Question

我是角度js的新手，我想使用php从数据库中获取数据。我尝试以下代码。

          <html>
           <head>
              <script src = "http://ajax.googleapis.com/ajax/libs/angularjs/1.2.15/angular.min.js"></script>
<script>
   var myApp = angular.module('myApp',[]);
   myApp.controller('studentController', ['$scope','$http', function ($scope,$http) {
                  var url = "ajax.php";
                  $http.get(url).success( function(response) {
                      $scope.students = response;
                      });
               }]);


              </script>
           </head>
           <body>
              <div ng-app = "myApp" ng-controller = "studentController">
                 <table>
                    <tr>
                       <th>Name</th>
                    </tr>

                    <tr ng-repeat = "student in students">
                       <td>{{ student.Name }}</td>
                    </tr>
                 </table>
              </div>
           </body>
        </html>

ajax.php 这段代码给我以下输出：

[{"Name":"Mahesh","Roll":"122222"},{"Name":"ajay","Roll":"444433"}]

但我无法在前端显示...谢谢。

<?php
    $con = mysqli_connect("localhost","root","","adworld");
    $res= mysqli_query($con,"SELECT * FROM students");
    $result = array();
    while($row=mysqli_fetch_assoc($res)){
        $result[] = $row;
    }   
    echo json_encode($result);
?>

Answer 1

工作代码：

<!DOCTYPE html>
<html >
<script src= "angular.js"></script>
<body>

<div ng-app="myApp" ng-controller="customersCtrl">

<table>
  <tr>
     <th>Name</th>
     <th>Roll</th>
  </tr>
  <tr ng-repeat="x in names">
    <td>{{ x.Name }}</td>
    <td>{{ x.Roll}}</td>
  </tr>
</table>

</div>

<script>
var app = angular.module('myApp', []);
app.controller('customersCtrl', function($scope, $http) {
   $http.get("ajax.php")
   .success(function (response) {$scope.names = response;});
});
</script>

</body>
</html>

ajax.php

<?php
    $conn = new mysqli("localhost","root","","adworld");
    $result = $conn->query("SELECT * FROM students");
    $arr = array();
    while($rs = $result->fetch_assoc()) 
    {
      $arr[] = $rs;
    }

    echo json_encode($arr);
    ?>

使用php从角度js中获取数据库中的数据

1 个答案: