17488 Posts in 4473 Topics by 1978 members
|
Page:
1
|
Go to End | |
| Author | Topic: | 1242 Views |
-
Counting Page to create Pagination

9 July 2008 at 4:13am
Is there a pagination system for the site content pages? I would like to calculate the number of submenu pages and have it generate the next and previous buttons. For example: My main menu section is titled ABOUT. Under the ABOUT section there are 12 pages... at the bottom of the first page I would like it to read
1 of 12 [next]
when you are on page 2 it will read
[prev] 2 of 12 [next]
and when you are on the last page it will read
[prev] 12 of 12
I would like this pagination system for each of the "main" sections I set up.
Is there something like this available? if not does anyone know how I can create it? I'm pretty new to SilverStripe and am trying to figure it all out.
============
I will need to...#1 calculate the number of submenus/pages under a main menu (how many Child pages are assigned to this (current) Parent page?)
#2 the current Child page that I am on is number ? (1-12) in the list (database) of Child pages for that Parent page.
I hope I'm explaining this well... its a bit confusing but I think figuring out these two things will then generate what I am looking for.
Thanks.
-
Re: Counting Page to create Pagination

9 July 2008 at 11:20am
This may or may not help, but the following variables are available within the page class:
$Pos (current iterator position) and $TotalItems (total items). They're inherited from the site tree so I think they should give you info about the current page position and sibling tiems.
-
Re: Counting Page to create Pagination

11 July 2008 at 1:21pm
in what files do I add in this code? would it be the Page.ss under the Layout directory?
| 1242 Views | ||
|
Page:
1
|
Go to Top |


