Django provides a few classes that help you manage paginated data – that is, data that’s split across several pages, with “Previous/Next” links. These classes live in django/core/paginator.py.
Give Paginator a list of objects, plus the number of items you’d like to have on each page, and it gives you methods for accessing the items for each page:
>>> from django.core.paginator import Paginator
>>> objects = ['john', 'paul', 'george', 'ringo']
>>> p = Paginator(objects, 2)
>>> p.count
4
>>> p.num_pages
2
>>> p.page_range
[1, 2]
>>> page1 = p.page(1)
>>> page1
<Page 1 of 2>
>>> page1.object_list
['john', 'paul']
>>> page2 = p.page(2)
>>> page2.object_list
['george', 'ringo']
>>> page2.has_next()
False
>>> page2.has_previous()
True
>>> page2.has_other_pages()
True
>>> page2.next_page_number()
3
>>> page2.previous_page_number()
1
>>> page2.start_index() # The 1-based index of the first item on this page
3
>>> page2.end_index() # The 1-based index of the last item on this page
4
>>> p.page(0)
Traceback (most recent call last):
...
EmptyPage: That page number is less than 1
>>> p.page(3)
Traceback (most recent call last):
...
EmptyPage: That page contains no results
Note
Note that you can give Paginator a list/tuple, a Django QuerySet, or any other object with a count() or __len__() method. When determining the number of objects contained in the passed object, Paginator will first try calling count(), then fallback to using len() if the passed object has no count() method. This allows objects such as Django's QuerySet to use a more efficient count() method when available.
Here's a slightly more complex example using Paginator in a view to paginate a queryset. We give both the view and the accompanying template to show how you can display the results. This example assumes you have a Contacts model that has already been imported.
The view function looks like this:
from django.core.paginator import Paginator, InvalidPage, EmptyPage
def listing(request):
contact_list = Contacts.objects.all()
paginator = Paginator(contact_list, 25) # Show 25 contacts per page
# Make sure page request is an int. If not, deliver first page.
try:
page = int(request.GET.get('page', '1'))
except ValueError:
page = 1
# If page request (9999) is out of range, deliver last page of results.
try:
contacts = paginator.page(page)
except (EmptyPage, InvalidPage):
contacts = paginator.page(paginator.num_pages)
return render_to_response('list.html', {"contacts": contacts})
In the template list.html, you'll want to include navigation between pages along with any interesting information from the objects themselves:
{% for contact in contacts.object_list %}
{# Each "contact" is a Contact model object. #}
{{ contact.full_name|upper }}<br />
...
{% endfor %}
<div class="pagination">
<span class="step-links">
{% if contacts.has_previous %}
<a href="?page={{ contacts.previous_page_number }}">previous</a>
{% endif %}
<span class="current">
Page {{ contacts.number }} of {{ contacts.paginator.num_pages }}.
</span>
{% if contacts.has_next %}
<a href="?page={{ contacts.next_page_number }}">next</a>
{% endif %}
</span>
</div>
The Paginator class has this constructor:
The total number of objects, across all pages.
Note
When determining the number of objects contained in object_list, Paginator will first try calling object_list.count(). If object_list has no count() method, then Paginator will fallback to using object_list.__len__(). This allows objects, such as Django's QuerySet, to use a more efficient count() method when available.
The page() method raises InvalidPage if the requested page is invalid (i.e., not an integer) or contains no objects. Generally, it's enough to trap the InvalidPage exception, but if you'd like more granularity, you can trap either of the following exceptions:
Both of the exceptions are subclasses of InvalidPage, so you can handle them both with a simple except InvalidPage.
You usually won't construct Pages by hand -- you'll get them using Paginator.page().
Jul 05, 2010