Timothy, et al -- ...and then Timothy Johnson said... % % David->"Hmmm... OK, so that explains it, but I still don't get it... So % the match is going to spit out a scalar but in order to use it you have to % capture it in a list context?" % % No, actually the opposite. The match returns a list, but split() was % reading the list as a scalar. Thus: % % ($fullpath =~ m:/mp3/(.+):)[0] % % is element 0 of the list created by evaluating the match. That way what % split sees is a scalar (the element), so it can split it accordingly.
I'm getting close; my head is starting to hurt :-) Does the =~ and/or the m// return a list, or is it because we've had to wrap that operation in () to bind it properly? Thanks one last time (I hope) & HAND :-D -- David T-G * It's easier to fight for one's principles (play) [EMAIL PROTECTED] * than to live up to them. -- fortune cookie (work) [EMAIL PROTECTED] http://www.justpickone.org/davidtg/ Shpx gur Pbzzhavpngvbaf Qrprapl Npg!
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