How to paginate Django with other get variables?

user_204349 0
Viewed 34698

I am having problems using pagination in Django. Take the URL below as an example:

http://127.0.0.1:8000/users/?sort=first_name

On this page I sort a list of users by their first_name. Without a sort GET variable it defaults to sort by id.

Now if I click the next link I expect the following URL:

http://127.0.0.1:8000/users/?sort=first_name&page=2

Instead I lose all get variables and end up with 

http://127.0.0.1:8000/users/?page=2

This is a problem because the second page is sorted by id instead of first_name.

If I use request.get_full_path I will eventually end up with an ugly URL:

http://127.0.0.1:8000/users/?sort=first_name&page=2&page=3&page=4

What is the solution? Is there a way to access the GET variables on the template and replace the value for the page?

I am using pagination as described in Django's documentation and my preference is to keep using it. The template code I am using is similar to this:

{% if contacts.has_next %}
    <a href="?page={{ contacts.next_page_number }}">next</a>
{% endif %}
pagination django
21 Answers

After some playing around I found a solution... although I don't know if it's really a good one. I'd prefer a more elegant solution.

Anyway I pass the request to the template and am able to access all the GET variables via request.GET. Then I loop through the GET dictionary and as long as the variable isn't page I print it.

{% if contacts.has_previous %}
    <a href="?page={{ contacts.previous_page_number }}{% for key,value in request.GET.items %}{% ifnotequal key 'page' %}&{{ key }}={{ value }}{% endifnotequal %}{% endfor %}">previous</a>
{% endif %}

<span class="current">
    Page {{ contacts.number }} of {{ contacts.paginator.num_pages }}.
</span>

{# I have all of this in one line in my code (like in the previous section), but I'm putting spaces here for readability.  #}
{% if contacts.has_next %}
    <a href="?page={{ contacts.next_page_number }}
        {% for key,value in request.GET.items %}
            {% ifnotequal key 'page' %}
                &{{ key }}={{ value }}
            {% endifnotequal %}
        {% endfor %}
    ">next</a>
{% endif %}
user_2043490

In your views.py you will somehow access the criteria on which you sort, e.g. first_name. You'll need to pass that value to the template and insert it there to remember it.

Example:

{% if contacts.has_next %}
    <a href="?sort={{ criteria }}&page={{ contacts.next_page_number }}">next</a>
{% endif %}
user_893910

@skoval00 's answer is the most elegant, however it adds duplicate &page= query parameters to the url.

Here is the fix:

from urllib.parse import urlencode
from django import template

register = template.Library()

@register.simple_tag(takes_context=True)
def url_replace(context, next_page):
    query = context['request'].GET.copy().urlencode()
    
    if '&page=' in query:
        url = query.rpartition('&page=')[0]
    else:
        url = query
    return f'{url}&page={next_page}'
user_107462240

My solution is based on this one above with the slight improvement to remove &page= from appearing multiple times. See this comment

    @register.simple_tag(takes_context=True)
    def url_replace(context, **kwargs):
        query = context['request'].GET.copy()
        query.pop('page', None)
        query.update(kwargs)
        return query.urlencode()

This line query.pop('page', None) silently removes the page from the url

user_115762120

Another slight modification to skoval00 and Reinstate Monica to fully get rid of duplication and avoid the ugly ?&page=1 part:

from urllib.parse import urlencode
from django import template

register = template.Library()

@register.simple_tag(takes_context=True)
def url_replace(context, next_page):
    if query.startswith('page') or not len(query):
        new_url = f'page={next_page}'
    elif '&page=' in query:
        get_params = query.rpartition('&page=')[0] # equivalent to .split('page='), except more efficient 
        new_url = f'{get_params}&page={next_page}'
    else:
        new_url = f'{query}&page={next_page}'
    return new_url
user_30489890

@Elrond Supports Monica

@register.simple_tag(takes_context=True)
def url_replace(context, **kwargs):
    query = context['request'].GET.copy()
    for key in kwargs:
        query[key] = kwargs[key]
    return query.urlencode()

Use in template

<a class="page-link" href="?{% url_replace p=1 q='bar'%}">
user_28199860

Every such link you put in your view has to be equipped with relevant parameters. There is no implicit magic that would convert:

http://127.0.0.1:8000/users/?page=2

into:

http://127.0.0.1:8000/users/?sort=first_name&page=2

So what you need is some Sorter object/class/function/snippet (whatever might fit here without overdoing it), that would act similarly to django.core.paginator.Paginator, but would handle sort GET parameter.

It could be as simple as this:

sort_order = request.GET.get('sort', 'default-criteria')

<paginate, sort>

return render_to_response('view.html', {
    'paginated_contacts': paginated_contacts,  # Paginator stuff
    'sort_order': sort_order if sort_oder != 'default-criteria' else ''
})

Then, in your view:

{% if contacts.has_next %}
    <a href="?page={{ contacts.next_page_number }}{%if sort_order%}&sort={{sort_oder}}{%endif%}">next</a>
{% endif %}

I could be made more generic, but I hope you get the concept.

user_1761860

your code should be like:

{% if contacts.has_next %}
<a href="?page={{ contacts.next_page_number }}{% for key,value in request.GET.items %}{% ifnotequal key 'page' %}&{{ key }}={{ value }}{% endifnotequal %}{% endfor %}">next</a>
{% endif %}
user_34696650

this works for me and i find it simpler

from django.http import HttpRequest


def get_query_params_href(req: HttpRequest):
    query_strings = req.GET.dict()
    string = '?'
    for i in query_strings:
        string += f'{i}={query_strings[i]}&'
return string[0:string.__len__()-1]
user_118806910
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