Perl RegExp：存在特定字符串时删除部分字符串

Question

如果字符串包含Potatoes或Peaches，如何应用Perl RegExp删除字符串的第一部分？

如果可能，请不要使用if / else条件，而只能使用RegExp。

Input:
Apples Peaches Grapes 
Spinach Tomatoes Carrots
Corn Potatoes Rice

Output:
Peaches Grapes 
Spinach Tomatoes Carrots 
Potatoes Rice

这是我的代码：

#! /usr/bin/perl
use v5.10.0;
use warnings;

$string1 = "Apples Peaches Grapes ";
$string2 = "Spinach Tomatoes Carrots";
$string3 = "Corn Potatoes Rice";

#Use RegExp to output strings with first word deleted  
#if it includes either Peaches or Rice.

$string1 =~ s///;
$string2 =~ s///;
$string2 =~ s///;


say $string1;
say $string2;
say $string3;

Answer 1

您可以使用以下表达式：

^(?=.*\bPeaches\b|.*\bPotatoes\b)\S+\s

• ^字符串的开头。
• (?=.*\bPeaches\b|.*\bPotatoes\b)正向查找，确保字符串中存在Peaches或Potatoes子字符串。
• \S+\s匹配所有非空白字符，后跟空白。

正则表达式演示here。

---

Perl 演示：

use feature qw(say);

$string1 = "Apples Peaches Grapes";
$string2 = "Spinach Tomatoes Carrots";
$string3 = "Corn Potatoes Rice";

$string1 =~ s/^(?=.*\bPeaches\b|.*\bPotatoes\b)\S+\s//;
$string2 =~ s/^(?=.*\bPeaches\b|.*\bPotatoes\b)\S+\s//;
$string2 =~ s/^(?=.*\bPeaches\b|.*\bPotatoes\b)\S+\s//;


say $string1;
say $string2;
say $string3;

打印：

Peaches Grapes
Spinach Tomatoes Carrots
Corn Potatoes Rice