删除包含多行字符串中的单词的内容答案

【问题标题】：Remove the stuff up to and including the word in a string with multiple lines删除包含多行字符串中的单词的内容
【发布时间】：2016-08-30 13:07:14
【问题描述】：

我有以下多行字符串。我不能让它成为一行，因为它来自命令输出。例如：

my $myString = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s,
when an unknown printer took a galley of type and scrambled it to make a 
type specimen book. It has survived not only five centuries, but also 
the leap into electronic typesetting, remaining essentially unchanged. 
It was popularised in the 1960s"

我想去掉包括“scrambled”这个词在内的东西

我在下面尝试过，但似乎不起作用。

if($myString =~ 's/.*(scrambled)//s')
{
  print "Match:   <$&>\n";
}

【问题讨论】：

标签： linux perl shell

【解决方案1】：

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

my $myString = "Lorem Ipsum is simply dummy text of the printing and typesetting industry. 
Lorem Ipsum has been the industry's standard dummy text ever since the 1500s,
when an unknown printer took a galley of type and scrambled it to make a 
type specimen book. It has survived not only five centuries, but also 
the leap into electronic typesetting, remaining essentially unchanged. 
It was popularised in the 1960s";

$myString =~ s/.*\bscrambled\b//s;

print $myString;

Demo

【讨论】：

如果字符串包含unscrambled...怎么办？更好的是我们将使用\bscrambled\b。

【解决方案2】：

尝试去掉引号：

if($myString =~ s/.*(scrambled)//s)

通过引号，它试图匹配文字 s/.*(scrambled)//s 而不试图改变任何东西。如果没有引号，将看到“s”替代品。

【讨论】：

【解决方案3】：

使用可以使用pre & post。

$myString =~ s/\bscrambled\b//ig && print $`.$';

否则

if($myString =~ s/\bscrambled\b//s)
{
    print "Pre: <$`>\n";
    print "Match:   <$&>\n";
    print "Post:   <$'>\n";
}

【讨论】：