Hi, Mark, :)
On Wed, 11 Dec 2002 [EMAIL PROTECTED] wrote:
> I have a list of names, separated by the '/' character in a
> variable, $list, and a name in the variable $name. If the name
> appears in the list, I want to remove it and clean up $list. It
> takes me 4 calls to s///, and it looks clunky. I have a feeling
> that someone out there on this list will know a cleaner way to do
> it.
> $list =~ s|$name||; # remove $name from the list if it's in the list
> $list =~ s|//|/|; # if $name was in the middle of the list, it would
> # have left behind two /s, remove one
> $list =~ s|^/||; # if $name was at the front of the list, remove the
> # leading /
> $list =~ s|/$||; # if $name was at the end of the list, remove the
> # trailing /
How about this:
use re 'eval'; # The (?{$num_slashes++}) construct requires this pragma.
$data =~ s{ (?:^|/)
(?(?<=/) (?{$num_slashes++}))
$name
(?(?=/) (?{$num_slashes++}))
(?:/|$)
}
{'/' if $num_slashes == 2}ex;
Basically, match $name and all surrounding /'s; if there were two
slashes in the match, then stick the joining slash in as the
replacement. That's it.
The main difference between this and Tanton's final result:
$data =~ s!(/)?\b$name\b(/)?!$2 ? ($1||'') : ($2||'')!e
Is that mine has no captures and no quaintifiers. With all of my
evaluation, though, I have no idea which one runs faster.
BTW, I think Tanton's RE could be written as:
$data =~ s!(/)?\b$name\b(/)?!$2 ? ($1||'') : ''!e;
or
$data =~ s! (/)? \b$name\b (/)? ! $1||'' if defined($2) !ex;
With no ill effects. (I find the trinary operator somewhat confusing
in a situation where you only want the 'if' portion and not the 'else'
portion.
Fun, fun!
---Jason
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