Display only some of the page numbers by django pagination

Question

I am using the django paginator in the template. Its working ok, but not good when there's large numbers of pages.

views.py:

def blog(request):
    blogs_list = Blog.objects.all()

    paginator = Paginator(blogs_list, 1)

    try:
        page = int(request.GET.get('page', '1'))
    except:
        page = 1

    try:
        blogs = paginator.page(page)
    except(EmptyPage, InvalidPage):
        blogs = paginator.page(page)
    return render(request, 'blogs.html', {
        'blogs':blogs
        })

snippet of the template:

  <div class="prev_next">

    {% if blogs.has_previous %}
      <a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
    {% endif %}
    {% if blogs.has_next %}
      <a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
    {% endif %}
    <div class="pages">
      <ul>
      {% for pg in blogs.paginator.page_range %}
        {% if blogs.number == pg %}
          <li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
        {% else %}
          <li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
        {% endif %}
      {% endfor %}
      </ul>
    </div>
    <span class="clear_both"></span>

  </div>

Now it looks like this:

enter image description here

What do I do to display only 7 page numbers and not all of it ranging from the current page number, like this:

Prev 1 (2) 3 4 5 Next

I hope I was clear, if not please ask. Your help and guidance will be very much appreciated. Thank you.

Rob L · Answer 1 · 2018-01-02T15:53:35.760

94

Gonna throw this in. I came up with it because it lets you know there are more pages on either side.

<ul class="pagination">

{% if page_obj.has_previous %}
    <li><a href="?page={{ page_obj.previous_page_number }}"><i class="fa fa-chevron-left" aria-hidden="true"></i></a></li>
{% else %}
    <li class="disabled"><span><i class="fa fa-chevron-left" aria-hidden="true"></i></span></li>
{% endif %}

{% if page_obj.number|add:'-4' > 1 %}
    <li><a href="?page={{ page_obj.number|add:'-5' }}">&hellip;</a></li>
{% endif %}

{% for i in page_obj.paginator.page_range %}
    {% if page_obj.number == i %}
        <li class="active"><span>{{ i }} <span class="sr-only">(current)</span></span></li>
    {% elif i > page_obj.number|add:'-5' and i < page_obj.number|add:'5' %}
        <li><a href="?page={{ i }}">{{ i }}</a></li>
    {% endif %}
{% endfor %}

{% if page_obj.paginator.num_pages > page_obj.number|add:'4' %}
    <li><a href="?page={{ page_obj.number|add:'5' }}">&hellip;</a></li>
{% endif %}

{% if page_obj.has_next %}
    <li><a href="?page={{ page_obj.next_page_number }}"><i class="fa fa-chevron-right" aria-hidden="true"></i></a></li>
{% else %}
    <li class="disabled"><span><i class="fa fa-chevron-right" aria-hidden="true"></i></span></li>
{% endif %}

</ul>

And it looks like this:

edited Jan 02 '18 at 15:53

answered Sep 20 '17 at 18:43

Rob L

3,634
2
19
38

1

Excellent answer !! Based on this answer I created a templatetag with the same result. – cwhisperer Feb 06 '19 at 12:38
Awesome! This is also working with generic class based views. – Tech Dec 30 '19 at 08:13
This is awesome! Thanks for sharing this, this saved me from using a package or having to wrap my brain around this. (Used on Django 2.2) – Kalob Taulien Jan 11 '20 at 05:57
This is awesome!! Thanks a lot! – jitesh2796 Jul 08 '21 at 17:24
how to responsive that for mobile? – Ebrahim Abdollahian Dec 09 '21 at 05:32
@EbrahimAbdollahian you could use a css framework, such as Bootstrap (v5 as of this comment), embed the above within `
` to get present responsively. The example used is very generic. – Carewen Jul 17 '22 at 03:49

score 54 · Answer 2 · edited May 27 '20 at 05:06

54

Another shorter solution with template is compare current forloop.counter with certain range.

with bootstrap I use this template

<nav aria-label="Page navigation">   <ul class="pagination">
{% if page_obj.has_previous %}
<li class="page-item">
  <a class="page-link" href="?page=1" aria-label="Previous">
    <span aria-hidden="true">&laquo;</span>
    <span class="sr-only">begin</span>
  </a>
</li>   {% endif %}

{% for n in page_obj.paginator.page_range %}
  {% if page_obj.number == n %}
    <li class="page-item active">
      <span class="page-link">{{ n }}<span class="sr-only">(current)</span></span>
    </li>
  {% elif n > page_obj.number|add:'-3' and n < page_obj.number|add:'3' %}
    <li class="page-item"><a class="page-link" href="?page={{ n }}">{{ n }}</a></li>
  {% endif %}
{% endfor %}

{% if page_obj.has_next %}
  <li class="page-item">
    <a class="page-link" href="?page={{ page_obj.paginator.num_pages }}" aria-label="Next">
      <span aria-hidden="true">&raquo;</span>
      <span class="sr-only">end</span>
    </a>
  </li>
  {% endif %}   </ul> </nav>

edited May 27 '20 at 05:06

Jason Liu

749
4
16
34

answered Aug 16 '17 at 15:26

Pavel1114

588
5
7

2

I liked this answer best, much shorter and easier to understand\ – ColinMasters Sep 02 '17 at 22:04
This is the simpliest approach.It helped me, Thank you very much – ahmet May 01 '18 at 17:25
(to avoid duplicate first and last page in between pages) Line 15 replace with: ` {% if n != 1 and n != page_obj.paginator.num_pages %}
{{ n } {% endif %} `

Piero

May 26 '20 at 01:50

this is wise amazing answer very very very simple solution to complex problem – soField Nov 03 '20 at 11:12

Simple approach helped me lot. Cheers! – Laban Nov 17 '20 at 06:03

Blackeagle52 · Accepted Answer · 2017-10-10T09:20:01.593

First of all I would change the following:

try:
    blogs = paginator.page(page)
except(EmptyPage, InvalidPage):
    blogs = paginator.page(page)  # Raises the same error

But you could pass a range within your context.

index = paginator.page_range.index(blogs.number)
max_index = len(paginator.page_range)
start_index = index - 3 if index >= 3 else 0
end_index = index + 3 if index <= max_index - 3 else max_index
page_range = paginator.page_range[start_index:end_index]

Now you should be able to loop over the range to construct the right links with ?page=.

=== Edit ===
So your view would be something like this:

def blog(request):
    paginator = Paginator(Blog.objects.all(), 1)

    try:
        page = int(request.GET.get('page', '1'))
    except:
        page = 1

    try:
        blogs = paginator.page(page)
    except(EmptyPage, InvalidPage):
        blogs = paginator.page(1)

    # Get the index of the current page
    index = blogs.number - 1  # edited to something easier without index
    # This value is maximum index of your pages, so the last page - 1
    max_index = len(paginator.page_range)
    # You want a range of 7, so lets calculate where to slice the list
    start_index = index - 3 if index >= 3 else 0
    end_index = index + 3 if index <= max_index - 3 else max_index
    # Get our new page range. In the latest versions of Django page_range returns 
    # an iterator. Thus pass it to list, to make our slice possible again.
    page_range = list(paginator.page_range)[start_index:end_index]

    return render(request, 'blogs.html', {
        'blogs': blogs,
        'page_range': page_range,
    })

So now we have to edit your template to accept our new list of page numbers:

<div class="prev_next">
    {% if blogs.has_previous %}
        <a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
    {% endif %}
    {% if blogs.has_next %}
        <a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
    {% endif %}
    <div class="pages">
        <ul>
        {% for pg in page_range %}
            {% if blogs.number == pg %}
                <li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
            {% else %}
                <li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
            {% endif %}
        {% endfor %}
        </ul>
    </div>
    <span class="clear_both"></span>
</div>

I am sorry, I didn't get you. Could you please elaborate a bit more. — Robin, Jun 16 '15 at 10:33
Exception: sequence index must be integer, not 'slice' on page_range = paginator.page_range[start_index:end_index] — Raymond Seger, Oct 09 '17 at 08:16
@RaymondSeger, this is because Django changed its functionality, in older functions paginator.page_range was a list, now it returns an iterator. Will update my answer. — Blackeagle52, Oct 10 '17 at 09:15

score 10 · Answer 4 · edited Apr 02 '20 at 21:41

I found the simplest thing to do was to create a pagination snippet that just shows the pages you want it to.

In my case I didn't want any previous or next links. I just wanted to always have a link to the first and last pages and then have the current page and the two pages either side of the current page.

My template snippet (uses variables from django-tables2 - variables will have slightly different names if you're using a Django Paginator directly)

{% load django_tables2 %}
{% load humanize %}
{% load i18n %}

{% if table.page %}
  {% with table.page.paginator.count as total %}
    {% with table.page.number as page_num %}
      {% with table.page.paginator.num_pages as num_pages %}
        {% block pagination %}
          <div class="row">
            <div class="col-md-12">    
              {% if table.paginator.num_pages > 1 %}
                <ul class="pagination pull-right">
                  {% for n in table.page.paginator.page_range %}
                    {% if table.page.number|add:'-3' == n %}
                      {# First page #}
                      <li><a href="{% querystring table.prefixed_page_field=1 %}">1</a></li>
                      {% if n != 1 %}
                        <li class="disabled"><a>&#8943;</a></li>
                      {% endif %}
                    {% elif table.page.number == n %}
                      {# Current page #}
                      <li class="active"><a href="#">{{ n }}</a></li>
                    {% elif table.page.number|add:'-3' < n and n < table.page.number|add:'3' %}
                      {# Pages around current page #}
                      <li><a href="{% querystring table.prefixed_page_field=n %}">{{ n }}</a></li>
                    {% elif table.page.number|add:'3' == n %}
                      {# Last page #}
                      {% if n != num_pages %}
                        <li class="disabled"><a>&#8943;</a></li>
                      {% endif %}
                      <li><a href="{% querystring table.prefixed_page_field=num_pages %}">{{ num_pages }}</a></li>
                    {% endif %}
                  {% endfor %}
                </ul>
              {% endif %}
            </div>
          </div>
        {% endblock pagination %}
      {% endwith %}
    {% endwith %}
  {% endwith %}
{% endif %}

Examples of what my pagination looks like at different pages

Credit: this was inspired by @Pavel1114's answer.

score 5 · Answer 5 · answered Mar 23 '21 at 23:42

5

Django 3.2 introduces a new feature on the Paginator class:

Paginator.get_elided_page_range(number, *, on_each_side=3, on_ends=2)

With the default values for on_each_side and on_ends, if the current page is 10 and there are 50 pages, the page range will be [1, 2, '…', 7, 8, 9, 10, 11, 12, 13, '…', 49, 50].

answered Mar 23 '21 at 23:42

jnns

5,148
4
47
74

Using `get_elided_page_range` as a template tag: https://stackoverflow.com/questions/69277936/how-to-use-get-elided-page-range-in-django-paginator – Somraj Chowdhury Nov 04 '21 at 11:24

score 4 · Answer 6 · answered May 24 '20 at 00:35

You can use the following code.

     <nav aria-label="Page navigation example">
        <ul class="pagination justify-content-center">

            <li class="page-item {% if not page_obj.has_previous %} disabled {% endif %}">
                <a class="page-link" href="?page=1" tabindex="-1">FIRST</a>
            </li>
            <li class="page-item {% if not page_obj.has_previous %} disabled {% endif %}">
                <a class="page-link" href="{% if page_obj.has_previous %}?page={{ page_obj.previous_page_number }}{% endif %} " tabindex="-1">Previous</a>
            </li>

            {% if page_obj.number|add:'-4' > 1 %}
                <li class="page-item disabled"><a class="page-link" href="?page={{ page_obj.number|add:'-5' }}">&hellip;</a></li>
            {% endif %}

            {% for i in page_obj.paginator.page_range %}
                {% if page_obj.number == i %}
                    <li class="active page-item disabled"><a class="page-link" href="?page={{ i }}">{{ i }}</a></li>
                {% elif i > page_obj.number|add:'-5' and i < page_obj.number|add:'5' %}
                    <li class="page-item"><a class="page-link" href="?page={{ i }}">{{ i }}</a></li>
                {% endif %}
            {% endfor %}

            {% if page_obj.paginator.num_pages > page_obj.number|add:'4' %}
                <li class="page-item disabled"><a class="page-link" href="?page={{ page_obj.number|add:'5' }}">&hellip;</a></li>
            {% endif %}

            <li class="page-item {% if not page_obj.has_next %} disabled {% endif %}">
                <a class="page-link" href="{% if page_obj.has_next %} ?page={{ page_obj.next_page_number }} {% endif %}">Next</a>
            </li>
            <li class="page-item {% if not page_obj.has_next %} disabled {% endif %}">
                <a class="page-link" href="{% if page_obj.has_next %} ?page={{ page_obj.paginator.num_pages }}  {% endif %}">LAST</a>
            </li>
        </ul>
    </nav>

Depending on the current page, number links with page number displayed is not constant. if current page is 1: 5 links are displayed, if current page is 5 or more 9 links are displayed. How to display the same number of links? — donmelchior, Jul 10 '20 at 13:01

score 3 · Answer 7 · answered Jun 07 '17 at 13:46

You could also extend Paginator class.

class BootstrapPaginator(Paginator):
    def __init__(self, *args, **kwargs):
        """        
        :param wing_pages: How many pages will be shown before and after current page.
        """
        self.wing_pages = kwargs.pop('wing_pages', 3)
        super(BootstrapPaginator, self).__init__(*args, **kwargs)

    def _get_page(self, *args, **kwargs):
        self.page = super(BootstrapPaginator, self)._get_page(*args, **kwargs)
        return self.page

    @property
    def page_range(self):
        return range(max(self.page.number - self.wing_pages, 1),
                     min(self.page.number + self.wing_pages + 1, self.num_pages + 1))

Then in template:

{% for num in action_list.paginator.page_range %}
    {% if action_list.number == num %}
         <li class="active"><a href="?page={{ num }}">{{ num }}</a></li>
    {% else %}
         <li><a href="?page={{ num }}">{{ num }}</a></li>
    {% endif %}
{% endfor %}

Now page_range will only consist of 7 items. wing_pages + current page + wing_pagescustom django paginator for bootstrap

Jason Liu · Answer 8 · 2020-04-02T16:00:30.810

Inspired by @Pavel1114 and @Inti. Removed the arrow at the beginning and at the end as they are not that useful. Added The first page and last page for faster accessing.

<ul>
  <li class="grp-results">
      <span>{{paginator.count}} total</span>
  </li>
  {% if data.number|add:'-4' > 1 %}
    <li><a href="?page=1">1</a></li>
    <li><a href="?page={{ data.number|add:'-5' }}">&hellip;</a></li>
  {% endif %}
  {% for i in data.paginator.page_range %}
    {% if data.number == i %}
        <li class="active"><span>{{ i }}</span></li>
    {% elif i > data.number|add:'-3' and i < data.number|add:'3' %}
        <li><a href="?page={{ i }}">{{ i }}</a></li>
      {% endif %}
  {% endfor %}
  {% if data.paginator.num_pages > data.number|add:'4' %}
    <li><a href="?page={{ data.number|add:'5' }}">&hellip;</a></li>
    <li><a href="?page={{ data.paginator.count }}">{{ data.paginator.count }}</a></li>
  {% endif %}
</ul>

score 1 · Answer 9 · answered Jun 08 '17 at 17:51

I did it only on templates with expressions:

{% if is_paginated %}
    <div class="text-center">
        <ul class="pagination pagination-sm">
            {% if page_obj.number >= 5 %}
            <li><a href="?page=1">1</a></li>
            <li><span>...</span></li>
            {% elif page_obj.number == 4 %}
            <li><a href="?page=1">1</a></li>
            {% endif %}
            {% if page_obj.number|add:"-2" >= 1  %}
            <li><a href="?page={{ page_obj.number|add:"-2" }}">{{ page_obj.number|add:"-2" }}</a></li>
            {% endif %}
            {% if page_obj.number|add:"-1" >= 1  %}
            <li><a href="?page={{ page_obj.number|add:"-1" }}">{{ page_obj.number|add:"-1" }}</a></li>
            {% endif %}
            <li class="active"><a href="?page={{ page_obj.number }}">{{ page_obj.number }}</a></li>
            {% if page_obj.number|add:"1" <= paginator.num_pages  %}
            <li><a href="?page={{ page_obj.number|add:"1" }}">{{ page_obj.number|add:"1" }}</a></li>
            {% endif %}
            {% if page_obj.number|add:"2" <= paginator.num_pages %}
            <li><a href="?page={{ page_obj.number|add:"2" }}">{{ page_obj.number|add:"2" }}</a></li>
            {% endif %}
            {% if page_obj.number|add:"2" <= paginator.num_pages|add:"-2" %}
            <li><span>...</span></li>
            <li><a href="?page={{ paginator.num_pages }}">{{ paginator.num_pages }}</a></li>
            {% elif page_obj.number|add:"1" <= paginator.num_pages|add:"-2" %}
            <li><a href="?page={{ paginator.num_pages }}">{{ paginator.num_pages }}</a></li>
            {% endif %}
        </ul>
    </div>
{% endif %}

I know django is like "dont write your code again" but i found this easier for me to understand right now. Hope i helped.

Display only some of the page numbers by django pagination

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