PHP AngularJS populate dynamic dropdown example

By Hardik Savani | April 18, 2018 | | 38000 Viewer | Category : PHP Bootstrap jQuery MySql AngularJS


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: JQuery - Display loading image on ajax call example

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

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<?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...



Hardik Savani
My name is Hardik Savani. I'm a full-stack developer, entrepreneur and owner of Aatman Infotech. I live in India and I love to write tutorials and tips that can help to other artisan. I am a big fan of PHP, Javascript, JQuery, Laravel, Codeigniter, VueJS, AngularJS and Bootstrap from the early stage.
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***Do you want me hire for your Project Work? Then Contact US.

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