1

I have a long list of categories in a specific category group, but only want to show a list of the categories from this group that currently have an entry associated with them.

I've been trying to implement the solution that I found at this thread: How can I output a list of categories that are not empty?

Only categories with related entries are showing. However, it's not being narrowed down to specific to a specific category group. Here's my code below:

{% set handshapeCategories = craft.categories.group('handshape').limit(null) %}
{% set entriesWithHandshapeCategories = craft.entries.relatedTo( handshapeCategories ).limit(null) %}
{% set handshapeCategoriesInUse = craft.categories.relatedTo(entriesWithHandshapeCategories) %}

{% for handshapeCategory in handshapeCategoriesInUse %}
    <h3>{{ handshapeCategory }}</h3>
{% endfor %}       

Any ideas why the shown categories are not being limited to categories from the 'handshape' category group?

  • If you got a solution to your problem, please consider accepting one of the answers below – Jérôme Coupé Jul 31 '15 at 20:02
5

I would reverse your logic:

  1. get all entries in whatever channel
  2. display only categories (from a certain group) that have entries fetched in step 1 attached to them

Something along the lines of:

{% set entries = craft.entries.section('myChannel').limit(null) %}
{% set activeCategories = craft.categories.group('myCatgroup').relatedTo(entries).find() %}

You can then use a simple for loop to display only the active categories

  • Yeah, this works really well and is easy to understand. – Justin K Aug 17 '15 at 17:42
  • Glad it helps. You might also be interested in a more advanced set of tricks I blogged about: webstoemp.com/blog/… – Jérôme Coupé Aug 19 '15 at 7:01
1

In order to limit the related categories to a specific category group I needed to add .group('desiredCategoryGroup') to the last line.

Here's an example of what I came up with:

{% set newsCategories = craft.categories.group('news').limit(null) %}
{% set soriesWithNewsCategory = craft.entries.relatedTo( newsCategories ).limit(null) %}
{% set newsCategoriesInUse = craft.categories.relatedTo(soriesWithNewsCategory).group('news') %}

<h2>List of News Categories in Use:</h2>
<ul>
    {% for category in newsCategoriesInUse %}
        <li>{{ category.title }}</li>
    {% endfor %}
</ul>
1

I don't know how it compares efficiency-wise, but another approach to this (which I personally find much easier to read) would be something like:

<ul>
  {% for handshapeCategory in craft.categories.group('handshape') %}
    {% if craft.entries.relatedTo(handShapeCategory).limit(1).getOne() is not empty %}
      <li>{{ category.title }}</li>
    {% endif %}
  {% endfor %}
</ul>

In other words, for each category in the group, if we can fetch an entry for that category, display the title.

  • This is not very effective as it will result in one db query for each category. – carlcs Jul 20 '15 at 18:04

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