PHP AngularJS Populate Dynamic Dropdown Example

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

Sometime, we may need to create dynamic dropdown in our angular application using PHP or using API. here i will give you small example for populate dropdown dynamically from database using php mysql. you can simply create country state city drop down list using angular.

Here, we will bind two select box state and city. so here we will create two tables called "demo_state" and "demo_cities". so here i will explain step by step. you need to just create two tables with bellow sql query and then add some dummy records.

So let's follow bellow three step to complete full example.

Step 1: Create Tables

now we need to create two tables "demo_state" and "demo_cities". you can follow both tables mysql query for create tables.

demo_state table query:

CREATE TABLE IF NOT EXISTS `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 query:

CREATE TABLE IF NOT EXISTS `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;

Step 2: Create index.php file

index.php

<!DOCTYPE html>
 <html>  
  <head>  
       <title>PHP AngularJS populate dynamic dropdown example - ItSolutionStuff.com</title>
       <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.5.7/angular.min.js"></script>  
       <link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.6/css/bootstrap.min.css" />  
  </head>  

  <body>  
     <div class="container" style="width:500px;">  
          <h3 align="center">PHP AngularJS populate dynamic dropdown example - ItSolutionStuff.com</h3>  
          <br />  
          <div ng-app="myapp" ng-controller="usercontroller" ng-init="loadState()">  
             <select name="state" ng-model="state" class="form-control" ng-change="loadCity()">  
                  <option value="">Select state</option>  
                  <option ng-repeat="state in states" value="{{state.id}}">{{state.name}}</option>  
             </select>  
             <br />  
             <select name="city" ng-model="city" class="form-control">  
                  <option value="">Select city</option>  
                  <option ng-repeat="city in cities" value="{{city.id}}">  
                       {{city.name}}  
                  </option>  
             </select>  
          </div>  
     </div>  
  </body>  

 </html>  

 <script>  

 var app = angular.module("myapp",[]);  

 app.controller("usercontroller", function($scope, $http){  

      $scope.loadState = function(){  
           $http.get("load_state.php")  
           .success(function(data){  
                $scope.states = data;  
           })  
      }  

      $scope.loadCity = function(){  
           $http.post("load_city.php", {'state_id':$scope.state})  
           .success(function(data){  
                $scope.cities = data;  
           });  
      }  

 });  

 </script>

Read Also: PHP AngularJS CRUD with Search and Pagination Tutorial

Step 3: Create API files

load_state.php

<?php  

	 $connect = mysqli_connect("localhost", "root", "password", "database_name");  

	 $output = [];  

	 $query = "SELECT * FROM demo_state ORDER BY name ASC";  

	 $result = mysqli_query($connect, $query);  

	 while($row = mysqli_fetch_array($result))  
	 {  
	      $output[] = $row;  
	 }  

	 echo json_encode($output);  

?>

load_city.php

Read Also: AngularJS Ajax Autocomplete Using Typeahead in PHP Example
<?php  

	 $connect = mysqli_connect("localhost", "root", "password", "database_name");   

	 $output = [];  

	 $data = json_decode(file_get_contents("php://input"));  

	 $query = "SELECT * FROM demo_cities WHERE state_id='".$data->state_id."' ORDER BY name ASC";  

	 $result = mysqli_query($connect, $query);  

	 while($row = mysqli_fetch_array($result))  
	 {  
	      $output[] = $row;
	 }  

	 echo json_encode($output);  
?>

I hope it can help you...

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