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):
             


        

Answer 1

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>  

Answer 2

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.

Answer 3

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>

Answer 4

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>

Answer 5

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

Answer 6

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>