Django - Django中的分页1.3

Question

I'm new to Django. I'm using the simple pagination that Django provides but I need to paginate pages like this:

Prev 1 2 3 4 5 6 ... 320

Or

Prev 120 121 122 123 Last

There are some code ready to reuse in Django 1.3 to achieve this?

Answer 1

这就是你要找的。

Answer 2

Let's supose we have a view:

dev showPage(request,pg):

where pg is the page number you are looking at this moment. Then we need a bit of code to get queryset (or objects) and create paginator object:

pg = int(pg)
objects = range(320)
p = Paginator(objects, 15)
page = p.page(pg)

Well, that you need to get:

Prev 1 2 3 4 5 6 ... 320

is send to template a list like pags :

pags = []   #var to be sent to template
if pg-1 in p.page_range: pags.append( ( 'Prev', p.page( pg - 1) , ) )
for n in range( pg-2, pg+2):
   if n in p.page_range: pags.append( ( n, p.page( pg - 2) , ) )
if p.end_index() not in range( pg-2, pg+2):
   pags.append( ( '...', None , ) )
   pags.append( ( p.end_index(), p.end_index(), ) )

now send pags to your template. And render to something like:

<ul>
{% for label, npag in pags %}
   <li>
   {%if npag %} <a href="asjflasdjf/{{npag}"}>{%endif%}
   label
   {%if npag %} </a> {%endif%}
   </li>
{% endfor %}
</ul>

for

Prev 120 121 122 123 124 Last

the solution is the same. Decorate a little with css and you get it.

Also, for example, you can assign a class for current page in order to get it in bold style:

{%if npag %} 
   <a href="asjflasdjf/{{npag}"    
      {% if npag == pg %} class="bold-style" {%endif%}
   }>
{%endif%}