in reply to Regular Expression Question

The . (dot) in your regular expression will match anything. You need to escape it, i.e.

$folder =~ /(\d{6}\.?\d{2}?\w?)/;

Also, this regular expression is matching for files which are of the form ######## (8 numbers without a dot), and files of the form ######a (6 numbers and the optional letter), as well as a few other forms that it sounds like you don't awnt to capture. Is the optional 2 digit number and letter dependent on whether there is a dot? If so, you are going to need a slightly more complex regular expression.

One last question, why do you even have the regular expression in there? Can't you use $folder instead of $1 without loss of generality? You aren't doing a conditional based on whether or not the regexp matches, so I assume all folders are meant to be replaced in...

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Re^2: Regular Expression Question
by MistaMuShu (Beadle) on Jul 22, 2004 at 19:00 UTC
    Can't you use $folder instead of $1 without loss of generality?

    Well, initially I was thinking that since the actual folder names are ######.##a_files (think html folders) that I would find the first part of the name and save it in $1

    Now that I look at it again with your replies, I see that it'd be a lot easier if I just negated "_files" and did /(^_files)/

    That was silly of me to not escape the dot :( And I'll have to read a bit more on the lookahead find because ?: still does not make perfect sense to me... Thanks!