How can I post data as form data instead of a requ

2018-12-31 03:21发布

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() {}
});

22条回答
旧人旧事旧时光
2楼-- · 2018-12-31 04:01

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
});
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公子世无双
3楼-- · 2018-12-31 04:03

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);
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刘海飞了
4楼-- · 2018-12-31 04:03

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
            }
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初与友歌
5楼-- · 2018-12-31 04:04

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'
            }
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裙下三千臣
6楼-- · 2018-12-31 04:05

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);
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梦该遗忘
7楼-- · 2018-12-31 04:06

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.

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