使用php從角度js的數據庫中獲取數據

我是角js中的新角色，我想使用php從數據庫中獲取數據。我嘗試下面的代碼。使用php從角度js的數據庫中獲取數據

  <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 這個代碼給我下面的輸出：

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

但我不能夠顯示在前端...謝謝。

<?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); 
?>

來源

2015-11-11 Mahesh Jagdale

你需要在你的應用程序註冊您的控制器 - 'myApp.controller（「studentController」，[ '$範圍' ，函數（$範圍）{ $ scope.greeting ='Hola！'; }]）;' – messerbill

https://docs.angularjs.org/guide/controller – messerbill

@messerbill：不工作.... –

工作代碼：

<!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); 
    ?>

來源

2015-11-12 05:08:46

使用php從角度js的數據庫中獲取數據

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