如果返回两个匹配项,我想只替换第一个实例。
示例:
$sen = "The quick brown fox jump over the lazy dog, fox is quick";
我怎样才能匹配第一个狐狸并将其替换为狼,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
感谢。
答案 0 :(得分:3)
使用全局选项/g
计算匹配项,将匹配列表分配给空列表以获取计数(优于临时变量)。第一次更换是基本的。第二个使用一个计数器,为每个匹配增加,并在适当的时间替换。
请注意使用单词边框\b
来防止错误匹配,例如firefox
,foxy lady
等。
<强>代码:强>
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;
}
<强>输出:强>
The quick brown wolf jump over the lazy dog, fox is quick
The quick brown fox jump over the lazy dog, wolf is quick
如果您希望有更多可重复使用的代码,可以从中创建一个子程序。
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;
}
然后你甚至可以使用sub进行两次替换:
my $first = replace_nr($sen, $fox, $wolf, 1);
my $second = replace_nr($sen, $fox, $wolf, 2);
答案 1 :(得分:2)
检查是否有(至少)两个并替换第一个:
# The quick brown *wolf* jump over the lazy dog, fox is quick
s/fox(?=.*fox)/wolf/s;
检查是否有(至少)两个并替换第二个:
# The quick brown fox jump over the lazy dog, *wolf* is quick:
s/fox.*\Kfox/wolf/s;
s/(fox.*)fox/${1}wolf/s; # Slower, but works pre 5.10
答案 2 :(得分:1)
默认情况下,替换运算符的确如此:匹配并替换第一次出现。
以下语句将为您提供第一个输出行:
s/wolf/fox/
答案 3 :(得分:0)
您的新问题是您想要点数
my $count = () = $text =~ /fox/g;
并替换第N个实例
my $n = 3;
my $nm1 = $n-1;
$text =~ s/(?:fox.*){$nm1}\Kfox/s;