CODEWITHSUNDEEP
htmljekyll
ustroetz
ustroetz

Reputation: 6302

Generating a list of pages (not posts) without the index file

I am using Jekyll as a static generator for a website (not a blog), and I want to have an automatically generated list of all pages on my index page. I get that to work with the following code for the sidebar.html file:

<ul>
  {% for page in site.pages %}
    <li><div class="drvce"><a href="{{ page.url }}">{{ page.title }}</a></div></li>
  {% endfor %}
</ul>

Now I would like that the index page is not shown in that list. Is there a way to do that?

enter image description here

Upvotes: 0

Views: 112

Answers (3)

Jackilyn
Jackilyn

Reputation: 1

If it's your main index page that you don't want to show up try this:

<ul>
  {% for page in site.pages and page.url != "/" %}
    <li><div class="drvce"><a href="{{ page.url }}">{{ page.title }}</a></div></li>
  {% endfor %}
</ul>

Upvotes: 0

Anthony
Anthony

Reputation: 37065

I have never used jekyll, but it's main page says that it uses Liquid, and according to their docs, I think the following should work:

<ul>
{% for page in site.pages %}
    {% if page.title != 'index' %}
    <li><div class="drvce"><a href="{{ page.url }}">{{ page.title }}</a></div></li>
    {% endif %}
{% endfor %}
</ul>

Upvotes: 3

zgames
zgames

Reputation: 71

I can offer you a jquery solution

add this in your <head></head> tag

<script type="text/javascript" src="http://code.jquery.com/jquery-1.10.2.min.js"></script>

add this after </ul>

 <script> $('ul li:first').remove(); </script>

Upvotes: 0

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