9.10.1. Problem

9.10.2. Solution

use File::Basename;

$base = basename($path);
$dir  = dirname($path);
($base, $dir, $ext) = fileparse($path);

9.10.3. Discussion

$path = "/usr/lib/libc.a";
$file = basename($path);    
$dir  = dirname($path);     

print "dir is $dir, file is $file\n";
# dir is /usr/lib, file is libc.a

$path = "/usr/lib/libc.a";
($name,$dir,$ext) = fileparse($path,'\..*');

print "dir is $dir, name is $name, extension is $ext\n";
# dir is /usr/lib/, name is libc, extension is .a

fileparse_set_fstype("MacOS");
$path = "Hard%20Drive:System%20Folder:README.txt";
($name,$dir,$ext) = fileparse($path,'\..*');

print "dir is $dir, name is $name, extension is $ext\n";
# dir is Hard%20Drive:System%20Folder, name is README, extension is .txt

To pull out just the extension, you might use this:

sub extension {
    my $path = shift;
    my $ext = (fileparse($path,'\..*'))[2];
    $ext =~ s/^\.//;
    return $ext;
}

When called on a file like source.c.bak, this returns an extension of "c.bak", not just "bak". If you want ".bak" returned, use '\.[^.]*' as the second argument to fileparse (this will, of course, leave the filename as source.c).

When passed a pathname with a trailing directory separator, such as "lib/", fileparse considers the directory name to be "lib/", whereas dirname considers it to be ".".