AJAX Call - recieve a Notice: Undefined index

Question

I am trying to refresh a table of members using AJAX, and I am receiving a Notice saying that the variable that I am passing via AJAX is not Undefined. I have two files, an index.php and load-members.php, the goal is to show the first 5 members in the index page then in each time I click a button a have more members shown so find bellow the script and the file. This is my Script

    <script>
$(document).ready(function() {
  var membersCount = 5;
    $("#btn").click(function() {
      membersCount = membersCount + 5;
      $("#members").load("load-members.php", {
          membersNew: membersCount
      });
    });  
});
</script>

The load-members.php file is :

    <?php
include 'includes/db.inc.php';
$membersNew = $_POST['membersNew'];
echo $membersNew;
 // e$membersNewcho mysqli_num_rows($query);
  $sql = "select * from users LIMIT $membersNew";
  $query = mysqli_query($conn,$sql);
   while ( $row_admin = mysqli_fetch_assoc($query)) {

  echo"<tr>
      <th scope='row'>".$row_admin['user_id']."</th>
      <td>".$row_admin['user_username']."</td>
      <td>".$row_admin['user_email']."</td>
      <td>".$row_admin['score']."</td>
      <td>".$row_admin['acc']."%</td>

    </tr>";

}
?>

`$().load` sounds like a GET request. So I guess it's in `$_GET['membersNew']` — Jeff, May 01 '18 at 19:05
do a `var_dump($_POST);` and `var_dump($_GET);` to check what you've got where! — Jeff, May 01 '18 at 19:06
This is the message Notice: Undefined index: membersNew in C:\xampp\htdocs\projectnew\load-members.php on line 3 — Mohdows, May 01 '18 at 19:07
Please open the developer tools and check the network tab when you call `load-members.php` - Check the request type (POST/GET) and the actual parameters passed in the request — Alon Eitan, May 01 '18 at 19:09
after @AlonEitan's test here's another possibility: check if you got smth in `var_dump(file_get_contents("php://input"));` — Jeff, May 01 '18 at 19:15

AJAX Call - recieve a Notice: Undefined index

0 Answers0