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Django Ajax Call and response with Render

I am new to Python and Django and I would like to do a task where the user have 2 drop downs. Item drop down and Machine drop down. The Machine drop down is filled depending on the selection of the Item drop down and at the same time a table is refreshed depending on the selection of the 2 drop downs.
I was thinking to do so, from JavaScript, onChange of Item drop down, I use an AJAX function which calls a function in view.py by providing Item selection as show in the Javascript part. On return of the Django function I use render return.
Both the JavaScript and def load_machines seems to work fine but the return render(request, 'home.html', {'machines': machines}) is calling home.html but machines is empty.
How shall I tackle such problem, any hint what to look at?
JavaScript part
<script>
$("#exampleFormControlSelect1").change(function () {
const url = $("#mainForm").attr("data-machines-url");
const itemId = $(this).val();
$.ajax({ // initialize an AJAX request
url: url,
data: {
'itemId': itemId // add the item id to the GET parameters
},
success: function (data) { // `data` is the return of the `load_cities` view function
}
});
});
</script>
Django Part
view.py
def load_machines(request):
item = request.GET.get('itemId')
machines = List.objects.filter(item=item).all()
print(machines) // working FINE
return render(request, 'home.html', {'machines': machines})
urls.py
from django.urls import path
from . import views
urlpatterns = [
path('', views.home, name ='home'),
path('ajax/load-machines/', views.load_machines, name='ajax_load_machines')
# AJAX
]

rather than using render
return render(request, 'home.html', {'machines': machines})
you should return JsonResponse
from django.http import JsonResponse
data = {'machines': machines}
return JsonResponse(data)
doc about JsonResponse is here. You can also check out some simple tutorials like this

Pass data from Django view to template

I have a very basic view that is supposed to render a page and pass some data to this page, here is how i do it:
def myview(request):
request = mydb.objects.filter(user=request.user)
return render(request,
"main/mytemplate.html",
context={"data":request})
When the page is loaded, the data is passed to the template, so to show that data, i'll only have to go to my html and add this:
{{data}}
But how can i do the same from a view that is not the same view that renders the page?
Let's say that this is a view that i can call with an Ajax request, so when the Ajax request is triggered, the view should send data in the same way and i want to be able to use it in the Django template language.
Here is an example:
def secondview(request):
request = mydb.objects.filter(item='free')
return HttpResponse(request)
This view is called from an Ajax request, it will send a response with the data, but i don't want to get the data in this format or in json format, instead i want to use it from the Django template language, just as i did with the first example. Is there any way to do it? Or can i only pass data to the template in the context?

1) Instead of returning HttpResponse in your secondview do this
def secondview(request):
from django.template.loader import render_to_string
x = 1
return render_to_string('template_name.jinja', locals())
2) If you want to display that response in your html, do this in your html,
<div id="x"> </div>
function someFunc(){
$.ajax({
url: "someurl",
type: 'GET',
success: function (response) {
document.getElementById("x").innerHtml = response
},
error: function (xhr, err) {
console.log(xhr.responseText);
},
cache: false,
contentType: false,
processData: false
});
I hope I've answered all of your questions, if not let me know.

def myview(request):
request = mydb.objects.filter(user=request.user)
context = {"data":request}
return render(request, "main/mytemplate.html", context)

Django render loading page then load actual page

My actual page that I want to load takes quite a bit of time because it querying an API for a lot of data (python is doing the backend querying). How can I render my loading page and then render my actual page when the data has been gathered.
What I am trying to do in my view.py
class Page(ListView):
def loading(request):
return render(request,'loading.html')
def viewProfile(request, player_name):
Page.loading(request)
context = {
'data1' : query_api(1),
'data2' : query_api(2),
'data3' : query_api(3),
}
return render(request, 'actualpage.html', context)

In your loading page, make an ajax request to the view which will query the api, and in the success callback set the html data in your template.
However, if the api takes a lot of time, I would suggest you to use celery for processing it asynchronously so that your user can navigate the website normally instead of waiting.
In your template -
$.ajax({
url: "<query_view_url>",
type: "GET",
dataType: 'json',
success: function (data){
$("#div_id").html(data.query_data);
},
error: function(e){
console.log(e);
}
});
In your views.py -
def view_of_query_url(request, <other parameters>):
<query_logic>
data = dict()
data['query_data'] = <your_queried_data> # In html format using render_to_string
return JsonResponse(data)

Django: How to Like an Object with Ajax

Here's my View,
class ObjLike(RedirectView):
def get_redirect_url(self, *args, **kwargs):
id = self.kwargs.get('id')
obj = get_object_or_404(Data, id=id)
user = self.request.user
if user.is_authenticated():
if user in obj.likes.all():
obj.likes.remove(user)
else:
obj.likes.add(user)
So after this view how can I redirect user to the same page?
I used "return redirect(request.META['HTTP_REFERER'])" but it gives an error "name 'request' is not defined"
I can't use the get absolute URL method, i'm using this view at several places.
So, how can I do that?

to like an object with ajax calls do this
first in html we want to make a like button:
<button id="like">Like!</button>
the add a script that contain the ajax:
<script>
$(document).ready(function() {
$("#like").click(function(event){
$.ajax({
type:"POST",
url:"{% url 'like' Obj.id %}",
success: function(data){
confirm("liked")
}
});
return false;
});
});
</script>
the we add the like url to the urlpatterns list:
url(r'like/obj/(?P<pk>[0-9]+)/', views.like, name="like"),
adding the view :
from django.views.decorators.csrf import csrf_exempt
#csrf_exempt
def like(request, pk)
obj = Obj.objects.get(id=pk)
obj.likes += 1
obj.save()
return HttpResponse("liked")
Note: you can customize the like view to check if user liked already

Django returns invalid ajax response

I have a ajax call in my django template file as:
$(document).ready(function () {
$("button#wdsubmit").click(function(){
$.ajax({
type: "post",
url: "/audit/addwd/",
data: $('form.wddetails').serialize(),
dataType: "json",
success: function(msg){
alert(msg);
alert('Added Successfully');
$("#newwd").modal('hide'); //hide popup
},
error: function(msg){
alert(msg.success);
}
});
});
});
Form:
class WDForm(ModelForm):
class Meta:
model = WDModel
fields = '__all__'
and view in django is :
def addwd(request):
if request.method == 'POST':
updated_request = request.POST.copy()
updated_request.update({'updated_by': request.user.username})
form = WDForm(updated_request)
if form.is_valid():
form.save()
response = simplejson.dumps({'success': True})
return HttpResponse(response, content_type="application/json", mimetype='application/json')
else:
response = simplejson.dumps({'error': True})
return HttpResponse(response , content_type="application/json")
Whenever I make a Ajax call it always returns error even though I have sent Success(Means the form is valid and data is successfully pushed to database).
I also tried to send response={'success':True} doesn't work.
Please help me to solve this issue.
Environment Details:
Python verion: 3.4
Django :1.7
Windows OS 8
I doubt on this line " response = simplejson.dumps({'success': success})
"

you can try JsonResponse objects.
from django.http import JsonResponse
return JsonResponse({'foo':'bar'})