Here's a short (44-line) Perl script that will do the job. It's not
flexible on the argument format - they have to be yyyy-mm-dd - and
it is Perl, but at least it doesn't use a zillion modules. The only
module it does use is POSIX, and that's only to get the floor() function;
if you aren't going to be dealing with negative years you can get rid
of the "use POSIX" line and replace floor(...) with int(...).
--
Mark REED | CNN Internet Technology
1 CNN Center Rm SW0831G | [EMAIL PROTECTED]
Atlanta, GA 30348 USA | +1 404 827 4754
--
If you are going to walk on thin ice, you may as well dance.
#!/usr/bin/perl
use POSIX;
# calculate the number of days between two Gregorian dates
if (@ARGV != 2)
{
die "Usage: $0 yyyy-mm-dd1 yyyy-mm-dd2\n";
}
my ($from, $to) = @ARGV;
my $from_jd = jd_from_gregorian(split('-',$from));
my $to_jd = jd_from_gregorian(split('-',$to));
print $to_jd - $from_jd, "\n";
exit(0);
# Calculate the astronomical Julian Day number as of noon on the given
# (year, month, day). In this program we don't really need the
# actual JD; any absolute day count would give us the same answer
# without the offset. But it doesn't hurt to use a number that's
# meaningful in other contexts.
sub jd_from_gregorian
{
my ($year, $month, $day) = @_;
my $elapsed_years = $year;
# We start counting from March to keep February from screwing
# up the math
my $months_into_year = $month - 3;
if ($months_into_year < 0)
{
$elapsed_years--;
$months_into_year += 12;
}
my $thirty1sts = floor((7*$months_into_year+7)/12);
# JD 1,721,120 began at noon UTC on March 1st, 0 AD (== 1 BC)
# in the retrojected Gregorian calendar.
return 1_721_120
+ $elapsed_years * 365
+ floor($elapsed_years/4)
- floor($elapsed_years/100)
+ floor($elapsed_years/400)
+ $months_into_year * 30
+ $thirty1sts
+ $day
- 1;
}
