Reply To: Show content if link field is not filled

• John Huebner

• July 3, 2020 at 10:00 pm

<?php 
  if (get_sub_field('type') == "Image") {
    ?>
      <div class="full-img">
        <?php
          $link = get_sub_field('link');
          $content = '<img src="'.get_sub_field('image').'">';
          if ($link) {
            $content = '<a href="'.esc_url($link).'">'.$content.'</a>';
          }
          echo $content;
        ?>
        <p class="caption"><?php the_sub_field('caption'); ?></p>
      </div>
    <?php 
  };
?>