Perl File::Find::Rule 排除单个目录

Question

我有如下所示的目录结构。

1. 我只想在我的@project 中获取项目名称。 @project=('project1','project2');
   
2. 我想排除OLD 目录及其在@project 中的子目录
   
3. 我想在@project 的所有项目的子目录中获取最新文件。即，对于project1，最新文件在子目录2014，即foobar__2014_0916_248.txt
   

如何制定规则来实现这一目标？

use strict;
use File::Find::Rule;
use Data::Dump;
my $output       = "/abc/def/ghi";
my @exclude_dirs = qw(OLD);
my @projects     = File::Find::Rule->directory->in("$output");
dd \@projects;

我的目录结构：

.
├── project1
│   ├── 2013
|        ├── file1_project1.txt
│   └── 2014
|         ├── foobar__2014_0912_255.txt
|         ├── foobar__2014_0916_248.txt
├── project2
│   ├── 2013
|        ├── file1_project2.txt
│   └── 2014
|         ├── foobarbaz__2014_0912_255.txt
|         ├── foobarbaz__2014_0916_248.txt
└── OLD
    └── foo.txt

Answer 1

按照 ikegami 的建议，只需分两步完成。

1. 首先找到您的项目名称
2. 二次查找最新文件

以下使用Path::Class 和Path::Class::Rule 执行此操作

use strict;
use warnings;
use autodie;

use Path::Class;
use Path::Class::Rule;

my $testdir = dir('testing');

for my $project ( $testdir->children ) {
    next if !$project->is_dir() || $project->basename eq 'OLD';

    my $newest;

    my $next = Path::Class::Rule->new->file->iter($project);
    while ( my $file = $next->() ) {
        $newest = $file if !$newest || $file->stat->mtime > $newest->stat->mtime;
    }

    print "$project - $newest\n";
}

输出：

testing/project1 - testing/project1/2014/foobar__2014_0916_248.txt
testing/project2 - testing/project2/2014/foobarbaz__2014_0916_248.txt