Building Pagination - pagination

Question

Though this question has been asked several times but I really couldn't find any clue from it.
I am working on a client project in which the page numbers are generated like this:
[1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] and so on ...
So if there are 100 pages, these numbers will count to ... [100].
Here is the code that I am using:
{foreach var=$page from=$pages counter=$counter}
<li>$counter
{/foreach}
I am looking to paginate them such that at most only 10 pages numbers are displayed per page.
Any help will be highly appreciate.
Thanks in advance.

Answer 1

First you have to remember page number of the current page. Then decide about the behavior of the pagination. Example current page, next four pages, three previous pages, first page and last page
So for page nr 9 it looks like this:
[1] ... [6] [7] [8] [bold 9] [10] [11] [12] [13] ... [100]
Next (hardest) thing is defining an efficient algorithm.
For example:
Define an empty list
Put current page on the list.
Add desired amount of elements before the current page (for example first element and m elements to the left of the current page)
Add desired amount of elements after the current page (for example n elements to the right of the page and the last element)
Iterate over the list to produce the pagination.
Iteration example:
foreach element on list:
if element.value == current_page
print_bold [element.value]
else
print [element.value]
nextElement = list.nextElement
if nextElement.value - element.value > 1:
print "..."

Answer 2

Here is an algorithm you might want to use, mostly because it is easy and actually works: (I just grabbed it from one of my projects)
Create an empty list. (This will be filled with with integers representing the page numbers)
Add page 1 and 2
Add the current page - 1
Add the current page
Add the current page + 1
Add the last page - 1
Add the last page
Only keep the unique records
Sort the list ascending
Filter each item below 0 or bigger than the last page
The last page can be calculated by: round(Total items / items per Page) + 1
You may add more items in the front, middle or end simply by adding more items with a greater offset like current page + 2.
You may also not use filters and only add items if they meet the condition upfront, but this is a very readable solution in my opinion.
Reference implementation:
$totalPages = ceil(count($items) / $perPage);
$pagination = array_unique([
1, 2,
$page - 1,
$page,
$page + 1,
$totalPages - 1,
$totalPages,
]);
sort($pagination);
$pagination = array_filter(
$pagination,
fn($page) => $page > 0 && $page <= $totalPages
);
If you want the dots between your list, you need to check on each iteration rendering an item if the difference between the last item and the current item is greater than one. Be sure to make sure a last item exists or to not run the code in the first iteration.
$page - $pagination[$i - 1] > 1
Good luck!

Answer 3

For large numbers of pages, consider "logarithmic" pagination, described here (PHP code included):
How to do page navigation for many, many pages? Logarithmic page navigation