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Kirby 4.4.1

$layouts

A collection of layouts

How to render layouts in templates?

Using the toLayouts field method, you can retrieve a Layouts collection - a collection of Kirby\Cms\Layout objects:

<?php foreach ($page->layout()->toLayouts() as $layout): ?>
<section class="grid" id="<?= $layout->id() ?>">
  <?php foreach ($layout->columns() as $column): ?>
  <div class="column" style="--span:<?= $column->span() ?>">
    <div class="blocks">
      <?= $column->blocks() ?>
    </div>
  </div>
  <?php endforeach ?>
</section>
<?php endforeach ?>

Calculate the column span value

Each column in a layout has a $column->width() method which will return the width defined in the blueprint. (i.e. 1/2) but for many grid systems you need to know how many columns the current column should span in the grid. This can be done with the $column->span() method. The method calculates with a 12-column grid by default. So for example, if your column width is 1/2 the span method would return a value of 6. If you are working with a different kind of grid system you can pass the number of columns like this: $column->span(6):

<?php foreach ($page->layout()->toLayouts() as $layout): ?>
<section class="6-column-grid" id="<?= $layout->id() ?>">
  <?php foreach ($layout->columns() as $column): ?>
  <div class="column" style="--span:<?= $column->span(6) ?>">
    <div class="blocks">
      <?= $column->blocks() ?>
    </div>
  </div>
  <?php endforeach ?>
</section>
<?php endforeach ?>

Working with individual blocks

In some cases, you might even want to controll the way blocks within layouts are rendered. $column->blocks() will return a blocks collection that you can work with and create another nested foreach loop.

<?php foreach ($page->layout()->toLayouts() as $layout): ?>
<section class="6-column-grid" id="<?= $layout->id() ?>">
  <?php foreach ($layout->columns() as $column): ?>
  <div class="column" style="--span:<?= $column->span(6) ?>">
    <div class="blocks">
      <?php foreach ($column->blocks() as $block): ?>
      <div class="block block-type-<?= $block->type() ?>">
        <?= $block ?>
      </div>
      <?php endforeach ?>
    </div>
  </div>
  <?php endforeach ?>
</section>
<?php endforeach ?>

Passing the layout object to the block snippet

If you need to access the layout object in a block snippet, you need to pass it to the snippet manually.

<?php foreach ($page->layout()->toLayouts() as $layout): ?>
<section class="6-column-grid" id="<?= $layout->id() ?>">
  <?php foreach ($layout->columns() as $column): ?>
  <div class="column" style="--span:<?= $column->span(6) ?>">
    <div class="blocks">
      <?php foreach ($column->blocks() as $block): ?>
      <div class="block block-type-<?= $block->type() ?>">
        <?php snippet('blocks/' . $block->type(), ['block' => $block, 'layout' => $layout]) ?>
      </div>
      <?php endforeach ?>
    </div>
  </div>
  <?php endforeach ?>
</section>
<?php endforeach ?>