[ajax] How can I post data as form data instead of a request payload?

In the code below, the AngularJS $http method calls the URL, and submits the xsrf object as a "Request Payload" (as described in the Chrome debugger network tab). The jQuery $.ajax method does the same call, but submits xsrf as "Form Data".

How can I make AngularJS submit xsrf as form data instead of a request payload?

var url = 'http://somewhere.com/';
var xsrf = {fkey: 'xsrf key'};

$http({
    method: 'POST',
    url: url,
    data: xsrf
}).success(function () {});

$.ajax({
    type: 'POST',
    url: url,
    data: xsrf,
    dataType: 'json',
    success: function() {}
});

For Symfony2 users:

If you don't want to change anything in your javascript for this to work you can do these modifications in you symfony app:

Create a class that extends Symfony\Component\HttpFoundation\Request class:

<?php

namespace Acme\Test\MyRequest;

use Symfony\Component\HttpFoundation\Request;
use Symfony\Component\HttpFoundation\ParameterBag;

class MyRequest extends Request{


/**
* Override and extend the createFromGlobals function.
* 
* 
*
* @return Request A new request
*
* @api
*/
public static function createFromGlobals()
{
  // Get what we would get from the parent
  $request = parent::createFromGlobals();

  // Add the handling for 'application/json' content type.
  if(0 === strpos($request->headers->get('CONTENT_TYPE'), 'application/json')){

    // The json is in the content
    $cont = $request->getContent();

    $json = json_decode($cont);

    // ParameterBag must be an Array.
    if(is_object($json)) {
      $json = (array) $json;
  }
  $request->request = new ParameterBag($json);

}

return $request;

}

}

Now use you class in app_dev.php (or any index file that you use)

// web/app_dev.php

$kernel = new AppKernel('dev', true);
// $kernel->loadClassCache();
$request = ForumBundleRequest::createFromGlobals();

// use your class instead
// $request = Request::createFromGlobals();
$response = $kernel->handle($request);
$response->send();
$kernel->terminate($request, $response);

As of AngularJS v1.4.0, there is a built-in $httpParamSerializer service that converts any object to a part of a HTTP request according to the rules that are listed on the docs page.

It can be used like this:

$http.post('http://example.com', $httpParamSerializer(formDataObj)).
    success(function(data){/* response status 200-299 */}).
    error(function(data){/* response status 400-999 */});

Remember that for a correct form post, the Content-Type header must be changed. To do this globally for all POST requests, this code (taken from Albireo's half-answer) can be used:

$http.defaults.headers.post["Content-Type"] = "application/x-www-form-urlencoded";

To do this only for the current post, the headers property of the request-object needs to be modified:

var req = {
 method: 'POST',
 url: 'http://example.com',
 headers: {
   'Content-Type': 'application/x-www-form-urlencoded'
 },
 data: $httpParamSerializer(formDataObj)
};

$http(req);

You can define the behavior globally:

$http.defaults.headers.post["Content-Type"] = "application/x-www-form-urlencoded";

So you don't have to redefine it every time:

$http.post("/handle/post", {
    foo: "FOO",
    bar: "BAR"
}).success(function (data, status, headers, config) {
    // TODO
}).error(function (data, status, headers, config) {
    // TODO
});

Use AngularJS $http service and use its post method or configure $http function.

If you do not want to use jQuery in the solution you could try this. Solution nabbed from here https://stackoverflow.com/a/1714899/1784301

$http({
    method: 'POST',
    url: url,
    headers: {'Content-Type': 'application/x-www-form-urlencoded'},
    transformRequest: function(obj) {
        var str = [];
        for(var p in obj)
        str.push(encodeURIComponent(p) + "=" + encodeURIComponent(obj[p]));
        return str.join("&");
    },
    data: xsrf
}).success(function () {});

Create an adapter service for post:

services.service('Http', function ($http) {

    var self = this

    this.post = function (url, data) {
        return $http({
            method: 'POST',
            url: url,
            data: $.param(data),
            headers: {'Content-Type': 'application/x-www-form-urlencoded'}
        })
    }

}) 

Use it in your controllers or whatever:

ctrls.controller('PersonCtrl', function (Http /* our service */) {
    var self = this
    self.user = {name: "Ozgur", eMail: null}

    self.register = function () {
        Http.post('/user/register', self.user).then(function (r) {
            //response
            console.log(r)
        })
    }

})

I'm currently using the following solution I found in the AngularJS google group.

$http
.post('/echo/json/', 'json=' + encodeURIComponent(angular.toJson(data)), {
    headers: {
        'Content-Type': 'application/x-www-form-urlencoded; charset=UTF-8'
    }
}).success(function(data) {
    $scope.data = data;
});

Note that if you're using PHP, you'll need to use something like Symfony 2 HTTP component's Request::createFromGlobals() to read this, as $_POST won't automatically loaded with it.

The continued confusion surrounding this issue inspired me to write a blog post about it. The solution I propose in this post is better than your current top rated solution because it does not restrict you to parametrizing your data object for $http service calls; i.e. with my solution you can simply continue to pass actual data objects to $http.post(), etc. and still achieve the desired result.

Also, the top rated answer relies on the inclusion of full jQuery in the page for the $.param() function, whereas my solution is jQuery agnostic, pure AngularJS ready.

http://victorblog.com/2012/12/20/make-angularjs-http-service-behave-like-jquery-ajax/

Hope this helps.

I took a few of the other answers and made something a bit cleaner, put this .config() call on the end of your angular.module in your app.js:

.config(['$httpProvider', function ($httpProvider) {
  // Intercept POST requests, convert to standard form encoding
  $httpProvider.defaults.headers.post["Content-Type"] = "application/x-www-form-urlencoded";
  $httpProvider.defaults.transformRequest.unshift(function (data, headersGetter) {
    var key, result = [];

    if (typeof data === "string")
      return data;

    for (key in data) {
      if (data.hasOwnProperty(key))
        result.push(encodeURIComponent(key) + "=" + encodeURIComponent(data[key]));
    }
    return result.join("&");
  });
}]);

These answers look like insane overkill, sometimes, simple is just better:

$http.post(loginUrl, "userName=" + encodeURIComponent(email) +
                     "&password=" + encodeURIComponent(password) +
                     "&grant_type=password"
).success(function (data) {
//...

The only thin you have to change is to use property "params" rather than "data" when you create your $http object:

$http({
   method: 'POST',
   url: serviceUrl + '/ClientUpdate',
   params: { LangUserId: userId, clientJSON: clients[i] },
})

In the example above clients[i] is just JSON object (not serialized in any way). If you use "params" rather than "data" angular will serialize the object for you using $httpParamSerializer: https://docs.angularjs.org/api/ng/service/$httpParamSerializer

You can try with below solution

$http({
        method: 'POST',
        url: url-post,
        data: data-post-object-json,
        headers: {'Content-Type': 'application/x-www-form-urlencoded'},
        transformRequest: function(obj) {
            var str = [];
            for (var key in obj) {
                if (obj[key] instanceof Array) {
                    for(var idx in obj[key]){
                        var subObj = obj[key][idx];
                        for(var subKey in subObj){
                            str.push(encodeURIComponent(key) + "[" + idx + "][" + encodeURIComponent(subKey) + "]=" + encodeURIComponent(subObj[subKey]));
                        }
                    }
                }
                else {
                    str.push(encodeURIComponent(key) + "=" + encodeURIComponent(obj[key]));
                }
            }
            return str.join("&");
        }
    }).success(function(response) {
          /* Do something */
        });

var fd = new FormData();
    fd.append('file', file);
    $http.post(uploadUrl, fd, {
        transformRequest: angular.identity,
        headers: {'Content-Type': undefined}
    })
    .success(function(){
    })
    .error(function(){
    });

Please checkout! https://uncorkedstudios.com/blog/multipartformdata-file-upload-with-angularjs

In your app config -

$httpProvider.defaults.transformRequest = function (data) {
        if (data === undefined)
            return data;
        var clonedData = $.extend(true, {}, data);
        for (var property in clonedData)
            if (property.substr(0, 1) == '$')
                delete clonedData[property];

        return $.param(clonedData);
    };

With your resource request -

 headers: {
                'Content-Type': 'application/x-www-form-urlencoded'
            }

Just set Content-Type is not enough, url encode form data before send. $http.post(url, jQuery.param(data))

This isn't a direct answer, but rather a slightly different design direction:

Do not post the data as a form, but as a JSON object to be directly mapped to server-side object, or use REST style path variable

Now I know neither option might be suitable in your case since you're trying to pass a XSRF key. Mapping it into a path variable like this is a terrible design:

http://www.someexample.com/xsrf/{xsrfKey}

Because by nature you would want to pass xsrf key to other path too, /login, /book-appointment etc. and you don't want to mess your pretty URL

Interestingly adding it as an object field isn't appropriate either, because now on each of json object you pass to server you have to add the field

{
  appointmentId : 23,
  name : 'Joe Citizen',
  xsrf : '...'
}

You certainly don't want to add another field on your server-side class which does not have a direct semantic association with the domain object.

In my opinion the best way to pass your xsrf key is via a HTTP header. Many xsrf protection server-side web framework library support this. For example in Java Spring, you can pass it using X-CSRF-TOKEN header.

Angular's excellent capability of binding JS object to UI object means we can get rid of the practice of posting form all together, and post JSON instead. JSON can be easily de-serialized into server-side object and support complex data structures such as map, arrays, nested objects, etc.

How do you post array in a form payload? Maybe like this:

shopLocation=downtown&daysOpen=Monday&daysOpen=Tuesday&daysOpen=Wednesday

or this:

shopLocation=downtwon&daysOpen=Monday,Tuesday,Wednesday

Both are poor design..

AngularJS is doing it right as it doing the following content-type inside the http-request header:

Content-Type: application/json

If you are going with php like me, or even with Symfony2 you can simply extend your server compatibility for the json standard like described here: http://silex.sensiolabs.org/doc/cookbook/json_request_body.html

The Symfony2 way (e.g. inside your DefaultController):

$request = $this->getRequest();
if (0 === strpos($request->headers->get('Content-Type'), 'application/json')) {
    $data = json_decode($request->getContent(), true);
    $request->request->replace(is_array($data) ? $data : array());
}
var_dump($request->request->all());

The advantage would be, that you dont need to use jQuery param and you could use AngularJS its native way of doing such requests.

This is what I am doing for my need, Where I need to send the login data to API as form data and the Javascript Object(userData) is getting converted automatically to URL encoded data

        var deferred = $q.defer();
        $http({
            method: 'POST',
            url: apiserver + '/authenticate',
            headers: { 'Content-Type': 'application/x-www-form-urlencoded' },
            transformRequest: function (obj) {
                var str = [];
                for (var p in obj)
                    str.push(encodeURIComponent(p) + "=" + encodeURIComponent(obj[p]));
                return str.join("&");
            },
            data: userData
        }).success(function (response) {
            //logics
            deferred.resolve(response);
        }).error(function (err, status) {
           deferred.reject(err);
        });

This how my Userdata is

var userData = {
                grant_type: 'password',
                username: loginData.userName,
                password: loginData.password
            }

Complete answer (since angular 1.4). You need to include de dependency $httpParamSerializer

var res = $resource(serverUrl + 'Token', { }, {
                save: { method: 'POST', headers: { 'Content-Type': 'application/x-www-form-urlencoded' } }
            });

            res.save({ }, $httpParamSerializer({ param1: 'sdsd', param2: 'sdsd' }), function (response) {

            }, function (error) { 

            });

There is a really nice tutorial that goes over this and other related stuff - Submitting AJAX Forms: The AngularJS Way.

Basically, you need to set the header of the POST request to indicate that you are sending form data as a URL encoded string, and set the data to be sent the same format

$http({
  method  : 'POST',
  url     : 'url',
  data    : $.param(xsrf),  // pass in data as strings
  headers : { 'Content-Type': 'application/x-www-form-urlencoded' }  // set the headers so angular passing info as form data (not request payload)
});

Note that jQuery's param() helper function is used here for serialising the data into a string, but you can do this manually as well if not using jQuery.

As a workaround you can simply make the code receiving the POST respond to application/json data. For PHP I added the code below, allowing me to POST to it in either form-encoded or JSON.

//handles JSON posted arguments and stuffs them into $_POST
//angular's $http makes JSON posts (not normal "form encoded")
$content_type_args = explode(';', $_SERVER['CONTENT_TYPE']); //parse content_type string
if ($content_type_args[0] == 'application/json')
  $_POST = json_decode(file_get_contents('php://input'),true);

//now continue to reference $_POST vars as usual