List subfolder elements in breadcrumb

Kirby 2 Solved ✅
1

Hello.

I try to modify a breadcrumb to list only elements in the current subfolder and it’s subfolders.

The original code below.

<nav class="breadcrumb">
  <ul>
    <?php $x = 1; foreach($site->breadcrumb() as $crumb_key => $crumb): ?>
    <li<?php e($crumb->isActive(), ' class="is-active"') ?>><a href="<?php echo $crumb->url() ?>"><?php echo $crumb->title()->html(); echo $x != count($site->breadcrumb()) ? ' &rsaquo;' : ''; $x++ ?></a></li>
    <?php endforeach; ?>
  </ul>
</nav>

How can I achieve this with code like this?

<?php if($crumb->is(page('blog'))) continue; ?>

2

The breadcrumb method returns a child collection, so you should be able to find a special page by uri within this children collection.

In this way you can check if the blog page is part of the breadcrumb and then display the breadcrumb (untested):

<?php if ($site->breadcrumb()->find('blog')): ?>
  <nav class="breadcrumb">
    <!-- your breadcrumb code -->
  </nav>
<?php endif; ?>
3

Hm, I’m not quite sure, looks like you want to exclude the blog page but display the children etc.?

Then you could use not() like this:

  <nav class="breadcrumb">
    <ul>
      <?php $x = 1; foreach($site->breadcrumb()->not('blog') as $crumb_key => $crumb): ?>
      <li<?php e($crumb->isActive(), ' class="is-active"') ?>><a href="<?php echo $crumb->url() ?>"><?php echo $crumb->title()->html(); echo $x != count($site->breadcrumb()) ? ' &rsaquo;' : ''; $x++ ?></a></li>
      <?php endforeach; ?>
    </ul>
  </nav>
4

2 posts were split to a new topic: Exclude pages in treemenu