AngularJS：从php脚本获取JSON

Question

I'm trying the following AngularJS tutorial at w3schools. 我正在w3schools尝试以下AngularJS教程。

Link to w3schools tutorial 链接到w3schools教程

I changed the script to make it work with one of my php script: 我更改了脚本以使其与我的PHP脚本之一一起工作：

<!DOCTYPE html>
<html >
<style>
table, th , td  {
  border: 1px solid grey;
  border-collapse: collapse;
  padding: 5px;
}
table tr:nth-child(odd) {
  background-color: #f1f1f1;
}
table tr:nth-child(even) {
  background-color: #ffffff;
}
</style>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.4.8/angular.min.js"></script>
<body>

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

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

</div>

<script>
var app = angular.module('myApp', []);
app.controller('customersCtrl', function($scope, $http) {
       $http.get("http://wirtschaftsinformatik.web2page.ch/selectModule.php")
//my url
       .then(function (response) {$scope.names = response.data.records;});
    });
    </script>

    </body>
    </html>

This is actually not working. 这实际上是行不通的。 I guess its because of the php file, here is the code: 我想这是因为php文件，这是代码：

<?PHP
    //Verbindung zur Datenbank
    include "inc_mysql.php";

    //SQL-String
    $select = "SELECT * FROM module ORDER BY name ASC";

    // Codierung
    mysql_set_charset("utf8");

    //SQL-Befehl in Variable speichern
    $sql = mysql_query($select);

    //Array erstellen
    $jsonArray = array();

    while ($row = mysql_fetch_array($sql)){ 
        $jsonArray[] = $row;
    }

    echo json_encode($jsonArray, JSON_UNESCAPED_UNICODE);   
?>

Do you figure out what is missing? 您知道丢失了什么吗？

Here is the JSON output 这是JSON输出

[{"0":"12","id":"12","1":"Betriebsbuchhaltung","name":"Betriebsbuchhaltung","2":"2016-10-19 15:44:16","timestamp":"2016-10-19 15:44:16"},{"0":"13","id":"13","1":"Datenbanken","name":"Datenbanken","2":"2016-10-28 20:35:49","timestamp":"2016-10-28 20:35:49"},{"0":"6","id":"6","1":"Logistik","name":"Logistik","2":"0000-00-00 00:00:00","timestamp":"0000-00-00 00:00:00"},{"0":"14","id":"14","1":"Statistik","name":"Statistik","2":"2016-11-02 15:13:13","timestamp":"2016-11-02 15:13:13"},{"0":"1","id":"1","1":"System Engineering","name":"System Engineering","2":"0000-00-00 00:00:00","timestamp":"0000-00-00 00:00:00"}]

Answer 1

HTML 的HTML

 <div ng-controller="listController">
    <table>
      <tr>
        <th>Id</th>
        <th>Name</th>
        <th>Time</th>
      </tr>
      <tr ng-repeat="person in names">
        <td>{{person.id}}</td>
        <td>{{person.name}}</td>
        <td>{{person.timestamp}}</td>
      </tr>
    </table>
  </div>

DEMO 演示

Answer 2

返回的json没有records属性，这是一个简单的解决方案：

echo json_encode(["records"=>$jsonArray], JSON_UNESCAPED_UNICODE);

AngularJS：从php脚本获取JSON

问题描述

2 个解决方案

解决方案1
1 2016-11-26 13:50:20

解决方案2
1 已采纳 2016-11-26 14:10:29

AngularJS：从php脚本获取JSON

问题描述

2 个解决方案

解决方案1 1 2016-11-26 13:50:20

解决方案2 1 已采纳 2016-11-26 14:10:29

解决方案1
1 2016-11-26 13:50:20

解决方案2
1 已采纳 2016-11-26 14:10:29