在 Django 中使用 AJAX POST 请求进行 CSRF 检查期间禁止 403

Question

I can't fix the problem with CSRF check. Could you please tell me what I need to do? This is my function from views.py:

@login_required
def like_day(request):
if request.method == 'POST':
    if 'day_id' in request.POST:
        day_id = request.POST['day_id']
        if day_id:
            day = Day.objects.get(id=int(day_id))
            likes = day.likes + 1
            day.likes = likes
            day.save()
            return HttpResponse(likes)

This is AJAX request:

$(document).ready( function(){
    $('#likes2').click(function(){
        var catid;
        protect
        catid = $(this).attr("data-catid");
        $.post('/friends_plans/like_day/', {day_id: catid}, function(data){
            $('#like_count').html(data);
            $('#likes2').hide();
        });
    });
});

I've tried to use @ensure_csrf_cookie decorator forthe function, but it didn't help. I've tried to add this to my code:

$.ajaxSetup({
    data: {csrfmiddlewaretoken: '{{ csrf_token }}' },
});

It didn't help either. I've done as it is described in the documentation: https://docs.djangoproject.com/en/1.7/ref/contrib/csrf/ but then there is a mistake $.cookie is not a function. To cope with this problem I downloaded csrf cookie plugin to my static directory or add this to my template <script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jquery-cookie/1.4.1/jquery.cookie.min.js"></script> , but still the same mistake.

Answer 1

You can add {% csrf_token %} template tag in your form or posting function the make posting , if you want to avoid the error you can use decorater @csrf_exempt

 from django.views.decorators.csrf import csrf_exempt
    @csrf_exempt
    def hello(request):
       if request.is_ajax():
          print "hello is working fine"

more details : https://docs.djangoproject.com/en/1.7/ref/contrib/csrf/

Answer 2

To account for CSRF using ajax, try following the Django docs suggestion of aquiring the token via JS: https://docs.djangoproject.com/en/1.9/ref/csrf/#ajax

Create a JS file called csrf.js using the following (assuming jQuery):

// using jQuery
function getCookie(name) {
    var cookieValue = null;
    if (document.cookie && document.cookie != '') {
        var cookies = document.cookie.split(';');
        for (var i = 0; i < cookies.length; i++) {
            var cookie = jQuery.trim(cookies[i]);
            // Does this cookie string begin with the name we want?
            if (cookie.substring(0, name.length + 1) == (name + '=')) {
                cookieValue = decodeURIComponent(cookie.substring(name.length + 1));
                break;
            }
        }
    }
    return cookieValue;
}
var csrftoken = getCookie('csrftoken');

Include that JS file on the template containing your ajax JS.

You should then be able to do what you need safely.

Answer 3

I was also facing the same issue, but it got resolved.

Information: I had a basic contact form which makes POST request to Django View '/contact-form'. But it was giving '403 Forbidden' error. The HTML is as follows:

<form method="post">
    {% csrf_token %}
    <input>...</input>
    <button id="contact-form" type="submit"> <span>Envoyer</span</button>
</form>

Reason: It says that the Django Url is protected by any forgery and you should include a CSRF token in you Ajax call.

Solution: 1) Get CSRF token before making Ajax request like this:

var csrftoken = jQuery("[name=csrfmiddlewaretoken]").val();

2) And then include csrftoken in you Ajax request header like this:

        jQuery.ajax({
            type: "POST",
            url: "/contact-form/",
             headers: {
                'X-CSRFToken': csrftoken
           },
            data: data,
            success: function() {
                alert('success');
            }
        });

Answer 4

When you use csrfmiddlewaretoken in your templates you use {% csrf_token %} ie include in somewhere in your html body. {% csrf_token %} is evaluated to

<input type="hidden" name="csrfmiddlewaretoken" value="somevalue">

You can then use jquery to lookup this particular hidden field and retrieve its value and pass it in your ajax call.