On Mon, May 30, 2005 at 11:48:49PM +0000, Christian Weisgerber wrote:
>
> I don't think there is a reliable solution without something like
> FreeBSD's -v or GNU's -d extensions.
>
If you only want yesterday then this should do (it is ugly but it has
been tested on Solaris/Linux/NetBSD):
#!/bin/sh
#
# A way of working out what the date was yesterday - portably.
#
isleapyear()
{
#
# Determining leap years is easy (sortof).
# returns 0(true) if leapyear, 1 otherwise
#
if [ `expr $1 % 4` -eq 0 ]
then
if [ `expr $1 % 100` -eq 0 ]
then
if [ `expr $1 % 400` -eq 0 ]
then
return 0
fi
else
return 0
fi
fi
return 1
}
#
# Return the last day of the month taking into account leap years, $1
# is the month
#
ldom() {
case $1 in
1|3|5|7|8|10|12)
day=31
;;
2)
if isleapyear $year
then
day=29
else
day=28
fi
;;
*)
day=30
;;
esac
return $day
}
#
# Calculate the date for yesterday. Takes three parameters, $1 is the day
# $2 is the month and $3 is the year
#
yesterday() {
day=$1
month=$2
year=$3
if [ $day -ne 1 ]
then
day=`expr $day - 1`
else
if [ $month -ne 1 ]
then
month=`expr $month - 1`
ldom $month
else
year=`expr $year - 1`
month=12
day=31
fi
fi
}
#
# This is just for testing...
#
while read day month year
do
yesterday $day $month $year
echo "Yesterday was $day/$month/$year"
done
--
Brett Lymn
