PHP Ajax Dependent Dropdown List Example

By Hardik Savani November 5, 2023 Category : PHP jQuery MySql Ajax

In this post, i am going to share with you how to make dynamic dependent dropdown list using jquery ajax with php MySQL. In this simple example through we understand how to work dependent dropdown in core PHP even if you beginner. I also posted for php laravel framework How to make simple dependent dropdown using jquery ajax in Laravel 5?.

We sometimes require to make dependent dropdown like when state select at that time bellow city drop down list should change, i mean related to selected state. In this example i have two tables and there are listed bellow:

1.demo_state

2.demo_cities

So, when user will change state at that time, dynamically change city drop down box from database. you can implement this example in your application by follow bellow few step.


Step 1: Create Tables and Dummy Data

In first step we require to create database and tables for dependent dropdown example. So first you require to create "test" database in your phpmyadmin. After created database successfully we require to create two new table "demo_state" and "demo_cities" using following bellow SQL Query.

demo_state table:

CREATE TABLE `demo_state` (
  `id` int(11) NOT NULL,
  `name` varchar(155) NOT NULL,
  `created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `updated_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00'
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

demo_cities table:

CREATE TABLE `demo_cities` (
  `id` int(11) NOT NULL,
  `state_id` int(12) NOT NULL,
  `name` varchar(155) NOT NULL,
  `created_at` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
  `updated_at` timestamp NOT NULL DEFAULT '0000-00-00 00:00:00'
) ENGINE=InnoDB DEFAULT CHARSET=latin1;

Now, in your database, you will have two tables "demo_state" and "demo_cities". Now we require to add some dummy data in both table like as bellow image.

demo_state

demo_cities

Step 2: Create index.php file

In this step we will create index.php file, in this file we write code for state drop-down and city drop-down. In this file layout will display so put bellow code on index.php file.

index.php

<!DOCTYPE html>
<html>
<head>
    <title>PHP - How to make dependent dropdown list using jquery Ajax?</title>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.1.1/jquery.js"></script>
    <script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
    <link rel="stylesheet" href="http://demo.itsolutionstuff.com/plugin/bootstrap-3.min.css">
</head>
<body>

<div class="container">
    <div class="panel panel-default">
      <div class="panel-heading">Select State and get bellow Related City</div>
      <div class="panel-body">
            <div class="form-group">
                <label for="title">Select State:</label>
                <select name="state" class="form-control">
                    <option value="">--- Select State ---</option>

                    <?php
                        require('db_config.php');
                        $sql = "SELECT * FROM demo_state"; 
                        $result = $mysqli->query($sql);
                        while($row = $result->fetch_assoc()){
                            echo "<option value='".$row['id']."'>".$row['name']."</option>";
                        }
                    ?>

                </select>
            </div>

            <div class="form-group">
                <label for="title">Select City:</label>
                <select name="city" class="form-control" style="width:350px">
                </select>
            </div>

      </div>
    </div>
</div>

<script>
$( "select[name='state']" ).change(function () {
    var stateID = $(this).val();

    if(stateID) {

        $.ajax({
            url: "ajaxpro.php",
            dataType: 'Json',
            data: {'id':stateID},
            success: function(data) {
                $('select[name="city"]').empty();
                $.each(data, function(key, value) {
                    $('select[name="city"]').append('<option value="'+ key +'">'+ value +'</option>');
                });
            }
        });

    }else{
        $('select[name="city"]').empty();
    }
});
</script>

</body>
</html>

Read Also: Display Loading Image when AJAX Call is in Progress Example

Step 3: Add DB Configuration File

Now we require to create db configuration file for database, that way you can set username, password, database and host. So let's create db_config.php file and put bellow code:

db_config.php 

<?php

define (DB_USER, "root");
define (DB_PASSWORD, "root");
define (DB_DATABASE, "test");
define (DB_HOST, "localhost");

$mysqli = new mysqli(DB_HOST, DB_USER, DB_PASSWORD, DB_DATABASE);

?>

Step 4: Add Ajax File

In last step, we need to create ajax pro file, in this file we write code for dynamic ajax city drop-down json data. So let's create ajaxpro.php file and put bellow code:

ajaxpro.php 

<?php

   require('db_config.php');

   $sql = "SELECT * FROM demo_cities
         WHERE state_id LIKE '%".$_GET['id']."%'"; 

   $result = $mysqli->query($sql);

   $json = [];
   while($row = $result->fetch_assoc()){
        $json[$row['id']] = $row['name'];
   }

   echo json_encode($json);
?>

Ok, now we are ready to run our example. So let's run bellow command on your root directory for quick run:

php -S localhost:8000

Now you can open bellow URL on your browser:

Read Also: How to Remove Multiple Keys from PHP Array?
http://localhost:8000

I hope it can help you...

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