AngularJS实现ajax请求的方法

时间:2020-05-15 栁罗风尘人气:0

本文实例讲述了AngularJS实现ajax请求的方法。分享给大家供大家参考，具体如下：

【HTML 代码】

<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <meta name="viewport" content="width=device-width,user-scalable=no, initial-scale=1">
  <link rel="stylesheet" type="text/css" href="" />
  <title>angularjs实现 ajax</title>
</head>
<body ng-app="HelloAjax">
  <div ng-controller="HelloAjax">
    <form>
      <input type="text" ng-model="username" />
      <input type="text" ng-model="email" />
    </form>
    <table>
     <tr ng-repeat="user in users">
       <td>{{user.username}}</td>
       <td>{{user.email}}</td>
     </tr>
    </table>
    <button ng-click="get_more();">get more</button>
  </div>
</body>
<script type="text/javascript" src="./js/angular.min.js" charset="utf-8"></script>
  <script type="text/javascript" src="ajax.js" charset="utf-8"></script>
</html>

【js代码 ajax.js】

var myModule = angular.module("HelloAjax",[]);
myModule.controller("HelloAjax",["$scope","$http",function HelloAjax($scope,$http){
  /*
  $scope.users=[{'username':"zhangsan","email":"zs@11.com"},
    {'username':"zhangsan2","email":"zs@22.com"},
    {'username':"zhangsan3","email":"zs@33.com"}];
  */
  $scope.get_more = function(){
    $http({
        method: "POST",
        url: "./ajax.php",
        data:{'username':$scope.username,
           'email':$scope.email
          }
      }).
      success(function(data, status) {
       //$scope.status = status;
        $scope.users = data;
      }).
      error(function(data, status) {
       //$scope.data = data || "Request failed";
       //$scope.status = status;
     });
   }
}]);

【PHP代码 ajax.php】

<?php
//获取参数
$data = file_get_contents("php://input");
$user = json_decode($data);
//查询数据库
$conn = mysql_connect("localhost","root","");
mysql_select_db("test");
$sql ="select username,email from users ";
$res = mysql_query($sql,$conn);
$users = array();
while($row = mysql_fetch_assoc($res)){
  $users[] = $row;
}
//当然这里简化了插入数据库
$users[] = array('username'=>$user->username,
         'email'=>$user->email);
//返回数据库
echo json_encode($users);

希望本文所述对大家AngularJS程序设计有所帮助。

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