如何从某个位置的字符串中删除字符？

Question

我有变量$pos,$toRemove和$line。我想从$pos位置删除此字符串$toRemove。

$line = "Hello kitty how are you kitty kitty nice kitty"; 
$toRemove = "kitty"; 
$pos = 30; # the 3rd 'kitty' 

我要检查，如果从30位有串kitty，我想正是删除此一个。

你能给我一个解决方案吗？我可以使用大量的循环和变量，但它看起来很奇怪，而且工作非常慢。

Answer 1

if (substr($line, $pos, length($toRemove)) eq $toRemove) { 
    substr($line, $pos, length($toRemove)) = ""; 
} 

Answer 2

$line = "Hello kitty how are you kitty kitty nice kitty"; 
$toRemove = "kitty"; 
$pos = 30; # the 3rd 'kitty' 

pos($line) = $pos; 
$line =~ s/\G$toRemove//gc; 
print $line; 

输出：

Hello kitty how are you kitty nice kitty 

Answer 3

还有一种方法：

$line = "Hello kitty how are you kitty kitty nice kitty"; 
$toRemove = "kitty"; 
$pos = 30; 

$line =~ s/(.{$pos})$toRemove/$1/; 
print $line; 

结果：

Hello kitty how are you kitty nice kitty 

Answer 4

$line =~ s/^.{30}\K$toRemove//; 

这将使用一个后置断言来匹配前30个字符，而不会将其包含在替换的模式部分中。

Answer 5

的[pos] [POS]运营商是一个左值只是这样的事情：

[POS]：

use strict; 
use warnings; 

my $line = "Hello kitty how are you kitty kitty nice kitty"; 
my $toRemove = "kitty"; 
my $pos = 30; 

pos($line) = $pos; 

$line =~ s/\G$toRemove//; 

print $line; 

输出

Hello kitty how are you kitty nice kitty