perl脚本以递归方式列出目录中的所有文件名 [英] perl script to recursively list all filename in directory

问题描述

我写了以下perl脚本，但问题是它总是出现在其他部分，而不报告文件.我确实在输入中提供的目录中有文件.我在这里做什么错了?

I have written following perl script but problem is its always going in else part and reporting not a file. I do have files in the directory which I am giving in input. What am I doing wrong here?

我的要求是递归访问目录中的每个文件，将其打开并以字符串形式读取.但是逻辑的第一部分失败了.

My requirement is to recursively visit every file in a directory, open it and read it in a string. But the first part of the logic is failing.

#!/usr/bin/perl -w
use strict;
use warnings;
use File::Find;

my (@dir) = @ARGV;
find(\&process_file,@dir);

sub process_file {
    #print $File::Find::name."\n";
    my $filename = $File::Find::name;
    if( -f $filename) {
        print " This is a file :$filename \n";
    } else {
        print " This is not file :$filename \n";
    }
}

推荐答案

$File::Find::name给出相对于原始工作目录的路径.但是，除非另有说明，否则 File :: Find 会不断更改当前工作目录.

$File::Find::name gives the path relative to original working directory. However, File::Find keeps changing the current working directory unless you tell it otherwise.

要么使用no_chdir选项，要么使用仅包含文件名部分的-f $_.我推荐前者.

Either use the no_chdir option, or use -f $_ which contains just the file name portion. I recommend the former.

#!/usr/bin/perl -w
use strict; 
use warnings;
use File::Find;

find({ wanted => \&process_file, no_chdir => 1 }, @ARGV);

sub process_file {
    if (-f $_) {
        print "This is a file: $_\n";
    } else {
        print "This is not file: $_\n";
    }
}

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