Develop Reference

Django pagination: EmptyPage: That page contains no results

When using Django CBV ListView with pagination:
class Proposals(ListView):
model = Proposal
ordering = "id"
paginate_by = 10
In the browser, if I provide a page that is out of range, I get an error:
I would like to have a different behaviour: to fallback to the last existing page if the provided page is out of range.
I dug into Django source code paginator.py file and was surprised to find some code that does exactly this:
So using paginator.get_page(page) (and not paginator.page(page)) would be the way to go. However, ListView does not use it as you can see here:
What is the best way to deal with this?
Thanks.

The only solution I found is by overriding the paginate_queryset method.
However I don't like it as I'm forced to rewrite the whole logic while I just want to change a single line.
Open to any better suggestion.
class PermissivePaginationListView(ListView):
def paginate_queryset(self, queryset, page_size):
"""
This is an exact copy of the original method, jut changing `page` to `get_page` method to prevent errors with out of range pages.
This is useful with HTMX, when the last row of the table is deleted, as the current page in URL is not valid anymore because there is no result in it.
"""
paginator = self.get_paginator(
queryset,
page_size,
orphans=self.get_paginate_orphans(),
allow_empty_first_page=self.get_allow_empty(),
)
page_kwarg = self.page_kwarg
page = self.kwargs.get(page_kwarg) or self.request.GET.get(page_kwarg) or 1
try:
page_number = int(page)
except ValueError:
if page == "last":
page_number = paginator.num_pages
else:
raise Http404(_("Page is not “last”, nor can it be converted to an int."))
try:
page = paginator.get_page(page_number)
return (paginator, page, page.object_list, page.has_other_pages())
except InvalidPage as e:
raise Http404(
_("Invalid page (%(page_number)s): %(message)s")
% {"page_number": page_number, "message": str(e)}
)

Next Page / Previous Page not working for Django pagination

I was hoping someone could help me with a pagination question.
I am trying to use Django pagination following the information on this page (https://docs.djangoproject.com/en/2.2/topics/pagination/). Whilst I have successfully displayed the correct number of items on the first page and the last page works, the next and previous pages keep taking me to the first page.
I think the issue may revolve around the ‘request’ element and I’m not sure if I am picking up an incorrect version. The example states:-
def listing(request):
contact_list = Contacts.objects.all()
paginator = Paginator(contact_list, 25) # Show 25 contacts per page
page = request.GET.get('page')
contacts = paginator.get_page(page)
return render(request, 'list.html', {'contacts': contacts})
The command:
page = request.GET.get(‘page’)
returns “AttributeError: 'Request' object has no attribute 'GET'”
By replacing this code with:
page = request.args.get('page', type=int)
the code successfully renders the first (and last) page but next and previous do not work.
As background I built my system on the Flask megatutorial but I have been unable to use that pagination, I understand because I haven’t used the Flask SQL Alchemy for creating and updating databases. My routes file has
from flask import request
Should I replace this with another utility's “request” and if so, which?

It seems the problem was in a missing () within the HTML file:-
next
listed within the example should have been:
next

How to implement pagination in a Django and React app without using the REST framework?

I am working on a website whose back-end is in Django and front-end is developed using React. I want to add the pagination feature in the website but don't know how it will be implemented in the frontend part.
I want to send the total page count to the react app.
I have written the following code in my views function till now
def index(request):
influencers = Influencer.objects.all()
paginator = Paginator(influencers,16)
page = request.GET.get('page')
paged_listings = paginator.get_page(page)
user_list = UserList.objects.all().filter(user_id = request.user.id)
queryset = list(chain(paged_listings,user_list),paginator.count)
ser_query = serializers.serialize('json', queryset)
return HttpResponse(ser_query)
Also I am not using REST framework to develop the back-end site.
I want to know what information I have to send to the React front-end site for this to work. How should I proceed?

I'd recommend you do use django-rest-framework because it's probably easier with pagination.
However, if you want to avoid that you could pass your own structure (obviously, you may need to pass other information such as next page/previous page etc).
The challenge here is ensuring everything is JSON serializable - that means for Queryset you have to use .values() and wrap the result with list(). For Paginator you need to wrap with list().
from django.http.response import JsonResponse
def index(request):
influencers = Influencer.objects.all().values()
paginator = Paginator(influencers,16)
page = request.GET.get('page')
paged_listings = paginator.get_page(page)
user_list = UserList.objects.all().filter(user_id = request.user.id).values()
queryset = list(chain(paged_listings,user_list),paginator.count)
content = {'total_page_count': paginator.num_pages, 'data': queryset}
return JsonResponse(content)

Django Pagination Current page in View.py

I am using Pagination in Django but using AJAX so I have to send all variables values from view to AJAX call. But For Current page there is no builtin variable available ?. As I saw official documentation. So how to Send this data already calculated in view.py ?
<span class="current">
Page {{ contacts.number }} of {{ contacts.paginator.num_pages }}.
</span>
Link referenceFollowing this example

The only sensible way to do pagination/ajax/django template tags, would be to generate the entire table + current page data + table navigation etc. in the django view. I.e. move all the variables for the table from your containing page to the view for the table.
Probably a better solution is to find yourself a javascript table component and have django serve it data...

Instead of creating two different views, you can deliver the paginated content from the same view by adding a GET parameter to the url, to check for the page number and sending the ajax request to the same view. This way, it'll be easier to manage one view instead of two for the same content. And if your view does much more than generating the particular content, as you are using ajax, you can easily split the view such that one view delivers only the related content.
For example, if the url of your view is \url-to-somepage
you can send the ajax request to \url-to-somepage?page=2
Then in your template file, say template.html, include another template, say __sub_template.html for the content which will be paginated. Like,
<div>
<!--
rest of the page
-->
{% include 'templates\__sub_template.html' %}
</div>
Then in your view,
.def your_view(request):
"""
Your code here
"""
paginator = Paginator(contacts, number)
page = request.GET.get('page')
try:
result_list = paginator.page(page)
except PageNotAnInteger:
result_list = paginator.page(1)
except EmptyPage:
result_list = []
if page:
return render(request, '__sub_template.html', {'contacts': result_list})
else:
return render(request, 'template.html', {'contacts': result_list})

Use Django endless pagination http://django-endless-pagination.readthedocs.io/en/latest/twitter_pagination.html#pagination-on-scroll

Django Pagination

I need to make real pagination instead of paginating on all retreived data. The example in Django documentation site, is like;
def listing(request):
contact_list = Contacts.objects.all()
paginator = Paginator(contact_list, 25) # Show 25 contacts per page
page = request.GET.get('page')
try:
contacts = paginator.page(page)
except PageNotAnInteger:
# If page is not an integer, deliver first page.
contacts = paginator.page(1)
except EmptyPage:
# If page is out of range (e.g. 9999), deliver last page of results.
contacts = paginator.page(paginator.num_pages)
return render_to_response('list.html', {"contacts": contacts})
This code is paginating records on all retreived records. But there is a trouble. Trying to retreive all record takes many time if there are so many records. I need a solution to retrieve the records page by page from database.
Is there another solution to do this in Django?

You make a false assumption. Django does not retrieve all objects when paginating: it slices the queryset appropriately, which uses LIMIT and COUNT on the SQL.

A QuerySet is a lazy object. When you assign contact_list = Contacts.objects.all(), Django will NOT hit the database. Until you call the methods such as count(), filter(), first(),..., or contact_list[1:5], Django will really retrieve data from the database. SQL statements that Django generate will correlate to each method and these SQL statments will be sent to the DB.
E.g: contact_list[1:5] generate a SQL statement have LIMIT and OFFSET clauses.
In Paginator class, the QuerySet will passed to the its constructor
paginator = Paginator(contact_list, 25)
When you call paginator.page(page), the statement is called:
return self._get_page(self.object_list[bottom:top], number, self)

Look inside Paginator class (django/core/paginator.py), it fetches only required pages. There is only one problem on big tables: if you want to show total page numbers you must make count(*) on entire table which can took a long time in some databases (i.e. postgresql, mysql with innodb).
BTW, try to use generic views in django, ListView would be fine here.

Here we using get_page() to get page wise data(1 page contain 25 data).I would suggest for this like :
def listing(request):
contact_list = Contacts.objects.all()
paginator = Paginator(contact_list, 25) # Show 25 contacts per page
page = request.GET.get('page')
contacts = paginator.get_page(page)
return render_to_response('list.html', {"contacts": contacts})