从PHP获取角度未处理错误中的数据

我犯了一个错误，这个错误没有告诉我什么。我是一个新手，所以这可能是一个非常愚蠢的错误，但我不明白

它说有一个语法错误：

SyntaxError: Unexpected token /
at Object.parse (native)
at fromJson (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:1252:14)
at defaultHttpResponseTransform (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:9414:16)
at https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:9505:12
at forEach (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:336:20)
at transformData (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:9504:3)
at transformResponse (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:10276:23)
at processQueue (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:14745:28)
at https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:14761:27
at Scope.$eval (https://cdnjs.cloudflare.com/ajax/libs/angular.js/1.4.7/angular.js:15989:28)

我调用控制器的html代码：

<div ng-controller="dbTestCtrl">
  <table>
    <tr ng-repeat="x in names">
      <td>{{ x.Name }}</td>
      <td>{{ x.Country }}</td>
    </tr>
  </table>
</div>

module.controller('dbTestCtrl', function($scope, $http) {
   $http.get(PHP_API + "datahandler/dbTest.php")
   .then(function (response){
      $scope.names = response.data.records;});
});

还有我的php文件，据我所知，它正在工作：

<?php
  header("Access-Control-Allow-Origin: *");
  header("Content-Type: application/json; charset=UTF-8");

  $conn = new mysqli("localhost", "root", "", "cmweb");

  $result = $conn->query("SELECT loginName, userID, createDate FROM users");

  $outp = "";
  while($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    if ($outp != "") {
        $outp .= ",";
    }
    $outp .= '{"Name":"'  . $rs["loginName"] . '",';
    $outp .= '"ID":"'   . $rs["userID"]        . '",';
    $outp .= '"Created":"'. $rs["createDate"]     . '"}'; 
  }
  $outp ='{"records":['.$outp.']}';
  $conn->close();

  echo($outp);
?>

正如我所说，我对angular是个新手，我自己就是搞不懂，所以如果有人能帮我，那就太好了：）

---

Thx可能是您的json格式错误。尝试使用json_encode避免自己编写所有这些引号和括号

$outp = array();

while ($rs = $result->fetch_array(MYSQLI_ASSOC)) {
    $outp[] = array('Name' => $rs["loginName"], 
                    'ID' => $rs["userID"], 
                    'Created' => $rs["createDate"]);
}

echo json_encode(array('records' => $outp));