使用 AngularJS JSONP 加载外部资源时出错
Error while loading external resources using AngularJS JSONP
我真的很困惑如何使用 $http.jsonp 加载外部资源。这是我的代码:
<!DOCTYPE html>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.4/angular.min.js"></script>
<body>
<div ng-app="myApp" ng-controller="customersCtrl">
<ul>
{{myData}}
</ul>
</div>
<script>
var app = angular.module('myApp', []);
app.controller('customersCtrl', function($scope, $http) {
var url = "https://angularjs.org/greet.php?callback=JSON_CALLBACK&name=Super%20Hero";
$http.jsonp(url).then(function (response) {
$scope.myData = response.data;
});
});
</script>
</body>
</html>
我得到了这样的错误:
Error: [$sce:insecurl]
我的代码有什么问题?谢谢。
通过将您的 angular 版本从 1.6.4
更新到 1.2.6
,工作正常
var app = angular.module('myApp', []);
app.controller('myCtrl', function($scope, $http) {
var url = "https://angularjs.org/greet.php?callback=JSON_CALLBACK&name=Super%20Hero";
$http.jsonp(url).then(function (response) {
console.log(response)
$scope.myData = response.data;
});
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.6/angular.min.js"></script>
<div ng-app="myApp" ng-controller="myCtrl"></div>
我真的很困惑如何使用 $http.jsonp 加载外部资源。这是我的代码:
<!DOCTYPE html>
<html>
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.6.4/angular.min.js"></script>
<body>
<div ng-app="myApp" ng-controller="customersCtrl">
<ul>
{{myData}}
</ul>
</div>
<script>
var app = angular.module('myApp', []);
app.controller('customersCtrl', function($scope, $http) {
var url = "https://angularjs.org/greet.php?callback=JSON_CALLBACK&name=Super%20Hero";
$http.jsonp(url).then(function (response) {
$scope.myData = response.data;
});
});
</script>
</body>
</html>
我得到了这样的错误:
Error: [$sce:insecurl]
我的代码有什么问题?谢谢。
通过将您的 angular 版本从 1.6.4
更新到 1.2.6
var app = angular.module('myApp', []);
app.controller('myCtrl', function($scope, $http) {
var url = "https://angularjs.org/greet.php?callback=JSON_CALLBACK&name=Super%20Hero";
$http.jsonp(url).then(function (response) {
console.log(response)
$scope.myData = response.data;
});
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.6/angular.min.js"></script>
<div ng-app="myApp" ng-controller="myCtrl"></div>