พี่ครับผม ให้ ajax ใช้เช็คชื่อซ้ำ แต่มันขึ้นว่า ใช้งานได้ตลอดเลยครับทั่งมีและไม่มีอ่ะครับ

<script type="text/javascript">
var HttPRequest = false;
     function doCallAjax() {
          HttPRequest = false;
          if (window.XMLHttpRequest) { // Mozilla, Safari,...
             HttPRequest = new XMLHttpRequest();
             if (HttPRequest.overrideMimeType) {
                HttPRequest.overrideMimeType('text/html');
             }
          } else if (window.ActiveXObject) { // IE
             try {
                HttPRequest = new ActiveXObject("Msxml2.XMLHTTP");
             } catch (e) {
                try {
                   HttPRequest = new ActiveXObject("Microsoft.XMLHTTP");
                } catch (e) {}
             }
          } 
          
          if (!HttPRequest) {
             alert('Cannot create XMLHTTP instance');
             return false;
          }
    
          var url = 'regitad.php';
          var pmeters = "adu=" + encodeURI( document.getElementById("adu").value);

            HttPRequest.open('POST',url,true);

            HttPRequest.setRequestHeader("Content-type", "application/x-www-form-urlencoded");
            HttPRequest.setRequestHeader("Content-length", pmeters.length);
            HttPRequest.setRequestHeader("Connection", "close");
            HttPRequest.send(pmeters);
            
            
            HttPRequest.onreadystatechange = function()
            {

                if(HttPRequest.readyState == 3)  // Loading Request
                {
                    document.getElementById("mySpan").innerHTML = "..";
                }

                if(HttPRequest.readyState == 4) // Return Request
                {
                    if(HttPRequest.responseText == 'Y')
                    {
                        window.location = 'AjaxPHPRegister3.php';
                    }
                    else
                    {
                        document.getElementById("mySpan").innerHTML = HttPRequest.responseText;
                    }
                }

                
            }
        }

</script>
<form action="insad.php?todo=insert" method="post" id="from" name="form1" onsubmit="JavaScript:return fncSubmit();">
                    <table class="table table-hover">
                        <tr>
                            <th>ชื่อ-สกุล:</th>
                            <td><input type="text" name="adn" onblur="checkText();"></td>
                        </tr>
                        <tr>
                            <th>ชื่อผู้ใช้:</th>
                            <td><input type="text" name="adu" id="adu" OnChange="JavaScript:doCallAjax();"><span id="mySpan"></span></td>
                        </tr>

                        <tr>
                            <th>รหัสผ่าน:</th>
                            <td><input type="password" name="adp"></td>
                        </tr>

                        <tr>
                            <th>ยืนยันรหัสผ่าน:</th>
                            <td><input type="password" name="cadp"></td>
                        </tr>
                        <tr>
                        <td></td>
                            <td><button  type="submit" class="btn btn-success">เพิ่มผู้ใช้งาน</button></td>
                        </tr>
                    </table>
                    </form>

$user = trim($_POST["adu"]);
  $strSQL = "SELECT * FROM tbl_admin WHERE ad_Name = '".$user."' ";
  $resultsql = mysqli_query($dblink, $strSQL);
  $rows = mysqli_fetch_array($resultsql);
  $num = mysqli_num_rows($resultsql);

  if($rows)
  { ?>
	  <i class="fa fa-times" style="color: #FF0000 " ></i><font style="color: #FF0000 ">มีชื่อผู้ใช้นี้อยู่แล้ว</font>
	  
 <? }else{ ?>
  <i  class="fa fa-check" style="color: #00CC00 "></i><font style="color: #00CC00 ">สามารถใช้ชื่อผู้ใช้นี้ได้</font>
<?  } ?>