在Perl中匹配和替换几个单词时,如何保留空白?

自从我使用Perl正则表达式以来已经有一段时间了,但是关于:

$match = s/(has\s+a\s+substring)/[$1]/ig

这将捕获单词之间零个或多个空格和换行符。它将用括号包装整个匹配,同时保持原始的分离。它不是自动的,但它确实起作用。

你可以用这个来玩游戏,比如拉绳子 "has a substring" 对它进行改造 "has\s*a\s*substring" 让它不那么痛苦。

编辑 :合并了Yth的注释,说明元字符与换行符和Hobbs对我的用法的更正相匹配。

此模式将匹配您要查找的字符串:

(has\s+a\s+substring)

因此,当用户输入搜索字符串时,将搜索字符串中的任何空白替换为 \s+ 你有你的模式。把每一场比赛都换成 [match starts here]$1[match ends here] 哪里 $1 是匹配的文本。

在正则表达式中,可以使用 + 意思是“一个或多个”,所以像这样

/has\s+a\s+substring/

比赛 has 后接一个或多个空格字符,后接 a 后接一个或多个空格字符,后接 substring .

将它与替换运算符组合在一起,可以说:

my $str = "here is some text that has     a  substring that I'm interested in embedded in it.";
$str =~ s/(has\s+a\s+substring)/\[match starts here]$1\[match ends here]/gs;

print $str;

输出为:

here is some text that [match starts here]has     a  substring[match ends here] that I'm interested in embedded in it.

很多人建议,使用 \s+ 以匹配空白。以下是您自动完成的方法:

my $original = "here is some text that has a substring that I'm interested in embedded in it.";
my $search = "has a\nsubstring";

my $re = $search;
$re =~ s/\s+/\\s+/g;

$original =~ s/\b$re\b/[match starts here]$&[match ends here]/g;

print $original;

输出:

here is some text that [match starts here]has a substring[match ends here] that I'm interested in embedded in it.

您可能希望转义字符串中的任何元字符。如果有人感兴趣,我可以补充一下。

这是一个如何做到这一点的例子。

#! /opt/perl/bin/perl
use strict;
use warnings;

my $submatch = "has a\nsubstring";

my $str = "
here is some
text that has
a substring that I'm interested in, embedded in it.
";

print substr_match($str, $submatch), "\n";

sub substr_match{
  my($string,$match) = @_;

  $match =~ s/\s+/\\s+/g;

  # This isn't safe the way it is now, you will need to sanitize $match
  $string =~ /\b$match\b/;
}

这当前执行任何操作以检查 $match 不安全字符的变量。