How to display URL result ouside search drop down?

JavaScript
3 
<?php $dbhost="localhost";
 $dbname="tutorial";
 $dbuser="root";
 $dbpass="root";
 global $tutorial_db;
 $tutorial_db=new mysqli();
 $tutorial_db->connect($dbhost,
$dbuser,
$dbpass,
$dbname);
 $tutorial_db->set_charset("utf8");
 if ($tutorial_db->connect_errno) {
  printf("Connect failed: %s\n", $tutorial_db->connect_error);
 exit();
}
$html='';
 $html .='<li class="result">';
 $html .='<a target="_blank" href="urlString">';
 $html .='<h3>nameString</h3>';
 $html .='<h4>functionString</h4>';
 $html .='</a>';
 $html .='</li>';
 $search_string=preg_replace("/[^A-Za-z0-9]/",
" ",
$_POST['query']);
 $search_string=$tutorial_db->real_escape_string($search_string);
 if (strlen($search_string) >=1 && $search_string !==' ') {
  $query='SELECT * FROM search WHERE function LIKE "%'.$search_string.'%" OR name LIKE "%'.$search_string.'%"';
  $result=$tutorial_db->query($query);
  while($results=$result->fetch_array()) {
    $result_array[]=$results;
  }
  if (isset($result_array)) {
    foreach ($result_array as $result) {
      $display_function=preg_replace("/".$search_string."/i", "<b class='highlight'>".$search_string."</b>", $result['function']);
      $display_name=preg_replace("/".$search_string."/i", "<b class='highlight'>".$search_string."</b>", $result['name']);
      $display_url='http://php.net/manual-lookup.php?pattern='.urlencode($result['function']).'&lang=en';
      $output=str_replace('nameString', $display_name, $html);
      $output=str_replace('functionString', $display_function, $output);
      $output=str_replace('urlString', $display_url, $output);
      echo($output);
    }
  }
  else {
    $output=str_replace('urlString', 'javascript:void(0);', $html);
    $output=str_replace('nameString', '<b>No Results Found.</b>', $output);
    $output=str_replace('functionString', 'Sorry :(', $output);
    echo($output);
  }
}
?>