I have a Django web application and I'm trying to make an ajax call for uploading a large image. If I can upload the image I am trying to upload, I'll read the image with the pythonocc library and return the volume information. Since this process takes a long time, I am trying to do it using the django-background-tasks library. According to what I'm talking about, the ajax request that I am trying to take the image and send it to the view is as follows.
var data = new FormData();
var img = $('#id_Volume')[0].files[0];
data.append('img', img);
data.append('csrfmiddlewaretoken', '{{ csrf_token }}');
$.ajax({
method: 'POST',
url: '{% url 'data:upload' %}',
cache: false,
processData: false,
contentType: false,
type: 'POST',
data: data,
}).done(function(data) {
});
my view that will work after Ajax request;
def uploadFile(request):
if request.method == 'POST':
file = request.FILES.get('img')
filename = file.name
key = 'media/' + filename
s3_resource = boto3.resource('s3')
bucket = s3_resource.Bucket('bucket')
bucket.put_object(Key=key, Body=file)
new_data = Data(author=request.user)
new_data.save()
data_id = new_data.id
initial = CompletedTask.objects.filter(verbose_name=verbose)
request.session['u_initial'] = len(initial)
verbose = str(request.user) + '_3D'
read_3D(data_id, key, filename, verbose_name = verbose) ###background-task function is called
return redirect('data:uploadStatus')
The view that calls this background task function should also redirect the view, which will show the status of the upload.
def upload_status(request):
customer= str(request.user)
verbose = customer + '_3D'
initial = request.session['u_initial']
if request.is_ajax():
running, completed = get_upload_status(verbose)
context = {
"initial": initial,
"running": running,
"completed": completed,
}
return JsonResponse(context)
return render(request, "file_pending.html")
However, when I made the Ajax request, the view gave The view data.views.uploadFile didn't return an HttpResponse object. It returned None instead. error because it doesn't enter if request.method == 'POST' condition. When I remove that condition line from view, it cannot get the img file. By the way, this uploadFile view was working without any problems. Strangely, when I added the upload_status view and redirect it, it stopped working. To revert, I deleted that part and reverted it back, but it still didn't work.
How can I fix this problem? Am I missing something?
A. You should restrict your view method to the actual method you are expecting using the require_http_methods decorator
from django.views.decorators.http import require_POST
#require_POST
def uploadFile(request):
file = request.FILES.get('img')
filename = file.name
key = 'media/' + filename
s3_resource = boto3.resource('s3')
# ....
return redirect('data:uploadStatus')
That way calls to this endpoint not using POST will receive a 405 Method not allowed response. It's easier to debug and clarifies what is going wrong.
B. Make sure that your Ajax call in the frontend follows the redirect.
The cache parameter is not necessary with POST (unless you plan for IE8)
About contentType=false and processData=false: it still looks to me like this is a regular multipart-formdata call. Don't do this unless you know why you are doing it. Your Django code looks like it will handle the regular form data just fine, no need to tinker with the default behaviour.
C. You have to implement an error callback, as well:
}).done(function(data) {
}).fail(function(xhr, text, error) {
console.log(text, error)
});
I suppose what happens is some error that you are not handling in the JS code.
You have not added your settings, so I cannot be sure, but I suppose that you have not setup your CSRF configuration for AJAX correctly. One way to fix this would be to add the CSRF cookie to the AJAX request, as well. Both the CSRF form field as well as the cookie have to be present.
See https://docs.djangoproject.com/en/3.1/ref/csrf/#ajax
I use the $.ajax to request the data, but how can I let the page switch as the same time?
in my js:
$.ajax({
type:'post',
url:'/api/server_payment',
...
success:success_func,
})
function success_func(response){
...
}
In my views.py:
def server_payment(request):
if request.method == 'POST':
# I don't know what to write here, because there I will switch the web page, and then fill the request data to the switched template.
EDIT
Because I want to pass data by the ajax to the views.py , and in the views.py I want to switch the url to a new url, and in the new url , I will render the passed data.
Because use the ajax requert I will get the response in the ajax callback function.
You just pass your data from ajax to the views with the post data:
file.js
data = {
'key': value,
'another_key': another_value,
// as many as you need
...
}
$.ajax({
type:'post',
url:'/api/server_payment',
data: data
...
success:success_func,
})
function success_func(response){
...
}
Now in your view server_payment store them in the session :
def server_payment(request):
if request.method == 'POST':
request.session['key'] = request.POST.get('key')
request.session['another_key'] = request.POST.get('another_key')
...
return HttpRedirectResponse('/other/url')
And now in your other views (corresponding to the one at '/other/url', you'll access to the data in request.session. In the other views, get data by poping them to empty the request.session dict :
views.py rendering the data
def another_view(request):
data = {}
data['key'] = request.session.pop('key', "NOT_FOUND") # this will prevent from raising exception
data['another_key'] = request.session.pop('another_key', "NOT_FOUND")
...
return render('/your/template.html', data)
I want to pass data by the ajax to the views.py , and in the views.py I want to switch the url to a new url, and in the new url , I will render the passed data.
The thing I don't understand is why you don't send your post data directly in the good view to render instead of having a view getting the post data and redirecting to another one.
I have an ajax call that sends a country label to my view function below:
views
...
posts = Post.objects.all()
if request.method == 'POST':
country = request.POST.get('country')
print('COUNTRY:', country) #successfully prints
posts = Post.objects.filter(country=country)
context = {
...
'posts': posts,
}
return render(request, 'boxes.html', context)
I successfully get the ajax data but what do I do after this in order to redirect to the same view with the new posts value?
If you are using Ajax.You have to use window.location.reload(); in success method of Ajax post call.
As i read that you are using Ajax:
The function reload #Himanshu dua said is ok
And i should check the way you use URL and VIEW
# The function reload will work well if you change the DATA
# and it will reload and return again the New value from data
Or you should try to replace the old with the new DATA you got from Server via ajax (just change the value with Jquery)
In your example the view returns a rendered template: boxes.html. For this to work, you would have to either
modify your view to use query parameters (eg /my/url?country=nowhere). This would then work with GET requests instead of posts, and you can easily call it via URL.
or use a html form instead of AJAX requests. A form can easily POST to an endpoint and it will load whatever your webserver returns.
Ajax calls are designed to exchange data with a server, not really for downloading whole webpages. So if you wanted to use ajax, you could make your view return a json/xml/whatever list of objects and then use js inject them into your page:
function successCallback(data) {
var elem = document.getElementById('myelement');
elem.innerHTML = data;
}
It doesn't work that way.
On the HTML page, using Javascript, you should:
(1) Send an AJAX call (GET/POST) to load the view function in the
backend
(5) Receive the output of the view function and do whatever you want with it.
On the Backend, using the view function, you should:
(2) Receive the GET/POST Data from the frontend (HTML)
(3) Do whatever you want with that.
(4) Return the output of the function as JSON using HttpResponse
Note the numbers next to each item. It happens according to that
sequence from 1 to 5.
Here is an example:
On the HTML Page (Javascript):
var posts;
$("button").click(function(){
$.post("/test/", function(data, status){
posts = data;
console.log(data);
});
});
On the Backend (Python):
import json
from django.http import HttpResponse
def test(request):
if request.method == 'POST':
country = request.POST.get('country')
posts = Post.objects.filter(country=country)
return HttpResponse( json.dumps(posts) )
else:
return HttpResponse( "error" )
The main page of my website has multiple buttons at the top. Whenever one of these buttons is pushed, a get request is sent to a django view, which is redirected and a queryset of django models is filtered and eventually displayed on the web page. I know that my ajax works because the terminal says the request is redirected properly. The function it redirects to also seems to be working, as it is quite simple and has not thrown any errors. However, my view remains the same and I'm not sure why.
urls.py
url(r'ajax_filter/', views.ajax_filter, name='ajax_filter'),
url(r'filter=(\w+)/$', views.filtered_index, name='filtered_index'),
views.py
def filtered_index(request, filter):
clothes = Clothes_Item.objects.filter(gender=filter)
if request.user.is_authenticated():
favorite_clothes_ids = get_favorite_clothes_ids(request)
return render(request, 'test.html', {'clothes': clothes, 'favorite_clothes_ids': favorite_clothes_ids})
else:
return render(request, 'test.html', {'clothes': clothes, })
def ajax_filter(request):
if request.is_ajax():
gender_filter = request.GET.get('gender_filter') #filter type
if gender_filter is not None:
return HttpResponseRedirect(reverse('filtered_index', args=[gender_filter]))
return HttpResponse('')
You can not use Django redirect in your case. When you send an ajax request you usually expect a json response, and based on that you can redirect the user via your JavaScript code.
$.ajax({
// You send your request here
}).done(function(data) {
// You can handle your redirection here
});
Here is how you can handle a redirect with your setup, you pass back a JsonResponse from django with the next page that you want to go to:
from django.http import JsonResponse
def ajax_filter(request):
if request.is_ajax():
gender_filter = request.GET.get('gender_filter') #filter type
if gender_filter is not None:
return JsonResponse({
'success': True,
'url': reverse('filtered_index', args=[gender_filter]),
})
return JsonResponse({ 'success': False })
In JS, you use done (or success) function to grab the URL from the passed back JsonResponse and redirect to that URL using window.location.href:
$.ajax({
// You send your request here
}).done(function (data) {
if (data.success) {
window.location.href = data.url;
}
});
I am new to Django and pretty new to Ajax. I am working on a project where I need to integrate the two. I believe that I understand the principles behind them both, but have not found a good explanation of the two together.
Could someone give me a quick explanation of how the codebase must change with the two of them integrating together?
For example, can I still use the HttpResponse with Ajax, or do my responses have to change with the use of Ajax? If so, could you please provide an example of how the responses to the requests must change? If it makes any difference, the data I am returning is JSON.
Even though this isn't entirely in the SO spirit, I love this question, because I had the same trouble when I started, so I'll give you a quick guide. Obviously you don't understand the principles behind them (don't take it as an offense, but if you did you wouldn't be asking).
Django is server-side. It means, say a client goes to a URL, you have a function inside views that renders what he sees and returns a response in HTML. Let's break it up into examples:
views.py:
def hello(request):
return HttpResponse('Hello World!')
def home(request):
return render_to_response('index.html', {'variable': 'world'})
index.html:
<h1>Hello {{ variable }}, welcome to my awesome site</h1>
urls.py:
url(r'^hello/', 'myapp.views.hello'),
url(r'^home/', 'myapp.views.home'),
That's an example of the simplest of usages. Going to 127.0.0.1:8000/hello means a request to the hello() function, going to 127.0.0.1:8000/home will return the index.html and replace all the variables as asked (you probably know all this by now).
Now let's talk about AJAX. AJAX calls are client-side code that does asynchronous requests. That sounds complicated, but it simply means it does a request for you in the background and then handles the response. So when you do an AJAX call for some URL, you get the same data you would get as a user going to that place.
For example, an AJAX call to 127.0.0.1:8000/hello will return the same thing it would as if you visited it. Only this time, you have it inside a JavaScript function and you can deal with it however you'd like. Let's look at a simple use case:
$.ajax({
url: '127.0.0.1:8000/hello',
type: 'get', // This is the default though, you don't actually need to always mention it
success: function(data) {
alert(data);
},
failure: function(data) {
alert('Got an error dude');
}
});
The general process is this:
The call goes to the URL 127.0.0.1:8000/hello as if you opened a new tab and did it yourself.
If it succeeds (status code 200), do the function for success, which will alert the data received.
If fails, do a different function.
Now what would happen here? You would get an alert with 'hello world' in it. What happens if you do an AJAX call to home? Same thing, you'll get an alert stating <h1>Hello world, welcome to my awesome site</h1>.
In other words - there's nothing new about AJAX calls. They are just a way for you to let the user get data and information without leaving the page, and it makes for a smooth and very neat design of your website. A few guidelines you should take note of:
Learn jQuery. I cannot stress this enough. You're gonna have to understand it a little to know how to handle the data you receive. You'll also need to understand some basic JavaScript syntax (not far from python, you'll get used to it). I strongly recommend Envato's video tutorials for jQuery, they are great and will put you on the right path.
When to use JSON?. You're going to see a lot of examples where the data sent by the Django views is in JSON. I didn't go into detail on that, because it isn't important how to do it (there are plenty of explanations abound) and a lot more important when. And the answer to that is - JSON data is serialized data. That is, data you can manipulate. Like I mentioned, an AJAX call will fetch the response as if the user did it himself. Now say you don't want to mess with all the html, and instead want to send data (a list of objects perhaps). JSON is good for this, because it sends it as an object (JSON data looks like a python dictionary), and then you can iterate over it or do something else that removes the need to sift through useless html.
Add it last. When you build a web app and want to implement AJAX - do yourself a favor. First, build the entire app completely devoid of any AJAX. See that everything is working. Then, and only then, start writing the AJAX calls. That's a good process that helps you learn a lot as well.
Use chrome's developer tools. Since AJAX calls are done in the background it's sometimes very hard to debug them. You should use the chrome developer tools (or similar tools such as firebug) and console.log things to debug. I won't explain in detail, just google around and find out about it. It would be very helpful to you.
CSRF awareness. Finally, remember that post requests in Django require the csrf_token. With AJAX calls, a lot of times you'd like to send data without refreshing the page. You'll probably face some trouble before you'd finally remember that - wait, you forgot to send the csrf_token. This is a known beginner roadblock in AJAX-Django integration, but after you learn how to make it play nice, it's easy as pie.
That's everything that comes to my head. It's a vast subject, but yeah, there's probably not enough examples out there. Just work your way there, slowly, you'll get it eventually.
Further from yuvi's excellent answer, I would like to add a small specific example on how to deal with this within Django (beyond any js that will be used). The example uses AjaxableResponseMixin and assumes an Author model.
import json
from django.http import HttpResponse
from django.views.generic.edit import CreateView
from myapp.models import Author
class AjaxableResponseMixin(object):
"""
Mixin to add AJAX support to a form.
Must be used with an object-based FormView (e.g. CreateView)
"""
def render_to_json_response(self, context, **response_kwargs):
data = json.dumps(context)
response_kwargs['content_type'] = 'application/json'
return HttpResponse(data, **response_kwargs)
def form_invalid(self, form):
response = super(AjaxableResponseMixin, self).form_invalid(form)
if self.request.is_ajax():
return self.render_to_json_response(form.errors, status=400)
else:
return response
def form_valid(self, form):
# We make sure to call the parent's form_valid() method because
# it might do some processing (in the case of CreateView, it will
# call form.save() for example).
response = super(AjaxableResponseMixin, self).form_valid(form)
if self.request.is_ajax():
data = {
'pk': self.object.pk,
}
return self.render_to_json_response(data)
else:
return response
class AuthorCreate(AjaxableResponseMixin, CreateView):
model = Author
fields = ['name']
Source: Django documentation, Form handling with class-based views
The link to version 1.6 of Django is no longer available updated to version 1.11
I am writing this because the accepted answer is pretty old, it needs a refresher.
So this is how I would integrate Ajax with Django in 2019 :) And lets take a real example of when we would need Ajax :-
Lets say I have a model with registered usernames and with the help of Ajax I wanna know if a given username exists.
html:
<p id="response_msg"></p>
<form id="username_exists_form" method='GET'>
Name: <input type="username" name="username" />
<button type='submit'> Check </button>
</form>
ajax:
$('#username_exists_form').on('submit',function(e){
e.preventDefault();
var username = $(this).find('input').val();
$.get('/exists/',
{'username': username},
function(response){ $('#response_msg').text(response.msg); }
);
});
urls.py:
from django.contrib import admin
from django.urls import path
from . import views
urlpatterns = [
path('admin/', admin.site.urls),
path('exists/', views.username_exists, name='exists'),
]
views.py:
def username_exists(request):
data = {'msg':''}
if request.method == 'GET':
username = request.GET.get('username').lower()
exists = Usernames.objects.filter(name=username).exists()
data['msg'] = username
data['msg'] += ' already exists.' if exists else ' does not exists.'
return JsonResponse(data)
Also render_to_response which is deprecated and has been replaced by render and from Django 1.7 onwards instead of HttpResponse we use JsonResponse for ajax response. Because it comes with a JSON encoder, so you don’t need to serialize the data before returning the response object but HttpResponse is not deprecated.
Simple and Nice. You don't have to change your views. Bjax handles all your links. Check this out:
Bjax
Usage:
<script src="bjax.min.js" type="text/javascript"></script>
<link href="bjax.min.css" rel="stylesheet" type="text/css" />
Finally, include this in the HEAD of your html:
$('a').bjax();
For more settings, checkout demo here:
Bjax Demo
AJAX is the best way to do asynchronous tasks. Making asynchronous calls is something common in use in any website building. We will take a short example to learn how we can implement AJAX in Django. We need to use jQuery so as to write less javascript.
This is Contact example, which is the simplest example, I am using to explain the basics of AJAX and its implementation in Django. We will be making POST request in this example. I am following one of the example of this post: https://djangopy.org/learn/step-up-guide-to-implement-ajax-in-django
models.py
Let's first create the model of Contact, having basic details.
from django.db import models
class Contact(models.Model):
name = models.CharField(max_length = 100)
email = models.EmailField()
message = models.TextField()
timestamp = models.DateTimeField(auto_now_add = True)
def __str__(self):
return self.name
forms.py
Create the form for the above model.
from django import forms
from .models import Contact
class ContactForm(forms.ModelForm):
class Meta:
model = Contact
exclude = ["timestamp", ]
views.py
The views look similar to the basic function-based create view, but instead of returning with render, we are using JsonResponse response.
from django.http import JsonResponse
from .forms import ContactForm
def postContact(request):
if request.method == "POST" and request.is_ajax():
form = ContactForm(request.POST)
form.save()
return JsonResponse({"success":True}, status=200)
return JsonResponse({"success":False}, status=400)
urls.py
Let's create the route of the above view.
from django.contrib import admin
from django.urls import path
from app_1 import views as app1
urlpatterns = [
path('ajax/contact', app1.postContact, name ='contact_submit'),
]
template
Moving to frontend section, render the form which was created above enclosing form tag along with csrf_token and submit button. Note that we have included the jquery library.
<form id = "contactForm" method= "POST">{% csrf_token %}
{{ contactForm.as_p }}
<input type="submit" name="contact-submit" class="btn btn-primary" />
</form>
<script src="https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js"></script>
Javascript
Let's now talk about javascript part, on the form submit we are making ajax request of type POST, taking the form data and sending to the server side.
$("#contactForm").submit(function(e){
// prevent from normal form behaviour
e.preventDefault();
// serialize the form data
var serializedData = $(this).serialize();
$.ajax({
type : 'POST',
url : "{% url 'contact_submit' %}",
data : serializedData,
success : function(response){
//reset the form after successful submit
$("#contactForm")[0].reset();
},
error : function(response){
console.log(response)
}
});
});
This is just a basic example to get started with AJAX with django, if you want to get dive with several more examples, you can go through this article: https://djangopy.org/learn/step-up-guide-to-implement-ajax-in-django
Easy ajax calls with Django
(26.10.2020)
This is in my opinion much cleaner and simpler than the correct answer. This one also includes how to add the csrftoken and using login_required methods with ajax.
The view
#login_required
def some_view(request):
"""Returns a json response to an ajax call. (request.user is available in view)"""
# Fetch the attributes from the request body
data_attribute = request.GET.get('some_attribute') # Make sure to use POST/GET correctly
# DO SOMETHING...
return JsonResponse(data={}, status=200)
urls.py
urlpatterns = [
path('some-view-does-something/', views.some_view, name='doing-something'),
]
The ajax call
The ajax call is quite simple, but is sufficient for most cases. You can fetch some values and put them in the data object, then in the view depicted above you can fetch their values again via their names.
You can find the csrftoken function in django's documentation. Basically just copy it and make sure it is rendered before your ajax call so that the csrftoken variable is defined.
$.ajax({
url: "{% url 'doing-something' %}",
headers: {'X-CSRFToken': csrftoken},
data: {'some_attribute': some_value},
type: "GET",
dataType: 'json',
success: function (data) {
if (data) {
console.log(data);
// call function to do something with data
process_data_function(data);
}
}
});
Add HTML to current page with ajax
This might be a bit off topic but I have rarely seen this used and it is a great way to minimize window relocations as well as manual html string creation in javascript.
This is very similar to the one above but this time we are rendering html from the response without reloading the current window.
If you intended to render some kind of html from the data you would receive as a response to the ajax call, it might be easier to send a HttpResponse back from the view instead of a JsonResponse. That allows you to create html easily which can then be inserted into an element.
The view
# The login required part is of course optional
#login_required
def create_some_html(request):
"""In this particular example we are filtering some model by a constraint sent in by
ajax and creating html to send back for those models who match the search"""
# Fetch the attributes from the request body (sent in ajax data)
search_input = request.GET.get('search_input')
# Get some data that we want to render to the template
if search_input:
data = MyModel.objects.filter(name__contains=search_input) # Example
else:
data = []
# Creating an html string using template and some data
html_response = render_to_string('path/to/creation_template.html', context = {'models': data})
return HttpResponse(html_response, status=200)
The html creation template for view
creation_template.html
{% for model in models %}
<li class="xyz">{{ model.name }}</li>
{% endfor %}
urls.py
urlpatterns = [
path('get-html/', views.create_some_html, name='get-html'),
]
The main template and ajax call
This is the template where we want to add the data to. In this example in particular we have a search input and a button that sends the search input's value to the view. The view then sends a HttpResponse back displaying data matching the search that we can render inside an element.
{% extends 'base.html' %}
{% load static %}
{% block content %}
<input id="search-input" placeholder="Type something..." value="">
<button id="add-html-button" class="btn btn-primary">Add Html</button>
<ul id="add-html-here">
<!-- This is where we want to render new html -->
</ul>
{% end block %}
{% block extra_js %}
<script>
// When button is pressed fetch inner html of ul
$("#add-html-button").on('click', function (e){
e.preventDefault();
let search_input = $('#search-input').val();
let target_element = $('#add-html-here');
$.ajax({
url: "{% url 'get-html' %}",
headers: {'X-CSRFToken': csrftoken},
data: {'search_input': search_input},
type: "GET",
dataType: 'html',
success: function (data) {
if (data) {
console.log(data);
// Add the http response to element
target_element.html(data);
}
}
});
})
</script>
{% endblock %}
I have tried to use AjaxableResponseMixin in my project, but had ended up with the following error message:
ImproperlyConfigured: No URL to redirect to. Either provide a url or define a get_absolute_url method on the Model.
That is because the CreateView will return a redirect response instead of returning a HttpResponse when you to send JSON request to the browser. So I have made some changes to the AjaxableResponseMixin. If the request is an ajax request, it will not call the super.form_valid method, just call the form.save() directly.
from django.http import JsonResponse
from django import forms
from django.db import models
class AjaxableResponseMixin(object):
success_return_code = 1
error_return_code = 0
"""
Mixin to add AJAX support to a form.
Must be used with an object-based FormView (e.g. CreateView)
"""
def form_invalid(self, form):
response = super(AjaxableResponseMixin, self).form_invalid(form)
if self.request.is_ajax():
form.errors.update({'result': self.error_return_code})
return JsonResponse(form.errors, status=400)
else:
return response
def form_valid(self, form):
# We make sure to call the parent's form_valid() method because
# it might do some processing (in the case of CreateView, it will
# call form.save() for example).
if self.request.is_ajax():
self.object = form.save()
data = {
'result': self.success_return_code
}
return JsonResponse(data)
else:
response = super(AjaxableResponseMixin, self).form_valid(form)
return response
class Product(models.Model):
name = models.CharField('product name', max_length=255)
class ProductAddForm(forms.ModelForm):
'''
Product add form
'''
class Meta:
model = Product
exclude = ['id']
class PriceUnitAddView(AjaxableResponseMixin, CreateView):
'''
Product add view
'''
model = Product
form_class = ProductAddForm
When we use Django:
Server ===> Client(Browser)
Send a page
When you click button and send the form,
----------------------------
Server <=== Client(Browser)
Give data back. (data in form will be lost)
Server ===> Client(Browser)
Send a page after doing sth with these data
----------------------------
If you want to keep old data, you can do it without Ajax. (Page will be refreshed)
Server ===> Client(Browser)
Send a page
Server <=== Client(Browser)
Give data back. (data in form will be lost)
Server ===> Client(Browser)
1. Send a page after doing sth with data
2. Insert data into form and make it like before.
After these thing, server will send a html page to client. It means that server do more work, however, the way to work is same.
Or you can do with Ajax (Page will be not refreshed)
--------------------------
<Initialization>
Server ===> Client(Browser) [from URL1]
Give a page
--------------------------
<Communication>
Server <=== Client(Browser)
Give data struct back but not to refresh the page.
Server ===> Client(Browser) [from URL2]
Give a data struct(such as JSON)
---------------------------------
If you use Ajax, you must do these:
Initial a HTML page using URL1 (we usually initial page by Django template). And then server send client a html page.
Use Ajax to communicate with server using URL2. And then server send client a data struct.
Django is different from Ajax. The reason for this is as follows:
The thing return to client is different. The case of Django is HTML page. The case of Ajax is data struct.
Django is good at creating something, but it only can create once, it cannot change anything. Django is like anime, consist of many picture. By contrast, Ajax is not good at creating sth but good at change sth in exist html page.
In my opinion, if you would like to use ajax everywhere. when you need to initial a page with data at first, you can use Django with Ajax. But in some case, you just need a static page without anything from server, you need not use Django template.
If you don't think Ajax is the best practice. you can use Django template to do everything, like anime.
(My English is not good)