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我想知道如何使 Django 分页搜索引擎友好,例如: object/224 而不是 object?page=224

此外,任何人都知道为什么默认情况下它对搜索引擎不友好!?

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1 回答 1

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调整您的网址:

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

然后调整您的视图(这里我使用的是文档中的示例):

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})
于 2012-12-02T05:51:27.450 回答