为星号代替粗话，并有星号的数量相匹配的粗话的字母数

Question

我有下面的代码，其正在努力从一个字符串过滤出粗话，并与askerisks取代他们，但是我想数的askerisks等于粗俗单词中的字母数量。例如，如果“屁股”一词被审查，那么它将被三个askerisks取代。我如何修改此代码以实现此目的？谢谢。

$naughtyWords = array("ahole","anus","ash0le","ash0les","asholes","ass"); //etc 

foreach ($naughtyWords as &$word) { 
    $word = ' '.$word.' '; 
} 

$string = str_replace($naughtyWords, " **** ", ' '.$string.' '); 

Answer 1

尝试：

$naughtyWords = array("ahole","anus","ash0le","ash0les","asholes","ass"); //etc 

foreach ($naughtyWords as $word) { 
    $replacement = str_repeat('*', strlen($word)); 
    $string = str_replace(' '.$word.' ', $replacement, $string); 
} 

Answer 2

试试这个：

$naughty_words = array('ahole', 'anus', 'ash0le', 'ash0les', 'asholes', 'ass'); 
$string = 'classical music ass dirty ass. molass'; 

foreach ($naughty_words as $naughty_word) { 
    $string = preg_replace_callback('#\b' . $naughty_word . '\b#i', function($naughty_word) {return str_repeat('*', strlen($naughty_word[0]));}, $string); 
}