Hansmagz Hansmagz - 1 year ago 48
CSS Question

What is the most efficient way to convert this jekyll code to PHP?

What is the most efficient way to convert this jekyll code to PHP?

{% for post in site.posts limit: 12 %}
<a href="{{post.url}}" style="background-image: url(assets/img/posts/thumbnails/{{post.thumbnail}})" class="academic-thumb">
<div class="academic-meta">
<div class="name">
{{post.title}}
</div>
</div>
</a>
{% endfor %}

Answer Source
<?php
  foreach ($site->getPosts() as $post) {
    echo '
      <a href="'. $post->getUrl() .'" style="background-image: url(assets/img/posts/thumbnails/'. $post->getThumbnail() .')" class="academic-thumb">
    <div class="academic-meta">
        <div class="name">
            '. $post->getTitle() .'
        </div>
    </div>
        </a>';
  }

This is will of course only work if the $post object is available and has gathers for the variables.

You should also listen to @borracciaBlu: study the ruby syntax first and the the php syntax