我在模板中使用django分页器。它可以正常工作,但是在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
        })

模板摘要:
  <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>

现在看起来像这样:

我该怎么做才能仅显示7个页码,而不是全部显示当前页码,如下所示:
Prev 1 (2) 3 4 5 Next

我希望我很清楚,否则请问。非常感谢您的帮助和指导。谢谢。

最佳答案

首先,我将更改以下内容:

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

但是您可以在上下文中传递范围。
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,您应该可以遍历整个范围,以使用?page=构建正确的链接。

===编辑===
因此,您的 View 将如下所示:
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,
    })

因此,现在我们必须编辑您的模板以接受我们的新页码列表:
<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>

关于django - 通过Django分页仅显示一些页码,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/30864011/

10-12 12:40
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