根据字母（AZ）分页

Question

I have a simple contacts application where I want to paginate the contact list based on first name alphabet.

Here is my views.py

def contacts(request):
    contact_list = Contact.objects.filter(user=request.user).order_by('first_name')
    return render(request, 'contact/contact.html', {'contact_list': contact_list})

Here is my template

<ul>
    {% for contact in contact_list %}
        <li><a href="/contacts/{{ contact.id }}/">{{ contact.first_name }}</a></li>
    {% endfor %}
</ul>

Is there a default way in Django that does this? There is Django pagination but I think that only splits data across pages. What would be the easiest way to do so?

Answer 1

It seems a letter-wise paginator does not exist by default. Pages like https://djangosnippets.org/snippets/1364/ <<< this one show hand-made implementations.

However, it's not so hard to implement: you can base yourself on startswith keyword and:

pages = [myQuerySet.filter(myfield__istartswith=i) for i in "ABC...XYZ"] #full alphabet here

(let myQuerySet be, actually, contact_list; let myfield be, actually, first_name).

Answer 2

I've done this once, by adding a letter field to my model and "paginating" manually. This raised a lot of interesting situation with foreign alphabets. Found this useful: https://pypi.python.org/pypi/Unidecode

Didn't find any "smarter" way to go about it. But I'd be interested in hearing about more "plug-n-play" solution...