CODEWITHSUNDEEP
phpsilverstripe
SIlverstripeNewbie
SIlverstripeNewbie

Reputation: 291

Paginated Menu List

How do a I create a paginated menu list of children (i.e. Menu(2)). I tried 

$list = Menu::get();

but class Menu does not exist. Is it best to iterate Menu(2) and assign it to a DataArray? I also tried 

$list = Page::get();

but it doesn't even show any pages?

Upvotes: 0

Views: 230

Answers (1)

3dgoo
3dgoo

Reputation: 15794

In SilverStripe 3.1 we can call $this->getMenu(2) in our controller to get the navigation menu of the given level (level 2 in this case).

We can then use PaginatedList to turn that menu into a paginated list. There are some great resources on how to create a paginated list such as:

  1. How to Create a Paginated List
  2. Lesson - Lists and Pagination

We can use these to create a PaginatedMenu function that will return a paginated list of our menu items:

class Page_Controller extends ContentController {

    public function PaginatedMenu($level = 1) {
        $paginatedMenu = PaginatedList::create(
            $this->getMenu($level), 
            $this->request
        );
        $paginatedMenu->setPageLength(5);
        $paginatedMenu->setPaginationGetVar('menu-start');

        return $paginatedMenu;
    }
}

setPageLength allows us to set the number of items to display per page.

Here is an example of how to use this in our template:

<% if $PaginatedMenu(2) %>
    <ul class="paginatedMenu">
        <% loop $PaginatedMenu(2) %>
        <li class="$LinkingMode"><a href="$Link">$MenuTitle</a></li>
        <% end_loop %>
    </ul>

    <% if $PaginatedMenu(2).MoreThanOnePage %>
        <div class="pagination">
            <% if $PaginatedMenu(2).NotFirstPage %>
            <a href="$PaginatedMenu(2).PrevLink" class="prev" aria-label="View the previous page">&larr;</a>
            <% end_if %>
            <span>
                <% loop $PaginatedMenu(2).PaginationSummary %>
                    <% if $CurrentBool %>
                    $PageNum
                    <% else_if $PageNum %>
                    <a href="$Link" class="pageLink" aria-label="View page number $PageNum">$PageNum</a>
                    <% end_if %>
                <% end_loop %>
            </span>
            <% if $PaginatedMenu(2).NotLastPage %>
            <a href="$PaginatedMenu(2).NextLink" class="next" aria-label="View the next page">&rarr;</a>
            <% end_if %>
        </div>

        <p>Page $PaginatedMenu(2).CurrentPage of $PaginatedMenu(2).TotalPages</p>
    <% end_if %>
<% end_if %>

Upvotes: 2

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