Following is my code, Every time I click Like (or Dislike) link, both Dislike and Like are incremented.
I am new to django, your help will be much appreciated. Thank you
Here is my models.py,
from django.db import models
from django.urls import reverse
# Create your models here.
class Movies(models.Model):
Director=models.CharField(max_length=30)
Cast_I=models.CharField(max_length=30)
Cast_II=models.CharField(max_length=30)
Name=models.TextField()
ReleaseYear=models.IntegerField()
ImdbRating=models.CharField(max_length=2)
Genre=models.TextField(null=True)
Language=models.CharField(max_length=20,null=True)
Like=models.IntegerField()
Dislike=models.IntegerField()
def like_this_movie(self):
self.Like+=1
self.save()
return reverse ('list',kwargs={})
def dislike_this_movie(self):
self.Dislike+=1
self.save()
return reverse ('list',kwargs={})
Here is the template,
{% block content %}
<p>
Like
Dislike
</p>
{% endblock %}
Here is my view class,
class MovieDetailView(DetailView):
template_name='movie/detail.html'
queryset=Movies.objects.all()
The templates in Django don't work like that, everything you put inside {{ }} is rendered on the server and then the page is sent to the client, so it is not possible to run a function like that.
You would have to do something like an AJAX request to another page which runs like_this_movie and takes the Movies id as a parameter. To send to AJAX request you can use Javascript with jQuery.
The problem is that each time you load the page both values are incremented because you call both functions in each call. This link will help you on how do it => How to make user to Like/Dislike post only for once in Django?
I have resolved my bug by adding those like_this_movie and dislike_this_movie function to my views.py and updating my urls.py and template. Thank you guys for your valuable answer.
views.py,
def like_this_movie(request,key):
obj=Movies.objects.get(pk=key)
obj.Like+=1
obj.save()
return redirect('/movie/list')
def dislike_this_movie(request,key):
obj=Movies.objects.get(pk=key)
obj.Dislike+=1
obj.save()
return redirect('/movie/list')
urls.py,
urlpatterns=[
path('list/<int:key>/like',like_this_movie,name='like'),
path('list/<int:key>/dislike',dislike_this_movie,name='dislike')
]
template,
Like
Dislike
Now I have resolved this by using jQquery AJAX, like_this_movie and dislike_this_movie function returns json value to ajax request. Now page doesn't reload every time I hit like or dislike button. Thank you guys for your valuable answer.
views.py,
def like_this_movie(request,key):
obj=Movies.objects.get(pk=key)
obj.Like+=1
obj.save()
return JsonResponse({'success':True,'content':'Like','Like':obj.Like})
def dislike_this_movie(request,key):
obj=Movies.objects.get(pk=key)
obj.Dislike+=1
obj.save()
return JsonResponse({'success':True,'content':'Dislike','Dislike':obj.Dislike})
urls.py,
urlpatterns=[
path('<int:key>/like',like_this_movie,name='like'),
path('<int:key>/dislike',dislike_this_movie,name='dislike')
]
detail.js
document.addEventListener('DOMContentLoaded',()=>{
document.querySelector('#likebtn').onclick=()=>{
const request=new XMLHttpRequest;
request.open('GET','like',true);
console.log(request);
request.onload=()=>{
const data = JSON.parse(request.responseText);
if ((data.success)&&(data.content=='Like')){
totalLike=data.Like;
likespan=document.querySelector('#likespan');
likespan.innerHTML=totalLike;
likebtn=document.querySelector('#likebtn');
likebtn.disabled = true;
dislikebtn.disabled=true;
}
}
request.send();
return false;
}
document.querySelector('#dislikebtn').onclick=()=>{
const request=new XMLHttpRequest;
request.open('GET','dislike',true);
request.onload=()=>{
const data = JSON.parse(request.responseText);
if ((data.success)&&(data.content=='Dislike')){
totalDislike=data.Dislike;
dislikespan=document.querySelector('#dislikespan');
dislikespan.innerHTML=totalDislike;
dislikebtn=document.querySelector('#dislikebtn');
dislikebtn.disabled = true;
likebtn.disabled=true;
}
}
request.send();
return false;
}
})
detail.html,
<h4>
<span id="likespan">{{object.Like}}</span>
<button id="likebtn" type="submit" class="btn btn-primary">Like</button>
<span id="dislikespan">{{object.Dislike}}</span>
<button id="dislikebtn" type="submit" class="btn btn-primary">Dislike</button>
</h4>
I'm following the suggestion at: Refresh <div> element generated by a django template
I'm passing along a few variables, a la:
url: '{% url 'search_results' 'sched_dep_local' flights|escapejs %}',
The problem is that 'flights' is a list of dicts that the search_results template needs access to, and it's pretty large and contains things like apostrophes
[{'foo': 'bar'}, {'foo': 'baz'}] and so on
So the only way I can use it with {% url %} appears to be with escapejs to get rid of the apostrophes, but then in views.py, I need it to be a list of dicts again, so I can do things like:
def search_results(request, sort_key, flights):
flights = search_utils.sort(flights, sort_key)
return render_to_response('search_results.html', { 'flights' : flights} )
Is there a simple way to do this? Alternatively, am I going about this whole thing all wrong?
ETA: See also (explains what I'm trying to do and why):
<script>
$(".sort").on("click", function() {
$.ajax({
url: '{% url 'search_results' 'sched_dep_local' flights|escapejs %}',
success: function(data) {
$('#search-results').html(data);
}
});
});
</script>
I have a template (in search_results.html) that prints some data for each flight in flights. I want to sort that data and rerender the template, but I can't figure out how.
This isn't the right way to deal with complex data. Rather than sending it via the URL, you should be using a POST and sending it in the body of the request: since you're using jQuery, you can just do method: "POST" in that call. In the backend, you can deserialize it from JSON.
However, it does seem a bit strange to do this at all; the data is evidently coming from the Django backend already, so it's not clear why you want to post it back there.
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 am currently making a website using django. Now I want to execute a python script from my template/view with a button on the website. It should be possible but to be honest I don't know how.
An example would be best.
Thanks for any help at all.
Well got it working now and just thought I would write my answer:
view.py:
import script as gh
def get_hostname(request):
gh.main()
return HttpResponseRedirect('/')
urls.py:
...
url(r'^get_hostname/$', 'thinco.views.get_hostname'),
...
somewhere in template:
...
<form action="/get_hostname/" method="GET">
<input type="submit" value="Liste der Thin Clients laden">
</form>
...
If you are using Django - best way, IMHO, is just to create the view that will handle your python code, and then access it at onclick event via ajax request.
yourapp/views.py
def your_python_script(request):
if request.is_ajax:
# do your stuff here
...
else:
return HttpRequest(status=400)
If you are using django also you should have jQuery, and in your template add javascript code, something like this:
$("#<your_button_id>").click( function() {
$.post("your_python_script_url", {<dict with any args you need>}, function () {
// What to do when request successfully completed
});
});
And not to forget about CRSF token if your are using it. How to handle it you can find in offical django documentation.
UPDATE
You can add csrf token to page template like:
<script>
var csrf_token = '{% csrf_token %}';
...
</script>
Next, you need to bind to the global jquery ajaxSend event, and add the token to any POST request.
$("body").bind("ajaxSend", function(elm, xhr, s) {
if (s.type == "POST") {
xhr.setRequestHeader('X-CSRF-Token', csrf_token);
}
});
Something like this should work.
Create a view function and do an #dajaxice_register decorator for it:
A silly example follows:
models.py:
class Funkyness(models.Model):
funktasm = models.CharField(max_length=128)
funfasticness = models.TextField()
urls.py:
url(r'^funk$', 'views.myFunkyView'),
views.py:
def myFunkyView(request)
render_to_request('index.htm', {'where': 'home'}, context_instance=RequestContext(request))
index.htm:
{% if where %}
You are {{ where }}
{% endif %}
When you go to http://yoursite.com/funk, you will get index.htm rendered and you will get a page that says "You are home."
Now, the dynamic part...
Write a view method as such:
from django.utils import simplejson
def getHowFunky(request, v):
html = """
<div class="my_message">
This is really funky, almost stinky...
<br />
""" + str(v) + """
</div>
"""
return simplejson.dumps({'message': html})
back in index.htm:
<script type="text/javascript>
/* first argument is return JS function, the second is the dictionary of data to sent to the python method. */
function init(){
Dajaxice.package.getHowFunky(showFunky, {'v': "Your's Truly... Fubar"});
}
function showFunky(data){
/* This is the data returned back from the AJAX (Dajaxice) call. */
document.write(data.message)
}
</script>
So, you build a python method that takes inputs and returns something. You register it with Dajaxice, you call it, passing a callback method. It runs and when it succeeds, it send the python return (possibly JSON object) to the callback method as an argument. That method then writes to the screen what it got from the Dajaxice call.
For more info on Dajaxice, go to: http://dajaxproject.com/
Props to Jorge Bastida, the sole developer of Dajax/Dajaxice!
What I could think of for an answer is to use:
Django + Dajax
Django link:
https://docs.djangoproject.com/en/1.4/
Dajax is actually ajax for django:
Visit their website and refer to their examples for quick start
http://www.dajaxproject.com/
You can create your button in django view and upon triggering your button, you can use run a python snippet, not script.
If you want to run a standalone script, you could probably check out djcelery.
http://pypi.python.org/pypi/django-celery