How to submit form without refreshing page using Django, Ajax, jQuery?

Question

I am newbie working with django. I need simple example. How to submit form (post) without refreshing page using Django, Ajax, jQuery?

This is my form, views and template:

views.py

from django.shortcuts import *
from django.template import RequestContext
from linki.forms import *

def advert(request):
    if request.method == "POST":
        form = AdvertForm(request.POST)

        if(form.is_valid()):
            print(request.POST['title'])
            message = request.POST['title']

        else:
            message = 'something wrong!'


        return render_to_response('contact/advert.html',
                {'message':message},
            context_instance=RequestContext(request))

    else:
        return render_to_response('contact/advert.html',
                {'form':AdvertForm()},
            context_instance=RequestContext(request))

forms.py (form using "ModelForm")

from django import forms
from django.forms import ModelForm
from linki.models import Advert


class AdvertForm(ModelForm):
    class Meta:
        model = Advert

TEMPLATES (form html code)

<html>
<head>

</head>
    <body>
    <h1>Leave a Suggestion Here</h1>
        {% if message %}
            {{ message }}
        {% endif %}
        <div>
            <form action="" method="post">{% csrf_token %}
                {{ form.as_p }}
                <input type="submit" value="Submit Feedback" />
            </form>
        </div>
    </body>
</html>

score 22 · Accepted Answer · answered Jul 27 '12 at 04:59

if you are planning to use an ajax submit with jquery you should not return html from your view.. I propose you to do this instead:

html:

<html>
<head>
</head>
<body>
    <h1>Leave a Suggestion Here</h1>
        <div class="message"></div>
        <div>
            <form action="" method="post">{% csrf_token %}
                {{ form.as_p }}
                <input type="submit" value="Submit Feedback" />
            </form>
        </div>
</body>
</html>

the js

$('#form').submit(function(e){
    $.post('/url/', $(this).serialize(), function(data){ ... 
       $('.message').html(data.message);
       // of course you can do something more fancy with your respone
    });
    e.preventDefault();
});

the views.py

import json
from django.shortcuts import *
from django.template import RequestContext
from linki.forms import *

def advert(request):
    if request.method == "POST":
        form = AdvertForm(request.POST)

        message = 'something wrong!'
        if(form.is_valid()):
            print(request.POST['title'])
            message = request.POST['title']

        return HttpResponse(json.dumps({'message': message}))

    return render_to_response('contact/advert.html',
            {'form':AdvertForm()}, RequestContext(request))

so that way you put the response in the message div. instead of returning plain html you should return json.

data.message in the JS section should be JSON.parse(data).message — roob, Jul 03 '14 at 22:55

score 7 · Answer 2 · edited Aug 09 '20 at 16:59

7

<script type="text/javascript">
$(document).ready(function() {
    $('#form_id').submit(function() { // On form submit event
        $.ajax({ // create an AJAX call...
            data: $(this).serialize(), // get the form data
            type: $(this).attr('method'), // GET or POST
            url: $(this).attr('action'), // the file to call
            success: function(response) { // on success..
                $('#success_div').html(response); // update the DIV
            },
            error: function(e, x, r) { // on error..
                $('#error_div').html(e); // update the DIV
            }
        });
        return false;
    });
});
</script>

edited Aug 09 '20 at 16:59

Programmer Gaurav

449
1
5
16

answered Jul 27 '12 at 05:50

Jubin Thomas

1,003
2
12
24

This is correctly firing the request for me and I've validated that the serialized data in the request is updated and accurate. It returns a 200 response. But when I refresh the page or look at the database the values have not been updated. Any suggestions? – James Parker Apr 29 '22 at 17:51

score 2 · Answer 3 · answered Jul 25 '12 at 10:32

2

$('#form-id').submit(function(e){
    $.post('your/url', $(this).serialize(), function(e){ ... });
    e.preventDefault();
});

answered Jul 25 '12 at 10:32

Alexander Larikov

2,328
15
15

score 0 · Answer 4 · answered Nov 14 '18 at 06:47

Here is a perfect tutorial for it. I'll include the important parts:

first add this jQuery scripts to main.js and link it to your page.

Add this codes to the main.js (I'll include my version for sending blog comment)

// Submit post on submit
$('#comment-form').on('submit', function(event){
    event.preventDefault();
    create_post();
});

// AJAX for posting
function create_post() {
    $.ajax({
        url : "/blogcomment/", // the endpoint
        type : "POST", // http method
        data : {
            blog_id: blog_id,
            c_name : $('#comment-name').val(),
            c_email:  $('#comment-email').val(),
            c_text:  $('#comment-text').val(),
        }, // data sent with the post request

        // handle a successful response
        success : function(json) {
            $('#comment-name').val(''); // remove the value from the input
            $('#comment-email').val(''); // remove the value from the input
            $('#comment-text').val(''); // remove the value from the input
            $('#comment-form').prepend("<div class='alert alert-success'><button type='button' class='close' data-dismiss='alert'>&times;</button>" + json.result +"</div>");
        },

        // handle a non-successful response
        error : function(xhr,errmsg,err) {
            $('#comment-form').prepend("<div class='alert alert-danger'><button type='button' class='close' data-dismiss='alert'>&times;</button>Oop! Error happend!</div>"); // add the error to the dom
            //console.log(xhr.status + ": " + xhr.responseText); // provide a bit more info about the error to the console
        }
    });
}

my views.py for getting comment looks like this:

def get_blog_comment(request):
    if request.method == 'POST':
        blog_id = request.POST.get('blog_id') 
        user = request.POST.get('c_name')
        email = request.POST.get('c_email')
        comment = request.POST.get('c_text')
        date = jdatetime.datetime.now().strftime("%Y-%m-%d");
        response_data = {}
        blogcomment = Comments(blog_id = blog_id, date = date, name = user, email = email, comment_text = comment)
        blogcomment.save()

        response_data['result'] = 'Success!!.'

        return HttpResponse(
            json.dumps(response_data),
            content_type="application/json"
        )
    else:
        return HttpResponse(
            json.dumps({"nothing to see": "this isn't happening"}),
            content_type="application/json"
        )

And finally the url part for urls.py which is not included in the original tutorial:

path('blogcomment/', views.get_blog_comment, name='get_blog_comment'),

score 0 · Answer 5 · answered Mar 03 '23 at 04:25

Don't forget to add csrf_token to the ajax post request.

This way we can submit the form securely with out reloading.

In Script.js

jQuery.noConflict()
jQuery(document).ready(($) => {
    $("#contactUsForm").on("submit", () => {
        var el = document.getElementsByName("csrfmiddlewaretoken");
        csrf_value = el[0].getAttribute("value");
        $.ajax({
            url: "/contactUs/",
            type: "POST",
            data: {
                userName: $("#userName").val(),
                companyName: $("#companyName").val(),
                emailAddress: $("#emailAddress").val(),
                message: $("#message").val(),
                'csrfmiddlewaretoken': csrf_value
            },
            success: function (json) {
                console.debug(json)
                console.debug("Successfully send contact form")
            },
            error: function (json) {
                console.error(json)
                console.error("Error occured while sending contact form")
            }
        })

        return false;
    })
})

In Views.py

def contactUs(request: HttpRequest):
    print("Contact Us")
    if request.method == "POST":
        username = request.POST.get("userName")
        companyName = request.POST.get("companyName")
        emailAddress = request.POST.get("emailAddress")
        message = request.POST.get("message")

        print(username)
        print(companyName)
        print(emailAddress)
        print(message)
        response_data = {}
        response_data['result'] = 'Success!!.'
        return HttpResponse(
            json.dumps(response_data),
            content_type="application/json"
        )
        pass
    else:
        logger.error(
            f"Contact US request method {request.method} is not Valid expected POST")
    
    return redirect("/")

How to submit form without refreshing page using Django, Ajax, jQuery?

5 Answers5

In Script.js

In Views.py

Linked