Jeff 'Japhy' Pinyan wrote:
>
> On May 13, Lee Johnson said:
> >
> >#
> ># End date is a year after start date
> >#
> >$edate = $sdate;
> >$edate =~ s/\///g;
> >$edate++;
> >$edate =~ s/.*(\d{4})$/01\/04\/$1/;
>
> Are you sure you want to just use April 1st (or January 4th) always? You
> don't want to use the day and month in $sdate?
>
> >where $sdate is a UK date of the type "dd/mm/yyyy"
> >
> >I've tried ($edate = $sdate) =~ s/(.*)(\d)$/$1($2++)/e; to no avail. I think
> >I'm misunderstanding the /e modifier here?
>
> A little bit, yes. The /e modifier makes the right-hand side Perl code,
> and you should recognize that
>
> $1($2++)
>
> isn't valid Perl code. You'd need
>
> $1 . ($2++)
>
> but that won't work for two reasons. First, $x++ returns its OLD value,
> and THEN increments it, so you wouldn't get a changed value. The second
> reason is that you can't modify the $DIGIT variables.
>
> Finally, your code is making a mistake in the regex. What if the year is
> 2009? You're only matching the LAST digit of the year, and adding one to
> it. Your string would end up being "...20010", which is wrong. Match the
> whole four-digit year, and add one to it. When it rolls around to 9999,
> then come back to me and complain about my bad code. ;)
Also, if the current date is 29/02/2004 and you increment the year to
2005...
John
--
use Perl;
program
fulfillment
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