通过django分页仅显示部分页码

Question

我在模板中使用django paginator.它工作正常,但是当有大量页面时效果不佳.

views.py:

def blog(request):
    blogs_list = Blog.objects.all()

    paginator = Paginator(blogs_list, 1)

    try:
        page = int(request.GET.get('page', '1'))
    except:
        page = 1

    try:
        blogs = paginator.page(page)
    except(EmptyPage, InvalidPage):
        blogs = paginator.page(page)
    return render(request, 'blogs.html', {
        'blogs':blogs
        })

模板的片段:

  <div class="prev_next">

    {% if blogs.has_previous %}
      <a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
    {% endif %}
    {% if blogs.has_next %}
      <a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
    {% endif %}
    <div class="pages">
      <ul>
      {% for pg in blogs.paginator.page_range %}
        {% if blogs.number == pg %}
          <li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
        {% else %}
          <li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
        {% endif %}
      {% endfor %}
      </ul>
    </div>
    <span class="clear_both"></span>

  </div> 

现在它看起来像这样:

我该怎么做只显示7个页码,而不是显示当前页码的全部,如下所示:

Prev 1 (2) 3 4 5 Next

我希望我很清楚,如果不是,请问.非常感谢您的帮助和指导.谢谢.

Answer 1

要把它扔进去.我想出来了,因为它让你知道两边都有更多的页面.

<ul class="pagination">

{% if page_obj.has_previous %}
    <li><a href="?page={{ page_obj.previous_page_number }}"><i class="fa fa-chevron-left" aria-hidden="true"></i></a></li>
{% else %}
    <li class="disabled"><span><i class="fa fa-chevron-left" aria-hidden="true"></i></span></li>
{% endif %}

{% if page_obj.number|add:'-4' > 1 %}
    <li><a href="?page={{ page_obj.number|add:'-5' }}">&hellip;</a></li>
{% endif %}

{% for i in page_obj.paginator.page_range %}
    {% if page_obj.number == i %}
        <li class="active"><span>{{ i }} <span class="sr-only">(current)</span></span></li>
    {% elif i > page_obj.number|add:'-5' and i < page_obj.number|add:'5' %}
        <li><a href="?page={{ i }}">{{ i }}</a></li>
    {% endif %}
{% endfor %}

{% if page_obj.paginator.num_pages > page_obj.number|add:'4' %}
    <li><a href="?page={{ page_obj.number|add:'5' }}">&hellip;</a></li>
{% endif %}

{% if page_obj.has_next %}
    <li><a href="?page={{ page_obj.next_page_number }}"><i class="fa fa-chevron-right" aria-hidden="true"></i></a></li>
{% else %}
    <li class="disabled"><span><i class="fa fa-chevron-right" aria-hidden="true"></i></span></li>
{% endif %}

</ul>

它看起来像这样:

Answer 2

首先,我会改变以下内容:

try:
    blogs = paginator.page(page)
except(EmptyPage, InvalidPage):
    blogs = paginator.page(page)  # Raises the same error

但是你可以在你的上下文中传递一个范围.

index = paginator.page_range.index(blogs.number)
max_index = len(paginator.page_range)
start_index = index - 3 if index >= 3 else 0
end_index = index + 3 if index <= max_index - 3 else max_index
page_range = paginator.page_range[start_index:end_index]

现在,您应该能够遍历范围以构建正确的链接?page=.

===编辑===
所以你的观点是这样的:

def blog(request):
    paginator = Paginator(Blog.objects.all(), 1)

    try:
        page = int(request.GET.get('page', '1'))
    except:
        page = 1

    try:
        blogs = paginator.page(page)
    except(EmptyPage, InvalidPage):
        blogs = paginator.page(1)

    # Get the index of the current page
    index = blogs.number - 1  # edited to something easier without index
    # This value is maximum index of your pages, so the last page - 1
    max_index = len(paginator.page_range)
    # You want a range of 7, so lets calculate where to slice the list
    start_index = index - 3 if index >= 3 else 0
    end_index = index + 3 if index <= max_index - 3 else max_index
    # Get our new page range. In the latest versions of Django page_range returns 
    # an iterator. Thus pass it to list, to make our slice possible again.
    page_range = list(paginator.page_range)[start_index:end_index]

    return render(request, 'blogs.html', {
        'blogs': blogs,
        'page_range': page_range,
    })

所以现在我们必须编辑您的模板以接受我们新的页码列表:

<div class="prev_next">
    {% if blogs.has_previous %}
        <a class="prev btn btn-info" href="?page={{blogs.previous_page_number}}">Prev</a>
    {% endif %}
    {% if blogs.has_next %}
        <a class="next btn btn-info" href="?page={{blogs.next_page_number}}">Next</a>
    {% endif %}
    <div class="pages">
        <ul>
        {% for pg in page_range %}
            {% if blogs.number == pg %}
                <li><a href="?page={{pg}}" class="btn btn-default">{{pg}}</a></li>
            {% else %}
                <li><a href="?page={{pg}}" class="btn">{{pg}}</a></li>
            {% endif %}
        {% endfor %}
        </ul>
    </div>
    <span class="clear_both"></span>
</div>

Answer 3

另一个使用模板的较短解决方案是将当前的forloop.counter与一定范围进行比较.

使用bootstrap我使用这个模板

<nav aria-label="Page navigation">   <ul class="pagination">
{% if page_obj.has_previous %}
<li class="page-item">
  <a class="page-link" href="?page=1" aria-label="Previous">
    <span aria-hidden="true">&laquo;</span>
    <span class="sr-only">begin</span>
  </a>
</li>   {% endif %}

{% for n in page_obj.paginator.page_range %}
  {% if page_obj.number == n %}
    <li class="page-item active">
      <span class="page-link">{{ n }}<span class="sr-only">(current)</span></span>
    </li>
  {% elif n > page_obj.number|add:'-3' and n < page_obj.number|add:'3' %}
    <li class="page-item"><a class="page-link" href="?page={{ n }}">{{ n }}</a></li>
  {% endif %}
{% endfor %}

{% if page_obj.has_next %}
  <li class="page-item">
    <a class="page-link" href="?page={{ page_obj.paginator.num_pages }}" aria-label="Next">
      <span aria-hidden="true">&raquo;</span>
      <span class="sr-only">end</span>
    </a>
  </li>
  {% endif %}   </ul> </nav>

截图

Answer 4

我发现最简单的方法是创建一个只显示您想要的页面的分页片段.

在我的情况下,我不想要任何上一个或下一个链接.我只是希望始终有一个指向第一页和最后一页的链接,然后将当前页面和当前页面两侧的两个页面放在一起.

我的模板片段(使用来自django-tables2的变量 - 如果你Paginator直接使用Django,变量的名称会略有不同)

{% load django_tables2 %}
{% load humanize %}
{% load i18n %}

{% if table.page %}
  {% with table.page.paginator.count as total %}
    {% with table.page.number as page_num %}
      {% with table.page.paginator.num_pages as num_pages %}
        {% block pagination %}
          <div class="row">
            <div class="col-md-12">    
              {% if table.paginator.num_pages > 1 %}
                <ul class="pagination pull-right">
                  {% for n in table.page.paginator.page_range %}
                    {% if table.page.number|add:'-3' == n %}
                      {# First page #}
                      <li><a href="{% querystring table.prefixed_page_field=1 %}">1</a></li>
                      {% if n != 1 %}
                        <li class="disabled"><a>&#8943;</a></li>
                      {% endif %}
                    {% elif table.page.number == n %}
                      {# Current page #}
                      <li class="active"><a href="#">{{ n }}</a></li>
                    {% elif table.page.number|add:'-3' < n and n < table.page.number|add:'3' %}
                      {# Pages around current page #}
                      <li><a href="{% querystring table.prefixed_page_field=n %}">{{ n }}</a></li>
                    {% elif table.page.number|add:'3' == n %}
                      {# Last page #}
                      {% if n != num_pages %}
                        <li class="disabled"><a>&#8943;</a></li>
                      {% endif %}
                      <li><a href="{% querystring table.prefixed_page_field=num_pages %}">{{ num_pages }}</a></li>
                    {% endif %}
                  {% endfor %}
                </ul>
              {% endif %}
            </div>
          </div>
        {% endblock pagination %}
      {% endwith %}
    {% endwith %}
  {% endwith %}
{% endif %}

我的分页在不同页面上看起来像的例子

信用:这是受@ Pavel1114的回答启发的

Answer 5

Django 3.2在 Paginator 类上引入了一个新功能：

\n

Paginator.get_elided_page_range(number, *, on_each_side=3, on_ends=2)\n

\n

\n

on_each_side使用和 的默认值on_ends，如果当前页为 10，并且有 50 页，则页面范围将为[1, 2, \'\xe2\x80\xa6\', 7, 8, 9, 10, 11, 12, 13, \'\xe2\x80\xa6\', 49, 50]。

\n

\n