无限滚动重复相同的结果

Question

我正在尝试为像facebook这样的用户朋友系统制作一个无限滚动页面。但我有一个问题。我的PHP代码重复相同的结果。我在这里想念的是谁能在这方面帮助我？

    <div id="dataList">
        <div data-Fid="1"></div>
        <div data-Fid="12"></div>
        <div data-Fid="9"></div>
        <div data-Fid="26"></div>
        <div data-Fid="27"></div> <-- lastID ==> friend_id
    </div>
<div class="p_user" data-kid="5"></div>

我的ajax代码：

$(window).scroll(function () { 
   if (scrollKLoad && $(window).scrollTop() >= $(document).height() - $(window).height() - 300) {
     scrollKLoad = false;
     var ID = $('#dataList').children().last().attr('data-Fid'); 
     var P_ID = $('.p_user').attr('data-kid');
     var newcontent = '<div class="postBody"><div class="glbD0w">No more user</div></div>';
     var dataString = "lastID=" + ID +"&pID=" + P_ID;
     if (ID) {
      $.ajax({
        type: "POST",
        url: '/requests/morefriend.php',
        data: dataString,
        cache: false,
        beforeSend: function() {
          // Do something
        },
        success: function(html) {
            if(html) {
              $("#dataList").append(html);
             } else {
            $('#dataList').append(newcontent);  
         }

        }
      });
    } else {
      $("#more").html('The End');
    }
    return false;
   }
});

和我的PHP代码：

     public function FolloWingUserList($uid,$lastID) {
         $uid=mysqli_real_escape_string($this->db,$uid);

         if($lastID) {
           $lastID = "AND F.friend_id > '".$lastID."'"; 
         }
         // The query to select for showing user details
         $query=mysqli_query($this->db,"SELECT U.username,U.name,U.uid,F.friend_id FROM users U, friends F WHERE U.status='1' AND U.uid=F.friend_two AND F.friend_one='$uid' AND F.friend_one='$uid' AND F.role='fri' $lastID ORDER BY U.uid DESC LIMIT " .$this->perpage)or die(mysqli_error($this->db));
           while($row=mysqli_fetch_array($query, MYSQLI_ASSOC)) {
             $data[]=$row;
           }
         if(!empty($data)) {
         // Store the result into array
           return $data;
        }     
     }