为什么PHP回调要求我提供请求参数2？

Question

我正在处理旧代码，却遇到了-失败：检查并替换字符串部分

 private function CleanHtml($string, $lenght)
{
  $string = strip_tags($string);                      
  $string = substr($string,0,$lenght);                    
  $search = array ("'<script[^>]*?>.*?</script>'si",   
                   "'<[\/\!]*?[^<>]*?>'si",            
                   "'([\r\n])[\s]+'",                 
                   "'([\r])+'",                        
                   "'([\n])+'",                        
                   "'&(quot|#34);'i",                 
                   "'&(amp|#38);'i",
                   "'&(lt|#60);'i",
                   "'&(gt|#62);'i",
                   "'&(nbsp|#160);'i",
                   "'&(iexcl|#161);'i",
                   "'&(cent|#162);'i",
                   "'&(pound|#163);'i",
                   "'&(copy|#169);'i",
                   "'&#(\d+);'e");                      

  $replace = array (
             "",
             "",
             "",
             "",
             "",
             "",
             "&",
             "<",
             ">",
             " ",
             chr(161),
             chr(162),
             chr(163),
             chr(169),
             "chr(\\1)");

  $description = preg_replace_callback($search, $replace, $string);


  return $description;
}

因为它给我错误“ preg_replace_callback（）：要求参数2'Array'是有效的回调”？