprintf("%02d", $mday ); # Add 2 decimal places, necessary if mday is 1-9.
return "${year}${mon}${mday}";
}
print "DATE: " . (&ydate($var1)) . "\n";
Hope it helps.
Wed Aug 19 08:56:25 2009: Subject: archanamangamuri
#!/usr/bin/perl
my $yesterday=`date -d 'yesterday' +%Y%m%d`;
chomp($yesterday);
print"date $yesterday\n";
Wed Aug 19 10:50:34 2009: Subject: TonyLawrence
Wow!
Useless use of Perl award, pointless use of "chomp", plus extra points for not reading, and more points for being Linux specific.
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