How to Auto populate dropdown with jQuery AJAX

Sometimes it requires auto-populate data on the element based on the selection of another element e.g. City names based on a state.

You can do this with only PHP but it required you to submit every time on selection.

This solves the problem but it is a little frustrating because it submits every time even if the selection is right or wrong.

For making it better you can use AJAX with jQuery which loads new data and removes the old one on each selection.

In the demonstration, I am creating a Department drop-down list, and based on the option selection show all existing users of that department on another Dropdown.

How to Auto populate dropdown with jQuery AJAX


  1. Table structure
  2. Configuration
  3. HTML
  4. PHP
  5. jQuery
  6. Demo
  7. Conclusion

1. Table structure

I am using 2 tables in the example –

department Table –

CREATE TABLE `department` (
  `depart_name` varchar(80) NOT NULL

users Table –

CREATE TABLE `users` (
  `username` varchar(80) NOT NULL,
  `name` varchar(80) NOT NULL,
  `email` varchar(80) NOT NULL,
  `department` int(11) NOT NULL

2. Configuration

Create a config.php to define database connection.

Completed Code

$host = "localhost"; /* Host name */
$user = "root"; /* User */
$password = ""; /* Password */
$dbname = "tutorial"; /* Database name */

$con = mysqli_connect($host, $user, $password,$dbname);
// Check connection
if (!$con) {
   die("Connection failed: " . mysqli_connect_error());


Creating two Dropdown elements one –

  • Fetch records from department table and use to add <option> in <select id='sel_depart'>  and
  • Another dropdown shows the users names which are filled with jQuery based on the department name selection from the first dropdown element.

Completed Code

include "config.php";

<div>Departments </div>
<select id="sel_depart">
   <option value="0">- Select -</option>
   // Fetch Department
   $sql_department = "SELECT * FROM department";
   $department_data = mysqli_query($con,$sql_department);
   while($row = mysqli_fetch_assoc($department_data) ){
      $departid = $row['id'];
      $depart_name = $row['depart_name'];
      // Option
      echo "<option value='".$departid."' >".$depart_name."</option>";
<div class="clear"></div>

<div>Users </div>
<select id="sel_user">
   <option value="0">- Select -</option>

4. PHP

  • Create getUsers.php file to handle AJAX requests.
  • Fetching users based on department selection from the users table.
  • Initialize $users_arr Array with userid and name.
  • Return $users_arr Array in JSON format.

Completed Code


include "config.php";

$departid = 0;

   $departid = mysqli_real_escape_string($con,$_POST['depart']); // department id

$users_arr = array();

if($departid > 0){
   $sql = "SELECT id,name FROM users WHERE department=".$departid;

   $result = mysqli_query($con,$sql);

   while( $row = mysqli_fetch_array($result) ){
      $userid = $row['id'];
      $name = $row['name'];

      $users_arr[] = array("id" => $userid, "name" => $name);
// encoding array to json format
echo json_encode($users_arr);

5. jQuery

Sending AJAX request when an option selected from the first drop-down. Pass the selected option value as data and on successful callback fill <select id='sel_user'> with response.

Completed Code


        var deptid = $(this).val();

            url: 'getUsers.php',
            type: 'post',
            data: {depart:deptid},
            dataType: 'json',

                var len = response.length;

                for( var i = 0; i<len; i++){
                    var id = response[i]['id'];
                    var name = response[i]['name'];
                    $("#sel_user").append("<option value='"+id+"'>"+name+"</option>");



6. Demo

View Demo

7. Conclusion

In the example, I used dropdown element for autofill but you can do this with any other element like TextBox, Textarea, etc. that you can fill with information according to input on another element.

You can view this tutorial to know to auto-populate dropdown with PDO and PHP.

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