regex - 如何匹配两个实例并在 perl 中使用正则表达式替换第一个实例？

Question

如果返回两个匹配项，我只想替换第一个实例。

例子:

$sen = "The quick brown fox jump over the lazy dog, fox is quick";

我怎样才能只匹配第一个 fox 并将其替换为 wolf，反之亦然。

输出:

The quick brown wolf jump over the lazy dog, fox is quick 

或:

The quick brown fox jump over the lazy dog, wolf is quick 

谢谢。

Answer 1

使用全局选项 /g 计算匹配项，将匹配项列表分配给空列表以获得计数(比临时变量更好)。首先替换是基本的。第二个使用一个计数器，该计数器在每次匹配时递增，并在适当的时间替换。

注意使用字边框\b防止误匹配，如firefox、foxy lady等

代码:

use strict;
use warnings;
use v5.10;   # only required for say, use print instead if your version is lower

my $fox  = 'fox';
my $wolf = 'wolf';
my $sen = "The quick brown fox jump over the lazy dog, fox is quick";

if ((()=$sen =~ /\b$fox\b/g) == 2) {   # counting matches 
    my $first = $sen;
    $first =~ s/\b$fox\b/$wolf/;       # replacing first
    my $second = $sen;
    my $i = 0;
    $second =~ s/\b$fox\b/ ++$i == 2 ? $wolf : $fox/eg;  # replacing second
    say for $first, $second;
}

输出:

The quick brown wolf jump over the lazy dog, fox is quick
The quick brown fox jump over the lazy dog, wolf is quick

如果你想要一个更可重用的代码，你可以用它做一个子程序。

my $second = replace_nr($sen, $fox, $wolf, 2);
...
sub replace_nr {
    my ($str, $find, $replace, $num) = @_;
    my $i = 0;
    $str =~ s/\b($find)\b/ ++$i == $num ? $replace : $find/eg;
    return $str;
}

然后您甚至可以将 sub 用于两个替换:

my $first  = replace_nr($sen, $fox, $wolf, 1);
my $second = replace_nr($sen, $fox, $wolf, 2);