下面是一个使用\K的正则表达式，它允许您替换第n次出现的字符串，而不会重复模式中的指针。如果您的搜索字符串是动态的，并且可能包含具有特殊含义的字符，则preg_quote()对于模式的完整性至关重要。
如果您想静态地将搜索字符串和第n个匹配项写入您的模式，可以是：

• (?:.*?\K ){8}
• 或者更有效地用于这种特殊情况：(?:[^ ]*\K ){8}

\K告诉正则表达式模式"忘记"全字符串匹配中任何先前匹配的字符。换句话说，"重新开始全字符串匹配"或"Keep from here"。在这种情况下，模式只保留第8个空格字符。
代码：（Demo）

function replaceNth(string $input, string $find, string $replacement, int $nth = 1): string {
    $pattern = '/(?:.*?\K' . preg_quote($find, '/') . '){' . $nth . '}/';
    return preg_replace($pattern, $replacement, $input, 1);
}

echo replaceNth($originalString, $findString, $newWord, $nthOccurrence);
// Hello world, what do you think of today's beautiful weather

另一个关于如何解决这个问题的观点是："如何在搜索字符串的第n个示例后插入新字符串？"这里有一个限制爆炸的非正则表达式方法，将新字符串添加到最后一个元素的前面，然后重新连接元素。（Demo）

$originalString = "Hello world, what do you think of today's weather"; 
$findString = ' ';
$nthOccurrence = 8;
$newWord = 'beautiful ';   // notice that leading space was removed

function insertAfterNth($input, $find, $newString, $nth = 1) {
    $parts = explode($find, $input, $nth + 1);
    $parts[$nth] = $newString . $parts[$nth];
    return implode($find, $parts);
}

echo insertAfterNth($originalString, $findString, $newWord, $nthOccurrence);
// Hello world, what do you think of today's beautiful weather

我在这里找到了答案-https://gist.github.com/VijayaSankarN/0d180a09130424f3af97b17d276b72bd

$subject = "Hello world, what do you think of today's weather"; 
$search = ' ';
$occurrence = 8;
$replace = ' nasty ';

/**
 * String replace nth occurrence
 * 
 * @param type $search      Search string
 * @param type $replace     Replace string
 * @param type $subject     Source string
 * @param type $occurrence  Nth occurrence
 * @return type         Replaced string
 */
function str_replace_n($search, $replace, $subject, $occurrence)
{

    $search = preg_quote($search);
    echo preg_replace("/^((?:(?:.*?$search){".--$occurrence."}.*?))$search/", "$1$replace", $subject);
}

str_replace_n($search, $replace, $subject, $occurrence);

$originalString = "Hello world, what do you think of today's weather"; 
$findString = ' ';
$nthOccurrence = 8;
$newWord = ' beautiful ';

$array = str_split($originalString);
$count = 0;
$num = 0;
foreach ($array as $char) {
    if($findString == $char){
        $count++;
    }
    $num++;
    if($count == $nthOccurrence){
        array_splice( $array, $num, 0, $newWord );
        break;
    }
}
$newString = '';
foreach ($array as $char) {
    $newString .= $char;
}

echo $newString;

我会考虑这样的事情：

function replaceNth($string, $substring, $replacement, $nth = 1){
  $a = explode($substring, $string); $n = $nth-1;
  for($i=0,$l=count($a)-1; $i<$l; $i++){
    $a[$i] .= $i === $n ? $replacement : $substring;
  }
  return join('', $a);
}
$originalString = 'Hello world, what do you think of today\'s weather';
$test = replaceNth($originalString, ' ', ' beautiful ' , 8);
$test2 = replaceNth($originalString, 'today\'s', 'good');

首先分解一个字符串的各个部分，然后将各个部分连接在一起，并与搜索字符串连接，但在特定的数字处与替换字符串连接（为了方便起见，这里的数字从0开始）：

function str_replace_nth($search, $replace, $subject, $number = 0) {
    $parts = explode($search, $subject);
    $lastPartKey = array_key_last($parts);
    $result = '';
    foreach($parts as $key => $part) {
        $result .= $part;
        if($key != $lastPartKey) {
            if($key == $number) {
                $result .= $replace;
            } else {
                $result .= $search;
            }
        }
    }
    return $result;
}

用法：

$originalString = "Hello world, what do you think of today's weather"; 
$findString = ' ';
$nthOccurrence = 7;
$newWord = ' beautiful ';
$result = str_replace_nth($findString, $newWord, $originalString, $nthOccurrence);