Fast and easy AJAX libraries for django applications.
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3.x
python_ >=3.5django_ >=2.02.x
python_ >=2.7django_ >=1.7,<2.0.. _python: http://www.python.org/
.. _django: https://djangoproject.com
.. _jQuery: http://jquery.com
Install django-ajax in your python environment
1- Download and install package:
.. code:: sh
$ pip install djangoajax
Through Github:
.. code:: sh
pip install -e git://github.com/yceruto/django-ajax#egg=djangoajax
or simply with:
.. code:: sh
$ python setup.py install
2- Add 'django_ajax' into the INSTALLED_APPS list.
3- Read usage section and enjoy this feature!
@ajax Decorator
.. code:: python
from django_ajax.decorators import ajax
@ajax
def my_view(request):
do_something()
When the view does not return anything, you will receive this response (JSON format):
.. code:: javascript
{"status": 200, "statusText": "OK", "content ": null}
**Sending content**
.. code:: python
@ajax
def my_view(request):
c = 2 + 3
return {'result': c}
The whole result is converted into a JSON format as part of the `content` element:
.. code:: javascript
{"status": 200, "statusText": "OK", "content": {"result": 5}}
**Combining with others decorators**
.. code:: python
from django.contrib.auth.decorators import login_required
from django_ajax.decorators import ajax
@ajax
@login_required
def my_view(request):
# if the request.user is anonymous then this view not proceed
return {'user_id': request.user.id}
The location or path of the redirection response will be given in the `content` item,
also the `status` and `statusText` will reflect what is going on:
.. code:: javascript
{"status": 302, "statusText": "FOUND", "content": "/login"}
**Template response**
.. code:: python
from django.shortcuts import render
from django_ajax.decorators import ajax
@ajax
def my_view(request):
return render(request, 'home.html')
The JSON response:
.. code:: javascript
{"status": 200, "statusText": "OK", "content": "<html>...</html>"}
**Catch exceptions**
.. code:: python
@ajax
def my_view(request):
a = 23 / 0 # this line throws an exception
return a
The JSON response:
.. code:: javascript
{"status": 500, "statusText": "INTERNAL SERVER ERROR", "content": "integer division or modulo by zero"}
AJAXMiddleware
~~~~~~~~~~~~~~
If you are using AJAX at all times in your project, we suggest you activate the AJAXMiddleware described below.
Add ``django_ajax.middleware.AJAXMiddleware`` to the ``MIDDLEWARE_CLASSES`` list in ``settings.py`` and all your responses will be converted to JSON whereas the request was made via AJAX, otherwise it will return a normal HttpResponse.
.. caution:: If this middleware is activated you cannot use the ``@ajax`` decorator. That will cause double JSON conversion.
AJAXMixin for class-based views
AJAXMixin is an object that call to AJAX decorator.
.. code:: python
from django.views.generic import TemplateView
from django_ajax.mixin import AJAXMixin
class SimpleView(AJAXMixin, TemplateView):
template_name = 'home.html'
The JSON response:
.. code:: javascript
{"status": 200, "statusText": "OK", "content": "<html>...</html>"}
Enjoy And Share!
FAQs
Powerful and easy AJAX framework for django applications.
We found that djangoajax demonstrated a healthy version release cadence and project activity because the last version was released less than a year ago. It has 1 open source maintainer collaborating on the project.
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