Reputation: 25584
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?
Upvotes: 0
Views: 473
Reputation: 51715
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%}
Upvotes: 1