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/