How to make ajax request with anti-forgery token in mvc

Question

I have problem with below details from MVC project.

When I am trying to use jquery ajax request with loading panel like spinning gif (or even text), I am getting err

Answer 1

Have you added the token to your View? Like this:

<form method="post" action="/my-controller/my-action">
    @Html.AntiForgeryToken()
</form>

Since your controller receiving the post is looking for the anti forgery token, you need to ensure that you add it to your form in the view.

EDIT:

Try building your data in json first:

var formData = $('form[action="/TransportJobAddress/Create"]').serialize();
$.extend(formData, {'__RequestVerificationToken': token });

//and then in your ajax call:
$.ajax({
    //...
    data:formData
    //...
});

Answer 2

I wanted to secure both Ajax and normal request, so here is what I came out with:

First using the excellent blog from haacked.com I Created the ConditionalFilterProvider as described.

Then I created all the classes as described on the blog from codethinked.

On my _layout page I added the piece with the $.ajaxPrefilter as described in the blog... This assures that all my Ajax-callback now send the Antiforgery token through header.

To glue all together I added this piece of code on my global.asax / Application_Start

(c, a) =>
                    (c.HttpContext.Request.IsAjaxRequest() &&
                     !string.Equals(c.HttpContext.Request.HttpMethod, "GET"))
                        ? new AjaxValidateAntiForgeryTokenAttribute()
                        : null,
(c, a) =>
                    (!c.HttpContext.Request.IsAjaxRequest() &&
                     string.Equals(c.HttpContext.Request.HttpMethod, "POST", StringComparison.OrdinalIgnoreCase))
                        ? new ValidateAntiForgeryTokenAttribute()
                        : null

Basically.. inject the attribute to all my Controllers that are not GET.

After that I just had to go to all my (Very few) forms and add the @Html.AntiForgeryToken().

To proof that all worked I just try to dens things with a form without the AntiForgeryToken and get the expected exception. And remove the $.ajaxPrefilter and create Ajax requests and the expected exception was received.

Answer 3

Rather than manually adding it to each request, I usually do something like this:

var token = $('input[name="__RequestVerificationToken"]').val();
$.ajaxPrefilter(function (options, originalOptions) {
  if (options.type.toUpperCase() == "POST") {
    options.data = $.param($.extend(originalOptions.data, { __RequestVerificationToken: token }));
  }
});

This will automatically add your token to any ajax POST you do.

Answer 4

Have you added your token to the header of the ajax call?

You need to add AntiForgeryToken in your message header in the ajax call:

var token = $('input[name="__RequestVerificationToken"]').val();

var headers = {};

headers['__RequestVerificationToken'] = token;

$.ajax({
        url: ... some url,
        headers: headers,
        ....
});

Try this in your code:

var token = $('input[name="__RequestVerificationToken"]').val();
var tokenadr = $('form[action="/TransportJobAddress/Create"] input[name="__RequestVerificationToken"]').val(); 

var headers = {};
var headersadr = {};
headers['__RequestVerificationToken'] = token;
headersadr['__RequestVerificationToken'] = tokenadr;

$('#submitaddress').click(function subaddr(event) {
    event.preventDefault();
    event.stopPropagation();
  //$('#addAddress').html('<img src="/img/animated-overlay.gif"> Sending...');
   // $('#addAddress').blur();
    //  $(this).bl
    if ($('#Jobid').val()!="") {
        $('#TransportJobId').val(parseInt($('#Jobid').val()));
        $.ajax(
              {
                  url: '/TransportJobAddress/create',
                  type: 'POST',
                  headers:headersadr, 
                  data: "__RequestVerificationToken=" + token + "" + $('form[action="/TransportJobAddress/Create"]').serialize(),
                  success: function poste(data, textStatus, jqXHR) { $('#addAddress').html(data); return false; },
                  error: function err(jqXHR, textStatus, errorThrown) { alert('error at address :' + errorThrown); }
              });
    }
    else {
        var transportid = 2;
        $.ajax({
            url: '/TransportJob/create',
            type: 'POST',
            headers:headers, 
            data: $('form[action="/TransportJob/Create"]').serialize(),
            success: function sfn(data, textStatus, jqXHR) {
                transportid = parseInt(data);
                $('#Jobid').val(data);
               // alert('inserted id :' + data);
                $('#TransportJobId').val((transportid));
                $.ajax(
         {

             url: '/TransportJobAddress/create',
             type: 'POST',
             //beforeSend: function myintserver(xhr){  
             //        $('#addAddress').html('<div id="temp_load" style="text-align:center">please wait ...</div>');
             //},
             headers:headers, 
             data: $('form[action="/TransportJobAddress/Create"]').serialize(),
             success: function poste(data, textStatus, jqXHR) {
                 $('#addAddress').html(data);

             },
             error: function err(jqXHR, textStatus, errorThrown) {
                 alert('error at address :' + errorThrown);
             }

         });
            },
            error: function myfunction(jqXHR, textStatus, errorThrown) {
                alert("error at transport :" + jqXHR.textStatus);
            },
            complete: function completefunc() {
              //  alert('ajax completed all requests');
                return false;
            }

        });
    }
});

Added headers line in your ajax call.

Answer 5

View or Layout:

< form id='_id' method='POST'> @html.antiforgeryToken(); < /form>

The ajax call function:

var data={...};
var token=$('#_id').serializeObject();
var dataWithAntiforgeryToken = $.extend(data,token);

$.ajax(
{
....,
data: dataWithAntiforgeryToken;
}
)