javascript - 请求的资源上不存在 'Access-Control-Allow-Origin' header 。我不知道该怎么办

Question

好吧，这条消息总是显示在我的控制台上..有人可以帮我解决这个问题吗？..顺便说一句，这是 ionic 的..这是我的controller.js

angular.module('directory.controllers', [])

.controller('EmployeeIndexCtrl', function ($scope, EmployeeService, $http) {

    $scope.searchKey = "";

    $scope.clearSearch = function () {
        $scope.searchKey = "";
        findAllEmployees();
    }

    $scope.search = function () {
        EmployeeService.findByName($scope.searchKey).then(function (employees) {
            $scope.employees = employees;
        });
    }
    var findAllEmployees = function() {
        $http({
          method: 'GET',
          url: 'https://miudr.000webhostapp.com'
        }).then(function (res){
          $scope.employees = res.data;
          console.log(res);
        })
    }
    findAllEmployees();

})

.controller('EmployeeDetailCtrl', function ($scope, $stateParams, EmployeeService) {
    EmployeeService.findById($stateParams.employeeId).then(function(employee) {
        $scope.employee = employee;
    });
})

.controller('EmployeeReportsCtrl', function ($scope, $stateParams, EmployeeService) {
    EmployeeService.findByManager($stateParams.employeeId).then(function(employees) {
        $scope.employees = employees;
    });
});

这是我的 json.php

<?php
header("Content-type:application/json");

 $connection = mysqli_connect("localhost", "root", "", "miudr") or die("Error " . mysqli_error($connection));

 $sql = "select * from data";
 $result = mysqli_query($connection, $sql) or die("Error " . mysqli_error($connection));

 while ($row = mysqli_fetch_assoc($result)) {
     $data[] = $row;
 }

 echo json_encode($data, JSON_PRETTY_PRINT);

 mysqli_close($connection);
 ?>

php 文件位于我的主机上..但无论如何我将我的用户名和密码隐藏为 root 和“”..我每次尝试显示此 json 时都不知道..它没有向我显示结果..但向我显示禁止“访问 blablabla”

顺便说一句，抱歉我的英语不好

Answer 1

您必须允许您的后端接受该请求。像这样的事情

header('Access-Control-Allow-Origin: *');
header('Access-Control-Allow-Methods: GET, POST, PATCH, PUT, DELETE, OPTIONS');
header('Access-Control-Allow-Headers: Origin, Content-Type, X-Auth-Token');

由于“*”，这将被所有人接受。如果您只想接受来自您的 IP 的请求，请将 * 替换为您的 IP。

否则你可以考虑使用 JSONP。只需将 dataType: 'JSONP' 添加到您的 $http 对象即可。