Question

我写了一个脚本，它读取目录中的每个文件，执行某些操作并将每个输入文件的结果输出到两个不同的文件，例如“outfile1.txt”和“outfile2.txt”。我希望能够将生成的文件链接到原始文件，因此如何将输入文件名（infile.txt）添加到生成的文件名中，以获得如下内容：

infile1_outfile1.txt，infile1_outfile2.txt

infile2_outfile1.txt，infile2_outfile2.txt

infile3_outfile1.txt，infile3_outfile2.txt ......？

感谢您的帮助！

Answer 1

使用替换从输入文件名中删除“.txt”。使用字符串连接来构建输出文件名：

my $infile = 'infile1.txt';

my $prefix = $infile;
$prefix =~ s/\.txt//;  # remove the '.txt', notice the '\' before the dot

# concatenate the prefix and the output filenames
my $outfile1 = $prefix."_outfile1.txt";
my $outfile2 = $prefix."_outfile2.txt";

Answer 2

use File::Basename;
$base = basename("infile.txt", ".txt");
print $base."_outfile1.txt";

Answer 3

简单字符串连接应该在这里工作，或者在cpan上查找适当的模块

Answer 4

如果我理解正确，你在找这样的东西吗？

use strict;
use warnings;

my $file_pattern = "whatever.you.look.for";
my $file_extension = "\.txt";

opendir( DIR, '/my/directory/' ) or die( "Couldn't open dir" );
while( my $name_in = readdir( DIR )) {
    next unless( $name_in =~ /$file_pattern/ );

    my ( $name_base ) = ( $name_in =~ /(^.*?)$file_pattern/ );
    my $name_out1 = $name_base . "outfile1.txt";
    my $name_out2 = $name_base . "outfile2.txt";
    open( IN,   "<", $name_in )   or die( "Couldn't open $name_in for reading" );
    open( OUT1, ">", $name_out1 ) or die( "Couldn't open $name_out1 for writing" );
    open( OUT2, ">", $name_out2 ) or die( "Couldn't open $name_out2 for writing" );

    while( <IN> ) {
        # do whatever needs to be done
    }

    close( IN );
    close( OUT2 );
    close( OUT1 );
}
closedir( DIR );

编辑：实现了扩展剥离，输入文件句柄已关闭，现在已经过测试。

Perl：将输入文件名添加到输出文件名

4 个答案: