Think it might be bust when year contains a 9.
e.g. year = 2009
$c = 9+1 = 10
qr/^\d*(?:${1}${2}${3}[$c-9]...
is:
qr/^\d*(?:200[10-9]...
and [10-9] matches any digit, so 2000, 2001 etc. would incorrectly pass.
Greg
-----Original Message-----
From: Jose Alves de Castro [mailto:[EMAIL PROTECTED]
Sent: Tuesday, June 01, 2004 10:53 AM
To: [EMAIL PROTECTED]
Subject: Regular expression for years after a given one
Yesterday, a friend of mine was toying around with some Perl code, when
he got to a function which would only accept a regular expression. He
said:
"How can I say that I want every single year after the one I have? It's
impossible..."
To which I replied:
"Nothing is impossible! Especially in Perl! ;-) " (ok, maybe these
weren't my words at the time, but they seem cool now, right? :-) )
So after a couple of minutes, this is what I came up with:
$year = 2004; # I'm stating it directly to improve clarity
$year =~/(\d)(\d)(\d)(\d)/;
($c,$d,$e,$f) = ($4+1,$3+1,$2+1,$1+1);
$regex =
qr/^\d*(?:${1}${2}${3}[$c-9]|${1}$2[$d-9]\d|$1[$e-9]\d{2}|[$f-9]\d{3})$/;
and that does the trick :-)
Thought I should share this with you guys :-) Thoughts are welcome :-)
BTW: This was made so it would work with $year consisting of four
digits... any care to make it generic? :-)
Regards,
jac
--
Jos� Alves de Castro <[EMAIL PROTECTED]>
Telbit - Tecnologias de Informa��o
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