如何修复不返回值的JS用户可用性

Hi all and thanks for yours time. I'm using a JS code that check users availablity. The code consists of two pages: index.php and username-check.php. After long attempts, now the code returns "checking...", but then does not return what is required in username-check.php. The writing remains on "checking ...". I've just included the library jquery-1.11.3 min, but doesn't work yet.

What can I do in order to fix this problem?

JS:

$(document).ready(function()

{

$("#name").keyup(function()

{
var name = $(this).val();

if(name.length > 3)
{
 $("#result").html('checking...');

 $.ajax({
  type : 'POST',
  url  : 'username-check.php',
  data : $(this).serialize(),
  success : function(data)
 {
   alert(data);
  $("#result").html(data);
 },
 error : function (j, t) { alert('ajax failed: ' + t); }
 });
 return false;

 }
 else
 {
 $("#result").html('');
 }
 });

});

HTML

          <div class="form-group">
            <div class="row" style="margin-left:40%;">Nome: <input class="form-control" type="text" name="name" id="name" placeholder="Inserisci Nome" style="width:40%!important;margin-left:5px;height:20px;font-size:small;" required><div class="disponibile"><span id="result"></span></div></div>
          </div>

PHP

$host="localhost";
$user="xxx";
$pass="";
$dbname="xxx";

$myconn = mysql_connect($host, $user, $pass) or die('Errore...');
mysql_select_db($dbname, $myconn) or die('Errore...');

  if($_POST)
  {
  $name     = strip_tags($_POST['name']);

  $stmt=$dbcon->prepare("SELECT * FROM utenti WHERE nome=:name");
  $stmt->execute(array(':name'=>$name));
  $count=$stmt->rowCount();

  if($count>0)
  {
      echo "<span style='color:brown;'>Il nome è già attualmente in uso.</span>";
  }
  else
  {
      echo "<span style='color:green;'>Il nome è disponibile!    
 </span>";
  }
 }