How can I show just the most recent post on my home page with jekyll?
It appears you can also just access the latest post via the first index of site.posts
as in:
{%- assign latest_post = site.posts[0] -%}
Latest post: <a href="{{ latest_post.url }}">{{ latest_post.title }}</a>
While site.posts.first
works too as mentioned by someone else, the above example also provides a consistent manner for accessing other indices besides just the first (not that you would ever need to). Also, I didn't have enough reputation to add this answer as a comment instead :)
Rather than create a loop, just assign the variable and move on...
{% assign post = site.posts.first %}
(Edit 2018) Since someone wanted to know how to iterate other posts after you've done this:
{% for post in site.posts offset:1 %}
... Show the next posts ...
{% endfor %}
If you got here for the question as stated in the title, "How can I show just the most recent post on my home page with jekyll?" and not "how do I show only the latest post in my template," the following might be helpful.
Given a brand new Jekyll version 3.7.3 install with the default minima theme, create a file, _layouts/home.html
with the following content:
---
layout: none
---
{{ site.posts.first }}
Causes Jekyll 3.7.3 to show the first post, using the post template, as the home page.
This can be accomplished through the use of limit
:
{% for post in site.posts limit:1 %}
... Show the post ...
{% endfor %}
You can also use limit
and offset
together to "feature" your most recent post:
<h1>Latest Post</h1>
{% for post in site.posts limit:1 %}
... Show the first post all big ...
{% endfor %}
<h1>Recent Posts</h1>
{% for post in site.posts offset:1 limit:2 %}
... Show the next two posts ...
{% endfor %}