AngularJS：如何使JSONP请求 [英] AngularJS: how to make a jsonp request

问题描述

我需要做一些跨站点脚本。低于code的块包含JSONP的方法，该方法返回，如果它失败了，但是当我改变这是一个GET请求，然后我有成功。我需要能够使用JSONP方法成功的响应。以下可以排除。响应有效JSON和这个参数是在url？回调= JSON_CALLBACK。下面是我从做http请求和code块执行此code接收JSON。

I need to do some cross site scripting. The block of code below contains the method of jsonp, the method returns as if it failed, but when I change it to be a get request I then have success. I need to be able to a successful response using the jsonp method. The following can be ruled out. The response is valid json and this param is in the url ?callback=JSON_CALLBACK. Here is the json I receive from doing the http request and the code block that executes this code.

HTTP响应状态code 200

http response status code 200

[{"cube":"1" ,"points":"160"},{"cube":"2","points":"690"},{"cube":"3","points":"331"}]

code座

 var myApp = angular.module('test', []);

    myApp.controller('UserCtrl', function($scope, users) {
        $scope.usersPerCube = users.getUsers();
    })

    myApp.factory('users', function($http) {
       return {
         getUsers: function() {
           var deferred = $q.defer();
           var url = "http://localhost/api/api/index.php/analytics/UsersPerCube?callback=JSON_CALLBACK";
         $http.get(url).success(function (data, status, headers, config) {
                console.log(data);
                deferred.resolve(data);
            }).error(function (data, status, headers, config) {
                //this always gets called
                console.log(status);
                deferred.reject(status);
            });
            return deferred.promise;

     }
   }

请注意，我已经编辑了我的服务器端code和现在接受

Note that I have edited my server side code and now receive

"angular.callbacks._1( {"cube":"1","points":"160"},{"cube":"2","points":"690"},{"cube":"3","points":"331"})"

更新的
以上是有效的，现在的成功方法正在执行。我只需要弄清楚如何解析的对象。一旦我找出了答案，我会再次发布。

UPDATE The above is valid and now the success method is executing. I just need to figure out how to parse the objects. I will post again once I figure out the answer.

推荐答案

我已经决定给出了如何做一个像我一样JSONP 要求，这样其他人就不会遇到同样的麻烦。

I have decided to give a detailed description of how to do a jsonp request so others will not run into the same troubles as I did.

myApp.factory('users', function($http) {
       return {
         getUsers: function() {
           var deferred = $q.defer();
           var url = "http://localhost/api/api/index.php/analytics/UsersPerCube?callback=JSON_CALLBACK";
         $http.get(url).success(function (data, status, headers, config) {
                console.log(data);
                deferred.resolve(data);
            }).error(function (data, status, headers, config) {
                //this always gets called
                console.log(status);
                deferred.reject(status);
            });
            return deferred.promise;

     }  

注意，网址中包含？回调= JSON_CALLBACK 。 <一href=\"http://stackoverflow.com/questions/12066002/parsing-jsonp-http-jsonp-response-in-angular-js\">Here是上一个不错的计算器。一旦你得到了响应，那么您会收到类似下面的一个JSON。

Notice that the url contains ?callback=JSON_CALLBACK. Here is a nice stackoverflow on that. Once you get the response then you'll receive a json like the one below.

"angular.callbacks._1( {"cube":"1","points":"160"},{"cube":"2","points":"690"},{"cube":"3","points":"331"})"

以下是关于这个主题一个不错的计算器

现在这让我的一部分是服务器有返回 GET 参数，回调。这里是一个很好的教程为。 <一href=\"http://niryariv.word$p$pss.com/2009/05/05/jsonp-quickly/\">http://niryariv.word$p$pss.com/2009/05/05/jsonp-quickly/因此，JSON看起来像上面的。

Now the one part that got me is that the server has to return the GET param, callback. Here is a good tutorial for that. http://niryariv.wordpress.com/2009/05/05/jsonp-quickly/ So the json looks like the one above.

嗯，我希望这可以帮助别人的未来。

Well, I hope this helps someone in the future.

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