Trimming Paths with regexes

TGI has asked for the wisdom of the Perl Monks concerning the following question:

As part of a project I'm working on, I need to play with a path-like data structure. It comes in the form '/foo/bar/baz'. I need to check the full path, then '/foo/bar', then '/foo' and finally '/'.

So I've put together a do {} while loop that terminates when the path variable is length 0. I'm using a regex to trim the path:

do
{   if (exists $o->{$path})
    {   # Do stuff
    }
    # trim path
    print "\t\t$path\n";
    $path =~ s:(^/$)|((?<=^/)[^/]+$)|(/[^/]+$)::x;

} while ( length $path > 0 );
[download]

I keep futzing with the regex and can't come up with something that works. I keep failing to get '/' processed.

Once again, here's the transformation I'm looking for:

 '/foo/bar'-> '/foo'
 '/foo' ----> '/'
 '/' -------> ''

TGI says moo

Comment on Trimming Paths with regexes Download Code

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Re: Trimming Paths with regexes by damian1301 (Curate) on Aug 30, 2001 at 01:15 UTC
I personally would find it much more easier to just split the string and then work with it from there. `$_.=($=+(6<<1));print(chr(my$a=$_));$^H=$_+$_;$_=$^H; print chr($_-39); # Easy but its ok.` [download]	[reply] [d/l]
Re: Trimming Paths with regexes by jryan (Vicar) on Aug 30, 2001 at 01:21 UTC
I had to deal with a similar situation one time. Instead of fiddling with a regex all day, I used `split` to split the path up, and then put it back together (minus the last part). It was similar to this: `my $oldpath = "/foo/bar/baz"; my @path = split(/\//, $oldpath); my $newpath = ""; for ($i=0; $i<(@path-1); $i++) { $newpath .= "/".$path[$i]; } print $newpath;` [download] It might not be as cleany coded as a regex (or as cool looking!), but it gets the job done. Sometimes the simple solutions are the best :)	[reply] [d/l] [select]
Re: Trimming Paths with regexes by demerphq (Chancellor) on Aug 30, 2001 at 05:24 UTC
`my $path='/foo/bar'; my @dirs=split(/\//,$path); $"='/'; $\="\n"; while (@dirs) { my $str="@dirs" \|\| "/"; print $str; pop @dirs; }` [download] outputs `/foo/bar /foo /` [download] Yves	[reply] [d/l] [select]
Re: Trimming Paths with regexes by dga (Hermit) on Aug 30, 2001 at 01:48 UTC
Returns the path shortened and handles paths with //'s together like ////usr/bin correctly. Returns undef in the special case of the root path. Does not handle relative paths `sub pathtrim { my($path)=@_; return undef if($path eq '/'); $path =~ s#/[^/]+$#/#; $path =~ s#(/[^/]*)/+$#$1#; return $path; }` [download]	[reply] [d/l]
Re: Trimming Paths with regexes by Hofmator (Curate) on Aug 30, 2001 at 14:43 UTC
Just to show that it's not too complicated with a regex: `do { # your stuff here print "\t\t$path\n"; } while $path =~ s#(?:(^/)\|/)[^/]+$#$1?$1:''#e;` [download] You can use a subroutine as well, then the regex is pretty simple if you handle the special case beforehand: `sub pathtrim { my ($path) = @_; return '/' if ($path =~ m#^/[^/]+$/); $path =~ s#/[^/]+$##; return $path; }` [download] Update: Fixed small bug in the subroutine. -- Hofmator	[reply] [d/l] [select]