CODEWITHSUNDEEP
javascriptjquerydjangopostrequest
premik91
premik91

Reputation: 89

Django jQuery post request

$.ajax({
    url:'/',
    type: "POST",
    data: {name: 'name', age: 'age'},
    success:function(response){},
    complete:function(){},
    error:function (xhr, textStatus, thrownError){}
});

And in views.py:

class SomeView(generic_views.TemplateView):
    template_name = 'something.html'

    def get(self, request, *args, **kwargs):
        ...something...
        return self.render_to_response(context)

    def post(self, request, *args, **kwargs):
        name = request.POST['name']
        age = request.POST['age']
        ...something...

And I get: [05/Oct/2012 12:03:58] "POST /something/ HTTP/1.1" 403 2294

I'd like to send this data(name and age) via jQuery to this post function in "SomeView". This is the same view as the loaded template, just the request type is different. On get() the template loads and on post, the post() function should be called. Is it possible? I've checked other questions and got this solution. It was supposed to be working. What am I doing wrong?

Upvotes: 8

Views: 25575

Answers (4)

Robert Johnstone
Robert Johnstone

Reputation: 5371

I read on another post that you can do something like this:

$.ajax({
    method: "POST",
    url: "{% url 'some_route' %}",
    headers: {'X-CSRFToken': '{{ csrf_token }}'},
    contentType: "application/json",
    dataType: 'json',
    data: {name:$('#id_name').val()}
})
.fail(function(message) {
    alert('error');
})
.done(function(data) {
    alert(data);
});

Upvotes: 1

Hoff
Hoff

Reputation: 39826

You need to include a CSRF (cross-site request forgery) token in your ajax call. This is well documented here: https://docs.djangoproject.com/en/dev/ref/contrib/csrf/

or, if you want to use a quick fix, use the @csrf_exempt decorator for your view:

from django.views.decorators.csrf import csrf_exempt

@csrf_exempt
def my_view(request):
   ...

Upvotes: 10

JxAxMxIxN
JxAxMxIxN

Reputation: 1731

Within a Django template you can add this...

{% csrf_token %}

Which will output something like this to your page html...

<input type="hidden" name="csrfmiddlewaretoken" value="ckhUdNOTj88A...hfTnREALlks2kz">

Then using Javascript you can simply find this input and get its value - something like this non jQuery example...

var el = document.getElementsByName("csrfmiddlewaretoken");
csrf_value = el[0].getAttribute("value");

Lastly add the csrf_value to your jQuery AJAX post form data line like so...

data: {name: 'name', age: 'age', csrfmiddlewaretoken: csrf_value},

Here's a working jsFiddle concept: https://jsfiddle.net/vdx1Lfpc/18/

Upvotes: 4

A.J.Rouvoet
A.J.Rouvoet

Reputation: 1213

The answer to your question what you are doing wrong, is: not much!

Django returns a 403 response (Forbidden) if an incoming POST request fails Csrf checks. You can do this through jQuery's ajaxSetup, code snippets are found here

The reason that this DOES work on a GET request, is simply that GET requests are not checked by the csrf middleware.

As it seems you are building a form here, another thing to consider is using class based forms. They handle get/post and also parameter validation for you. Very neat. Especially when you are making forms to edit/create/delete model instances, in which case you can embrace the power of ModelForms and CreateViews. Very neat.

It might take some time to get the hang of those generic class based views. But it's very well worth it.

Upvotes: 12

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