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multiple select dropdown list with ajax in php but keep 2 value selected for 1 result

I want to pass the value from first and second selects as a condition for the last select, but I can't. I tried creating a function, but it didn't work. or if anyone knows another way to do it, please let me know because I've been trapped here for a week. Any help would be appreciated, thank you!


script

$('#levelk').change(function() {
   var id_levelk = $(this).val();
     $.ajax({
     type: "POST",
     url: "ajax_db.php",
     data: {id:id_levelk,function:'levelk'},
     success: function(data){
        $('#levelr').html(data); 
     }
   });
 });
  $('#levelr').change(function() {
   var id_levelr= $(this).val();
     $.ajax({
     type: "POST",
     url: "ajax_db.php",
     data: {id:id_levelr,function:'levelr'},
     success: function(data){
       $('#kon').html(data);
     }
   });
}); 

ajax_db

function levelkadd($add) {
    global $levelk_value;
    $levelk_value = $add;
  }
function levelkSend() {
      global $levelk_value;
      return $levelk_value;
    }
if (isset($_POST['function']) && $_POST['function'] == 'levelk') {
    $id = $_POST['id'];
    levelkadd($id);
    $sql = "SELECT major.name,major.id FROM `major` 
    JOIN faculy
    ON faculy.Id = major.faId 
    WHERE faculy.Id = '$id'";
    $query = mysqli_query($con, $sql);
    echo '<option value="" selected disabled>-กรุณาเลือกสาขา-</option>';
    foreach ($query as $value2) {
      echo '<option value="'.$value2['id'].'">'.$value2['name'].'</option>';    
    }
  }
 if (isset($_POST['function']) && $_POST['function'] == 'levelr') {
    $id = $_POST['id'];
    $sql = "SELECT * FROM `admisinfo` 
    JOIN major
    ON major.Id = admisinfo.maId 
    WHERE major.Id = '$id'";
    $query3 = mysqli_query($con, $sql);
    $result = mysqli_fetch_assoc($query3);
    echo 'จำนวนนักศึกษาที่รับ'.strlen(levelkSend()).''.$result['unit'].'';
    exit();
  }
about 4 years ago · Juan Pablo Isaza
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