为什么这个程序无法复制文件？

Question

今天早上，我和我的朋友讨论并编写了以下代码。这个 Perl 脚本背后的想法是创建目录结构并将文件复制到相应的目录。

#!/usr/bin/perl
use File::Path;
use File::Copy;
use Path::Class;
use File::Basename qw/dirname/;
my $src = "/Vijay/new.txt";
unless (open(MYFILE, "file1")) {
    die ("cannot open input file file1\n");
}
$line = <MYFILE>;
while ($line ne "") {
    print ($line);
    mkdir_and_copy($src,$line);
    $line = <MYFILE>;
}
sub mkdir_and_copy {
    my ($from, $to) = @_;
    my($directory, $filename) = $to =~ m/(.*\/)(.*)$/;
    print("creating dir $directory");
    system "mkdir -p $directory";
    print("copying file $from to $to");
    system "cp -f $from $to";
    return;
}

上面这段代码创建了目录结构，但是无法将文件复制到相应的目录。请告诉我们，我们到底哪里错了？

file1 的内容：

test/test1/test2/test.txt

new.txt 的内容：

Shell/Test/test1/test1.txt
Shell/Test/test2/test2.txt
Shell/Test/test3/test3.txt

输出：

> ./mypgm.pl
test/test1/test2/test.txt
creating dir test/test1/test2/copying file /Vijay/new.txt to     test/test1/test2/test.txt
cp: cannot access /Vijay/new.txt: No such file or directory
>

目录Vijay有文件new.txt，上面的内容。

提前致谢，

维杰

---

大家好，
我刚刚修改了我的代码。请参考下面的代码部分。

#!/usr/bin/perl        
use File::Path;    
use File::Copy;    
use File::Basename qw/dirname/;    

my $src = "./Vijay/new.txt";       
unless (open(MYFILE, "file1"))    
{    
die ("cannot open input file file1\n");    
}

$line = ;
while ($line ne "")
{
print ($line); print("\n");
mkdir_and_copy($src,$line);
$line = ""; }

sub mkdir_and_copy        
{      
my ($from, $to) = @_;    
my($directory, $filename) = $to =~ m/(.\/)(.)$/;    
$temp = $directory.$filename;    
print("Creating dirrectory $directory \n");    
if(! -d $directory)    
{    
mkpath($directory) #or die "Failed to create path";    
}    
printf("From: $from \n");    
printf("To: $temp \n");    
copy($from,$temp) or die "Failed to Copy";    
return;    
}    

现在，它会创建准确的目录结构并将文件复制到相应的目录。你能告诉我，上面的代码是否正确？

Answer 1

我不清楚你的目标，但也许这会帮助你解决问题：

# Perl scripts should always include this.
# Your original script was generating some useful warnings.
use strict;
use warnings;

my $src = "/Vijay/new.txt";
my $f1  = 'file1';

# This is the recommended way to open a file --
# that is, using a lexical file handle.
open(my $file_handle, '<', $f1) or die "open() failed : $f1 : $!";

# This is the typical way of iterating over the lines in a file.
while (my $line = <$file_handle>){
    # You probably want to remove the newline
    # before passing the line to mkdir_and_copy()
    chomp $line;

    mkdir_and_copy($src, $line);
}

sub mkdir_and_copy {
    my ($from, $to) = @_;
    my ($directory, $filename) = $to =~ m/(.*\/)(.*)$/;

    # When writing a script that makes system() calls,
    # start by simply printing them. After everything
    # looks good, convert the print commands to system() calls.
    print "system(): mkdir -p $directory", "\n";
    print "system(): cp -f $from $to",     "\n";

    # The return is not needed.
}

当我使用您提供的输入运行脚本时，输出如下：

system(): mkdir -p test/test1/test2/
system(): cp -f /Vijay/new.txt test/test1/test2/test.txt

这不是你的意图。特别是，当 file1 只包含一行时，为什么要迭代它？也许您打算遍历new.txt？

Answer 2

如果某件事“不起作用”，首先要做的是捕捉错误并查看它们。然后调查变量的内容。在您的情况下，变量 $to 仅包含文件名，因此脚本将其复制到当前工作目录中，我想，而不是新创建的目录中。

但是，您用来完成工作的方法并不是最好的。实际使用File::Path 和File::Copy 会更好，尤其是您在第一个斜杠处将路径拆分为目录和文件名的方式绝非一般。这种事情应该在库中完成，Perl 有很多。

Answer 3

我敢打赌你的$line 变量仍然附加了一个换行符。从文件句柄输入运算符 (<MYFILE>) 返回的输入包括记录分隔符（通常是操作系统的换行符）。试试这个：

$line = <MYFILE>;
chomp($line);