3

该Path::Class模块是一个顶级模块，在 cpan.org上有很好的评论。我想知道Path::Class:Dir模块是否可以创建这样的目录树*：

/home/scottie/perl/lib/Path-Class-0.25
  -> lib/
     -> Path/
        -> Class/
           -> Dir.pm
           -> Entity.pl
           -> File.pl
        -> Class.pl
  -> t/
     -> 01-basic.t
     -> 02-foreign.t
     -> 03-filesystem.t
     -> 04-subclass.t
     -> 05-traverse.t
     -> author-critic.t
  -> Build.PL
  -> Changes
  -> dist.ini
  -> INSTALL
  -> LICENSE
  -> Makefile.PL
  -> MANIFEST
  -> META.yml
  -> README
  -> SIGNATURE

*) 来源：Path-Class-0.25.tar.gz文件和字符串末尾的“/”表示目录（如ls -p在 *nix 系统中）

而不是这样（从根目录的完整路径）：

  /home/scottie/perl/lib/Path-Class-0.25
  -> /home/scottie/perl/lib/Path-Class-0.25/lib/
     -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/
        -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/Class/
           -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/Class/Dir.pm
           -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/Class/Entity.pl
           -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/Class/File.pl
        -> /home/scottie/perl/lib/Path-Class-0.25/lib/Path/Class.pl
  -> /home/scottie/perl/lib/Path-Class-0.25/t/
     -> /home/scottie/perl/lib/Path-Class-0.25/t/01-basic.t
     -> /home/scottie/perl/lib/Path-Class-0.25/t/02-foreign.t
     -> /home/scottie/perl/lib/Path-Class-0.25/t/03-filesystem.t
     -> /home/scottie/perl/lib/Path-Class-0.25/t/04-subclass.t
     -> /home/scottie/perl/lib/Path-Class-0.25/t/05-traverse.t
     -> /home/scottie/perl/lib/Path-Class-0.25/t/author-critic.t
  -> /home/scottie/perl/lib/Path-Class-0.25/Build.PL
  -> /home/scottie/perl/lib/Path-Class-0.25/Changes
  -> /home/scottie/perl/lib/Path-Class-0.25/dist.ini
  -> /home/scottie/perl/lib/Path-Class-0.25/INSTALL
  -> /home/scottie/perl/lib/Path-Class-0.25/LICENSE
  -> /home/scottie/perl/lib/Path-Class-0.25/Makefile.PL
  -> /home/scottie/perl/lib/Path-Class-0.25/MANIFEST
  -> /home/scottie/perl/lib/Path-Class-0.25/META.yml
  -> /home/scottie/perl/lib/Path-Class-0.25/README
  -> /home/scottie/perl/lib/Path-Class-0.25/SIGNATURE

我试图这样做，但我的代码有问题：

#------------------8<------------------
my $dir = Path::Class::Dir->new('/home/scottie/perl/lib/Path-Class-0.25');

my $nfiles = $dir->traverse(sub {
    my ($child, $cont) = @_;
    return if -l $child; # don't follow symlinks

    #print Dumper($child);
    #print "$child\n";

    print $child->{'dir'}{'dirs'}[-1];
    print " -> ";
    print $child->{'file'};
    print "\n";
    return $cont->();
});
#------------------8<------------------

您能否看一下我的代码并告诉我如何在没有完整路径的情况下制作目录树$child？

非常感谢！

最好的问候，
斯科蒂

4

1 回答 1

4

您需要使用 的basename方法Class::Path::File（也可以作为 的方法使用Class::Path::Dir）来仅提取路径的最后一个元素。

您还必须使用回调参数来跟踪当前的缩进级别。

该程序似乎可以满足您的要求。

use strict;
use warnings;

use Path::Class;

my $dir = dir('/home/scottie/perl/lib/Path-Class-0.25');

$dir->traverse(sub {

  my ($child, $cont, $indent) = @_;
  $indent //= 0;

  if ($indent == 0) {
    print $child, "\n";
  }
  else {
    print '   ' x ($indent - 1), '-> ', $child->basename;
    print '/' if $child->is_dir;
    print "\n";
  }

  $cont->($indent + 1);
});
于 2012-05-15T00:50:52.520 回答