如何使Django分页搜索引擎友好？

Question

I am wondering how can I make Django pagination search engine friendly, like: object/224 instead of object?page=224

Also, anyone has an idea why it's not by default search engine friendly!?

Answer 1

Adjust your URL:

(r'object/(?P<page>\d+)/$','listing')

Then adjust your view (here I am using the sample from the documentation ):

def listing(request,page):
    contact_list = Contacts.objects.all()
    paginator = Paginator(contact_list, 25) # Show 25 contacts per page

    # page = request.GET.get('page') not needed
    try:
        contacts = paginator.page(page)
    except PageNotAnInteger:
        # If page is not an integer, deliver first page.
        contacts = paginator.page(1)
    except EmptyPage:
        # If page is out of range (e.g. 9999), deliver last page of results.
        contacts = paginator.page(paginator.num_pages)

    return render_to_response('list.html', {"contacts": contacts})