如何在Django中创建分页？

我遵循djangoproject.com的以下文档：。但这太简单了。这只是下一个和上一个按钮

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

现在我想创建具有更多功能的分页，例如

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

这是代码：

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

View.py

def hire(request):

hire_article_list = hire_article.objects.all().order_by('-id')
#hire_article_list = hire_article.objects.order_by("-publication_date")

paginator = Paginator(hire_article_list, 2) # Show 25 contacts per page

page = request.GET.get('page')
try:
    hire_article_s = paginator.page(page)

except PageNotAnInteger:
    # If page is not an integer, deliver first page.
    hire_article_s = paginator.page(1)
except EmptyPage:
    # If page is out of range (e.g. 9999), deliver last page of results.

    hire_article_s = paginator.page(paginator.num_pages)

#return render_to_response('hire/list.html', {"page_list": page_list})
context = {'hire_article_s': hire_article_s}
return render(request, 'hire/list.html', context)

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

list.html

{% for j in hire_article_s %}
    {# Each "j" is a page_list  model object. #}
    <li><a href="/hire/{{ j.slug }}-{{j.id}}">{{ j.hiring}}</a></li>
{% endfor %}

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

{j在hire\u article\u s%}
{#每个“j”都是一个页面列表模型对象。#

{%endfor%}

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

{%if hire\u article\u s.has\u previous%}
{%endif%}
第{hire_article_s.number}页，共{{hire_article_s.paginator.num_pages}页。
{%如果美国有下一篇文章{u%}
{%endif%}

我上周也有类似的需求，发现这个非常有用的要点（）很好用（尽管我不知道为什么在我将请求放入函数参数之前它不起作用）。它是django的Paginator函数的一个子类。您可以将其放入实用程序文件中，并在需要对范围使用Paginate时调用它

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

例如，我在一个名为utils.py的文件中有我的，它在我的核心应用程序中

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

视图.py

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

from core.utils import paginate

def hire(request):

    hire_article_list = hire_article.objects.all().order_by('-id')
    '''
    Show 25 contacts per page, with a page range of 5, which means if you are
    on page 8, it shows links to pages 6,7,8,9,10.
    '''
    hire_article_s = paginate(request, hire_article_list, 25, 5) 

    context = {'hire_article_s': hire_article_s}
    return render(request, 'hire/list.html', context)

list.html

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

{% if hire_article_s.has_previous %}
     <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
{% endif %}

{% for range in hire_article_s.neighbor_range %}
    {% if range == hire_article_s.number %}
       <li class="pagination__item active ">{{ range }}</li>
    {% else %}
       <li class="{% if range == hire_article_s.number %}active {% endif %}"><a href="?page={{ range }}">{{ range }}</a></li>
    {% endif %}
{% endfor %} 

{% if hire_article_s.has_next %}
    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
{% endif %}

{%if hire\u article\u s.has\u previous%}
{%endif%}
{租用物品的范围为%u美国邻居的范围为%u}
{%if range==hire\u article\u.number%}
{{range}
{%else%}

{%endif%}
{%endfor%}
{%如果美国有下一篇文章{u%}
{%endif%}

希望这有帮助

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

更新

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>

---

上面的代码已经编辑了一些。我已经添加了上下文和模板格式。请注意，我正在使用循环遍历

{{hire\u article\u s.neighbor\u range}

并打印页码。我还做了一个检查以突出显示当前页面的编号。这应该行得通，因为这几乎是我自己的代码，有你自己的变量名。

你可以发布你尝试使用更多功能进行分页的代码吗。谢谢你的回复。我补充道，如果这不起作用，我会使用django无止境分页库，但我不想使用djangopackage。只是python代码。对我来说似乎很难找到解决办法谢谢你，但当我努力但没有工作的时候。你能把你的代码发给我吗？@user3863069我已经用一些额外的说明更新了代码。如果这不起作用，你需要告诉我你得到了什么错误。输出看起来像是前一个1 | 2 | 3 | 4 | 5 | 6 | 7当我单击数字时，什么都没有发生\n它正在生成页码，这意味着它可能只是HTML标记中的一个错误。你能发布生成的html吗？

                {% if hire_article_s.has_previous %}
                    <a href="?page={{ hire_article_s.previous_page_number }}">previous</a>
                {% endif %}

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

                {% if hire_article_s.has_next %}
                    <a href="?page={{ hire_article_s.next_page_number }}">next</a>
                {% endif %}

        </span>
</div>