Re: Simple Date Validation

[johngg]

I have the following which I converted from C a long time ago when I was very new to Perl. It was set up for the UK so the Julian/Gregorian change is in 1752.

#   ------
sub isLeap
#   ------
{
    my $year = shift
       or croak "isleap(): no year supplied\n";
    croak "isLeap(): year not numeric\n"
       unless $year =~ /^\d+$/;
    return 0 if $year % 4;
    return 1 if $year < 1753;
    return 1 if $year % 100;
    return 1 unless $year % 400;
    return 0;
}

#   -------
sub valDate
#   -------
{
    my($year, $month, $day) = @_;
    return 0 unless
       $year =~ /^\d+$/
       and $month =~ /^\d+$/
       and $day =~ /^\d+$/;
    my $daysinm = [
       [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31],
       [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]];
    return 0 if $year < 1 or $year > 9999;
    return 0 if $month < 1 or $month > 12;
    return 0 if $day < 1 or
       $day > $daysinm->[isLeap($year)]->[$month - 1];
    return 0 if $year == 1752 and $month == 9 and
       ($day > 2 and $day < 14);
    return 1;
}

It works for any year from 1 to 9999 (not y10k compliant I'm afraid ;-)

I hope this is of use.

Cheers,

JohnGG

[Not_a_Number]

not y10k compliant I'm afraid ;-)

Not Julian-compliant either!

You need to change

return 1 if $year < 1753

to

return 1 if $year < 1753 and not $year %4;

</nitpick>

Update: Memo to /me: Test before you post! (See below.)

[johngg]

No, that's not necessary at all.

By the time we reach that line in the subroutine we are only left with years that are divisible by 4 because of the line above, which is

return 0 if $year % 4;

so your amendment, and not $year %4, is superfluous.

Cheers,

JohnGG