Please bottom post....
> Use the following:
>
> $option = "0" . $option if ($option / 10 < 1 && $option !~ /^0/);
>
Why divide by 10 first? Why not just check less than 10? Of course I
would probably drop the maths completely and use C<sprintf>,
perldoc -f sprintf
$option = sprintf('%02d', $option);
Of course I am assuming there has also been,
unless ($option =~ /^\d{2}$/ and $option <= 31) {
die "I said a date you wacko";
}
And I won't even go into the point that some months only have 28, 29, or
30 days (damn leap years).
http://danconia.org
> ~Sid
>
> On Thu, 9 Sep 2004 08:58:36 -0500, Errin Larsen
<[EMAIL PROTECTED]> wrote:
> > Hi All,
> >
> > I have a variable I'm reading off of the command line:
> >
> > my $option = shift;
> >
> > That variable should hold a number between 1 and 31 (yes, a day of the
> > month) I am checking to make sure that the number does indeed lie in
> > that range.
> >
> > However, I need to pass that variable to another system command which
> > expects any "day" value less than 10 to have a leading zero, so 7th
> > day of the month should say '07'. How can I check for that leading
> > zero? If it's missing, I know I could easily:
> >
> > $option = "0".$option;
> >
> > But I can't seem to figure out an easy, clean way to check for it.
> >
> > --Thanks,
> >
> > --Errin
> >
> > --
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> >
>
>
>
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