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)
I'm using curl to submit form data to my website.
curl -F some_file=#file.txt -F name=test_01 https://localhost:8000
It's not an API but I have a requirement for a single endpoint that behaves as an API. I'm a little out of my depth here, so I'm hoping someone can help me.
I've got the model set up and working and the CreateView, as well:
class CreateFile(CreateView):
model = SomeFile
fields = ['name', 'some_file', . . .]
When I send a POST request with curl as above to the specified URL (/file/request), the object is created in the DB and I get a response (eg, /thanks now which is an HTTP response from template view). But since a non-browser will be sending this request, I was hoping to respond with some JSON. Maybe with the object's name, status, etc.
I've tried a few things with mixed results... For example, if I use View instead of CreateView, I can return JSON but I really like the ease and convenience of the CreateView CBV, so I'm hoping I can do what I want this way.
How can I do this? I found a SO question that gave some clues: How do I return JSON response in Class based views, instead of HTTP response
But this deals with the typical form/view model in the browser. If I have to override the post method, what's the best way to get the form data so I can create the object? Do I need a form class even though I'm not processing a rendered form?
I ended up going with something from the Django docs:
from django.http import JsonResponse
class JSONResponseMixin:
"""
A mixin that can be used to render a JSON response
"""
def render_to_json_response(self, context, **response_kwargs):
return JsonResponse(self.get_data(context), **response_kwargs)
def get_data(self, context):
return context
Then I used this in a DetailView, overriding both get and post methods.
class FileRequest(JSONResponseMixin, DetailView):
def get:
. . .
return self.render_to_response(response_data)
def post:
. . .
return self.render_to_response(response_data)
def render_to_response(self, context, **response_kwargs):
return self.render_to_json_response(context, **response_kwargs)
I would like calls to /contacts/1.json to return json, 1.api to return browsableAPI, and calls with format=None aka /contacts/1/ to return a template where we call render_form. This way end-users can have pretty forms, and developers can use the .api format, and ajax/apps etc use .json. Seems like a common use case but something isn't clicking for me here in DRF...
Struggling with how DRF determines the Renderer used when no format is given. I found and then lost some info here on stack exchange that basically said to split the responses based on format. Adding the TemplateHTMLRenderer caused all sorts of pain. I had tried to split based on format but that is giving me JSON error below.
I don't understand the de facto way to define what renderer should be used. Especially when no format is provided. I mean, it "just works" when using Response(data). And I can get the TemplateHTMLRenderer to work but at the cost of having no default Renderer.
GET /contacts/1/ Gives the error:
<Contact: Contact object> is not JSON serializable
Using this code:
class ContactDetail(APIView):
permission_classes = (permissions.IsAuthenticatedOrReadOnly,
IsOwnerOrReadOnly,)
queryset = Contact.objects.all()
renderer_classes = (BrowsableAPIRenderer, JSONRenderer, TemplateHTMLRenderer,)
"""
Retrieve, update or delete a contact instance.
"""
def get_object(self, pk):
try:
return Contact.objects.get(pk=pk)
except Contact.DoesNotExist:
raise Http404
def get(self, request, pk, format=None):
contact = self.get_object(pk)
serializer = ContactSerializer(contact)
if format == 'json' or format == "api":
return Response(serializer.data)
else:
return Response({'contact': contact, 'serializer':serializer}, template_name="contact/contact_detail.html")
But GET /contacts/1.json , 1.api, or 1.html ALL give me the correct output. So it seems that I have created an issue with the content negotiation for the default i.e. format=None
I must be missing something fundamental. I have gone through the 2 tutorials and read the Renderers docs but I am unclear on what I messed up here as far as the default. I am NOT using the DEFAULT_RENDERERS in settings.py, didn't seem to make a difference if in default or inside the actual class as shown above.
Also if anyone knows a way to use TemplateHTMLRenderer without needing to switch on format value, I'm all ears.
EDIT: IF I use
if format == 'json' or format == "api" or format == None:
return Response(serializer.data)
else:
return Response({'contact': contact, 'serializer':serializer},
Then I am shown the browsable API by default. Unfortunately, what I want is the Template HTML view by default, which is set to show forms for end users. I would like to keep the .api format for developers.
TL; DR: Check the order of your renderers - they are tried in order of declaration until a content negotiation match or an error occurs.
Changing the line
renderer_classes = (BrowsableAPIRenderer, JSONRenderer, TemplateHTMLRenderer, )
to
renderer_classes = (TemplateHTMLRenderer, BrowsableAPIRenderer, JSONRenderer, )
Worked for me. I believe the reason is because the content negotiator starts at the first element in the renderer classes tuple when trying to find a renderer. When I have format==None, I'm thinking there is nothing else for DRF to go on, so it assumes I mean the "default" which seems to be the first in the tuple.
EDIT: So, as pointed out by #Ross in his answer, there is also a global setting in the settings.py for the project. If I remove my class level renderer_classes declaration and instead use this in settings.py
# ERROR
REST_FRAMEWORK = {
'DEFAULT_RENDERER_CLASSES': (
'rest_framework.renderers.BrowsableAPIRenderer',
'rest_framework.renderers.JSONRenderer',
'rest_framework.renderers.TemplateHTMLRenderer',
)
}
Then I get a (different) JSON error. However, as long as
'rest_framework.renderers.BrowsableAPIRenderer',
is not listed first, for example:
# SUCCESS, even though JSON renderer is checked first
'DEFAULT_RENDERER_CLASSES': (
'rest_framework.renderers.JSONRenderer',
'rest_framework.renderers.TemplateHTMLRenderer',
'rest_framework.renderers.BrowsableAPIRenderer',
)
So if we hit BrowsableAPIRenderer before we try TemplateHTMLRenderer then we get an error - whether or not we are relying on renderer_classes or DEFAULT_RENDERER_CLASSES. I imagine it passes through JSONRenderer gracefully but for whatever reason BrowsableAPIRenderer raises an exception.
So I have simplified my view code after analyzing this...
def get(self, request, pk, format=None):
contact = self.get_object(pk)
serializer = ContactSerializer(contact)
if format == None:
return Response({'contact': contact, 'serializer':serializer}, template_name="contact/contact_detail.html")
else:
return Response(serializer.data)
..which better reflects what I was originally trying to do anyway.
When I look at the source code, the priority seems to be the order of the renderers specified in the DEFAULT_RENDERER_CLASSES parameter in settings.py:
REST_FRAMEWORK = {
'DEFAULT_RENDERER_CLASSES': (
'rest_framework.renderers.JSONRenderer',
'rest_framework.renderers.TemplateHTMLRenderer',
),
'DEFAULT_PARSER_CLASSES': (
'rest_framework.parsers.JSONParser',
'rest_framework.parsers.TemplateHTMLRenderer',
)
}
So, if you specify a bunch of renderer classes, the first renderer that is valid will be selected based on if it is valid for the request given the .json/.api/.html extension and the Accept: header (not content-type, as I said in the comment on your question).
What I want is, if a user goes to this URL:
/user/22
then the front-end should render an HTML page for the user who's pk value is 22. This is my URLs.py:
router = routers.DefaultRouter()
router.register(r'users', views.UserViewSet)
urlpatterns = [
url(r'^user/(?P<pk>[0-9]+)$', views.UserPageView.as_view()),
url(r'^', include(router.urls)),
]
And this is UserPageView:
class UserPageView(generics.RetrieveAPIView):
queryset = User.objects.all()
renderer_classes = (TemplateHTMLRenderer,)
def get(self, request, *args, **kwargs):
self.user = self.get_object()
return Response({'user': self.user}, template_name='user.html')
And this is my UserViewSet:
class UserViewSet(viewsets.ModelViewSet):
queryset = User.objects.all()
serializer_class = UserSerializer
permission_classes = (IsCreationOrAuthenticated, IsWatchOrOwnerOrReadOnly,)
Assuming 'user' is a JS object, I plan on using my DRF API by sending a get request using AngularJS to the following URL:
("/users/" + user.pk) // which leads to UserViewSet (which serializes the user object)
to get the information for that specific user. However, user is not a JSON object, it is a Django variable which I can use in the template using the Django template tags, like so: {{ user }}.
How do I get my DRF view to return JSON to the HTML template as well? What's the best way for me to do this? Thanks in advance!
I'm not exactly sure if there's any chance that it will work the way you're currently doing at, as the first met URL will always direct you to the UserPageView.
However, perhaps you should take a look at DRF's format_suffixes.
With an additional format you can specify if you want to get the JSON or HTML representation of the user object.
Does anyone know how can I successfully retrieve the object count of a model, in JSON format, and how I need to configure my routing? I'm trying to achieve this using a APIView and returning a Response formatted by JSONRenderer.
UPDATE:
#api_view(['GET'])
#renderer_classes((JSONRenderer, JSONPRenderer))
def InfluenciasCountView(request, format=None):
influencia_count = Influencia.objects.count()
content = {'influencia_count': influencia_count}
return Response(content)
Here's the route I'm using:
url(r'^influencias/count/$', views.InfluenciasCountView, name='influencias-count')
Check out this snippet of code (the second one). If this does not suit your need, please add some of your code (for better understanding).
UPDATE
For routing, DRF offers a default router for each view. This means that you can have the following configuration in your urls.py: (using the example from the previous link)
url(r'^users/count/$', views. UserCountView.as_view(), name='users-count')
Then, when you access the URL your_base_url/users/count/ you will see something like {'user_count': 10}.
UPDATE 2
The entire code should look like this:
class UserCountView(APIView):
"""
A view that returns the count of active users.
"""
renderer_classes = (JSONRenderer, )
def get(self, request, format=None):
user_count = User.objects.count()
content = {'user_count': user_count}
return Response(content)
I am using routers from REST Framework to build my URLs. I tried above code but doesn't get it working. One of the problems is I cannot get /count/ into the router endpoints.
I checked DRF document (3.8.2) and found that there is a (new?) #action decorator (I was using 3.7.7 and it doesn't have it). So, here is my full solutions:
Upgrade DRF to 3.8.2 (or above) in requirements.txt (or PipFile if you using that).
Add a new action count method to the ModelViewSet
Update get_permissions to include the newly added action count
Here is my views.py:
from rest_framework.decorators import action
from rest_framework.response import Response
class PostViewSet(viewsets.ModelViewSet):
"""
API endpoint that allows recommend to be viewed or edited.
"""
model = Post
queryset = Post.objects.filter(is_active=True)
serializer_class = serializers.PostSerializer
filter_backends = (filters.SearchFilter, DjangoFilterBackend,)
search_fields = ('title', 'body',)
filter_fields = ('status', 'type')
def get_permissions(self):
if self.action in ('list', 'retrieve', 'create', 'count'):
return (AllowAny()),
if self.action in ('update', 'partial_update'):
return (IsAdminUser()),
return (IsAdminUser()),
#action(detail=False)
def count(self, request):
queryset = self.filter_queryset(self.get_queryset())
count = queryset.count()
content = {'count': count}
return Response(content)
To query count of posts: /api/posts/count/?format=json
To query count of published: /api/posts/count/?format=json&status=published
One of the important thing here is to use the queryset from filter_queryset(...), rather than Post.objects.all().
UPDATE
Since count is common, I created a mixin for that.
from rest_framework.decorators import action
from rest_framework.response import Response
class CountModelMixin(object):
"""
Count a queryset.
"""
#action(detail=False)
def count(self, request, *args, **kwargs):
queryset = self.filter_queryset(self.get_queryset())
content = {'count': queryset.count()}
return Response(content)
To use it, just add CountModelMixin to your ModelViewSet (also support nested ModelViewSet).
class PostViewSet(viewsets.ModelViewSet, CountModelMixin):
If you use permissions, also add 'count' to the list of granted action.