On Sep 17, Jason Tiller said:
>Assuming this is the way I want to go, here's what I've cooked up so
>far:
It's pretty good. The only change I would make is to structure it in a
slightly more idiomatic fashion:
my %paths;
$paths{$_}++ for split /:/, $ENV{PATH};
$ENV{PATH} = join ':', keys %paths;
>This works, but it feels really clunky. In my exploration of perl
>RE's, I can "sense" that there might be a RE solution to this problem
>without having to go to the hassle of creating a bunch of arrays, a
>hash, and then looping through the entries.
Well, regexes are A way, but they're not necessarily the best
approach. You'll end up having to do a lot of textual comparisons
(blegh) and the like.
$ENV{PATH} =~ s{
(?<! [^:] ) # make sure there's not a non-: behind us
([^:]+) : # match the path to \1, and match a :
(?= # if it's followed by...
(?: [^:]* : )* # any other paths
\1 # and then itself
(?! [^:] ) # and is NOT followed by a non-:
)
}{}gx; # remove it (and the colon that came after it)
The look-behind and look-ahead have double-negatives in them. Let me
explain. We want to match strings of non-:'s. These denote paths. But
we don't want to match the "oobar" of "foobar" in "abc:foobar:this:oobar"
because that would make a false positive.
So we create our own colon-boundary assertion by saying "be sure we are
NOT preceded by a non-colon." That means we're either preceded by a colon
or the beginning of the string. It could have been written as:
(?: ^ | (?<= : ) )
The same logic is used for (?! [^:] ). It could have been written as
(?: $ | (?= : ) )
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
RPI Acacia brother #734 http://www.perlmonks.org/ http://www.cpan.org/
** Look for "Regular Expressions in Perl" published by Manning, in 2002 **
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