perl - 使用通配符移动文件 [英] perl - moving files using wildcard

问题描述

是否可以使用 File::Copy 模块中的 perl 的 move 函数来使用通配符移动多个具有相同文件扩展名的常见文件?到目前为止，如果我明确命名文件，我只能让 move 工作.

Is it possible to use perl's move function from File::Copy module to use a wildcard to move a number of common files with the same file extension? So far, I can only get move to work if I explicitly name the files.

例如，我想做这样的事情:

For example, I wanted to do something like so:

my $old_loc = "/share/cust/abc/*.dat";
my $arc_dir = "/share/archive_dir/";

现在，我可以像这样处理一个文件:

Right now, I can do one file like so:

use strict;
use warnings;
use File::Copy;

my $old_loc = "/share/cust/abc/Mail_2011-10-17.dat";
my $arc_dir = "/share/archive_dir/Mail_2011-10-17.dat";
my $new_loc = $arc_dir;

#archive
print "Moving files to archive...\n";
move ($old_loc, $new_loc) || die "cound not move $old_loc to $new_loc: $!\n";

在我的 perl 程序结束时我想做什么，将所有这些名为 *.dat 的文件移动到一个存档目录.

What I want to do at the end of my perl program, move all these files named *.dat to an archive directory.

推荐答案

你可以使用 Perl 的 glob 操作符来获取你需要打开的文件列表:

You can use Perl's glob operator to get the list of files you need to open:

use strict;
use warnings;
use File::Copy;

my @old_files = glob "/share/cust/abc/*.dat";
my $arc_dir = "/share/archive_dir/";

foreach my $old_file (@old_files)
{
    my ($short_file_name) = $old_file =~ m~/(.*?\.dat)$~;
    my $new_file = $arc_dir . $short_file_name;

    move($old_file, $new_file) or die "Could not move $old_file to $new_file: $!\n";
}

这样做的好处是不依赖系统调用，因为系统调用不可移植、依赖于系统，而且可能很危险.

This has the benefit of not relying on a system call, which is unportable, system-dependent, and possibly dangerous.

EDIT:更好的方法是提供新目录而不是完整的新文件名.(抱歉没有早点想到这个！)

EDIT: A better way to do this is just to supply the new directory instead of the full new filename. (Sorry for not thinking of this earlier!)

    move($old_file, $arc_dir) or die "Could not move $old_file to $new_file: $!\n";
    # Probably a good idea to make sure $arc_dir ends with a '/' character, just in case

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