How do I calculate the previous business day in ksh shell script ?

Question

What is the most elegant way to calculate the previous business day in shell ksh script ?

What I got until now is :

#!/bin/ksh

set -x

DAY_DIFF=1
case `date '+%a'` in
"Sun")
   DAY_DIFF=2
   ;;
"Mon")
   DAY_DIFF=3
   ;;
esac

PREV_DT=`perl -e '($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)=localtime(time()-${DAY_DIFF}*24*60*60);printf "%4d%02d%02d",$year+1900,$mon+1,$mday;'`

echo $PREV_DT

How do I make the ${DAY_DIFF} variable to be transmitted as value and not as string ?

Answer 1

Well, if running Perl counts as part of the script, then develop the answer in Perl. The next question is - what defines a business day? Are you a shop/store that is open on Sunday? Saturday? Or a 9-5 Monday to Friday business? What about holidays?

Assuming you're thinking Monday to Friday and holidays are temporarily immaterial, then you can use an algorithm in Perl that notes that wday will be 0 on Sunday through 6 on Saturday, and therefore if wday is 1, you need to subtract 3 * 86400 from time(); if wday is 0, you need to subtract 2 * 86400; and if wday is 6, you need to subtract 1 * 86400. That's what you've got in the Korn shell stuff - just do it in the Perl instead:

#!/bin/perl -w
use strict;
use POSIX;
use constant SECS_PER_DAY => 24 * 60 * 60;
my(@days) = (2, 3, 1, 1, 1, 1, 1);
my($now) = time;
my($sec,$min,$hour,$mday,$mon,$year,$wday,$yday,$isdst)=localtime($now);
print strftime("%Y-%m-%d\n", localtime($now - $days[$wday] * SECS_PER_DAY));

This does assume you have the POSIX module; if not, then you'll need to do roughly the same printf() as you used. I also use ISO 8601 format for dates by preference (also used by XSD and SQL) - hence the illustrated format.

Answer 2

Here is a solution that doesn't use Perl. It works both with ksh and sh.

#!/bin/ksh

diff=-1
[ `date +%u` == 1 ] && diff=-3

seconds=$((`date +%s` + $diff * 24 * 3600))
format=+%Y-%m-%d

if date --help 2>/dev/null | grep -q -- -d ; then
    # GNU date (e.g., Linux)
    date -d "1970-01-01 00:00 UTC + $seconds seconds" $format
else
    # For BSD date (e.g., Mac OS X)
    date -r $seconds $format
fi

Answer 3

#!/bin/ksh
# GNU date is a veritable Swiss Army Knife...
((D=$(date +%w)+2))
if [ $D -gt 3 ]; then D=1; fi
PREV_DT=$(date -d "-$D days" +%F)

Answer 4

This should work for Solaris and Linux. It's realy complicating on Unix that you can not use the same commandline arguments on all Unix derivates.

On Linux you can use date -d '-d24 hour ago' to get the last day on Solaris its TZ=CET+24 date. I guess other UNIX'es works the same way as Solaris does.

#!/usr/bin/ksh

lbd=5               # last business day (1=Mon, 2=Thu ... 6=Sat, 7=Sun)
lbd_date=""         # last business day date

function lbdSunOS
{
    typeset back=$1
    typeset tz=`date '+%Z'`     # timezone

    lbd_date=`TZ=${tz}+$back date '+%Y%m%d'`
}

function lbdLinux
{
    typeset back=$1

    lbd_date=`date -d "-d$back hour ago"`
}

function calcHoursBack
{
    typeset lbd=$1
    typeset dow=`date '+%u'`    # day of the week

    if [ $dow -ge $lbd ]
    then
        return $(((dow-lbd)*24))
    else
        return $(((dow-lbd+7)*24))
    fi
}

# Main

calcHoursBack $lbd
lbd`uname -s` $?

echo $lbd_date