Question

如何将多行字符串转换为数组？

my $text= " ads da
sda
s 
da
d
as

das
d a as dasd
\n

";

注意：我不想删除或删除换行符？

Answer 1

目前的问题可能更明确。

my @text = split "\n", $text;

Answer 2

您可以使用^元字符和m regexp修饰符（让^匹配行的开头而不仅仅是行的开头）在行的开头拆分字符串）：

split /^/m, $text

实际上，在这种情况下，您可以省略m，因为split会为您提供。来自perldoc -f split：“”/ ^ /“的模式被视为”/ ^ / m“，因为它没有多大用处。”

使用$text的值，此代码：

use Data::Dumper;
$Data::Dumper::Useqq=1;
print Data::Dumper->Dump([[split /^/, $text]], ["*text"]);

打印出来：

@text = (
          " ads da\n",
          "sda\n",
          "s \n",
          "da\n",
          "d\n",
          "as\n",
          "\n",
          "das\n",
          "d a as dasd\n",
          "\n",
          "\n",
          "\n"
        );

Answer 3

请记住，split的第一个参数是一种模式：

#!/usr/bin/perl

use strict; use warnings;
use YAML;

my $text = " ads da
sda
s
da
d
as

das
d a as dasd
\n

";

print Dump [ split /(\n)/, $text ];

输出：

---
- ' ads da'
- "\n"
- sda
- "\n"
- s
- "\n"
- da
- "\n"
- d
- "\n"
- as
- "\n"
- ''
- "\n"
- das
- "\n"
- d a as dasd
- "\n"
- ''
- "\n"
- ''
- "\n"
- ''
- "\n"

Answer 4

我的感觉是你专注于错误的问题。

不是试图将标量多行字符串常量转换为列表，也许您的问题应该是“我如何将多行字符串启动到Perl列表或数组中？”

查看List value constructors中的Perl Perldata。

对您的问题的特殊适用性是如何使用heredoc来启动具有多行字符串的数组：

#!/usr/bin/perl
use strict; use warnings;
use YAML;

my @text= <<END =~ m/(^.*\n)/mg;
 ads da
sda
s 
da
d
as

das
d a as dasd
\n

END

print Dump \@text;

打印：

---
- " ads da\n"
- "sda\n"
- "s \n"
- "da\n"
- "d\n"
- "as\n"
- "\n"
- "das\n"
- "d a as dasd\n"
- "\n"
- "\n"
- "\n"

使用成语Luke！

Answer 5

我很乐意把这个放在一起： Voila！你的字符串现在是一个数组而没有 split - ting 它：

use strict qw<subs vars>;
use warnings;

@{" ads da
sda
s 
da
d
as

das
d a as dasd
\n
"} = 1..3
;

将多行字符串转换为数组

5 个答案: