Answers.Body: <p>By far, the easiest way to do this is to use the <a href="https://craftcms.com/docs/templating/nav#showing-children" rel="nofollow"><code>{% children %}</code></a> tag which is available inside a <code>{% nav %}{% endnav %}</code> tag pair, i.e. something like this:</p>
  
  <pre><code>&lt;ul&gt;
  {% nav category in craft.categories.group('products') %}
      &lt;li&gt;
        &lt;a href="{{ category.url }}"&gt;{{ category.title }}&lt;a&gt;
        {% ifchildren %}
          &lt;ul&gt;
              {% children %}
          &lt;/ul&gt;
        {% endifchildren %}
      &lt;/li&gt;
  {% endnav %}
  &lt;/ul&gt;
  </code></pre>
  
  <p>Note: The above will print a recursive, unordered list of all your top level <code>product</code> categories (i.e. "Akana", "Tama" etc) and their children.</p>
  
  <p>In your current code, you won't actually be able to use the <code>{% children %}</code>, because you have a <code>slug('akana')</code> parameter on your <code>craft.categories</code> query for some reason. Using the <code>slug()</code> parameter means that Craft will <em>only pull categories with that particular slug</em> – in other words, the child category "Ona" won't be returned by your query and the <code>{% children %}</code> tag will output nothing.</p>
  
  <p>If you have to use the <code>slug</code> parameter, there's no real use to the `{% nav %} tag (it's designed for recursive, hierarchical lists).</p>
  
  <p>For a more "manual" approach, you can also get to a categories children by doing something like this:</p>
  
  <pre><code>{% set category = craft.categories.group('products').slug('akana').first() %}
  &lt;a href="{{ category.url }}"&gt;{{ category.title }}&lt;/a&gt;
  {% set children = category.children %}
  {% for child in children %}
      &lt;a href="{{ category.url }}"&gt;{{ category.title }}&lt;/a&gt;
  {% endfor %}
  </code></pre>
  
  <p>One gotcha to note is that <code>category.children</code> will pull all your category's children, from all levels. To get children for different levels, you can use the <code>level</code> parameter:</p>
  
  <pre><code>{% set childrenA = category.children.level(2) %}
  {% set childrenB = category.children.level(3) %)    
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-08-05T09:02:20.387Z
Answers.LastActivityDate: 2016-08-05T09:02:20.387Z
Answers.UserDisplayName: Mats Mikkel Rummelhoff
Body: <p>So I have a categories with sub categories. </p>
  
  <pre><code>Anakena (Wine Brand Name)
  -Ona (Wine line)
  
  --product one
  --product two
  
  -Tama (Wine Line)
  
  --product one
  --product two
  </code></pre>
  
  <p>Im trying to make links to the specific lines. So I want to add Anakena under a field, and have it display an ul of Ona, Tama, etc. all the subcategories within Anakena. Once I click on this link, will take me to the category page with all of Ona's produtcs, etc.. </p>
  
  <pre><code>{% nav category in craft.categories.group('products').slug('anakena') %}
     &lt;a href="{{ category.url }}"&gt;{{ category.title }}&lt;a&gt;
  {% endnav %}
  </code></pre>
  
  <p>With this it gives me the main category, Anakena, how can I display the child categories assigned to Anakena.? </p>
  
  <p>Thanks</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-08-07T07:04:57.257Z
Comments.Text: Great,! 
  Thanks Mats.. One more thing now, Is there a way to replace the .slug name with a user defined something.? each entry will have a different slug, can I have the name filled by an entry.? 
  
  `{% set category = craft.categories.group('products').slug('{{ user defines entry?? }}').first() %}`
  
  Thanks
Comments.UserDisplayName: fidel
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-08-07T10:34:52.620Z
Comments.Text: Sure, just do `{% set category = craft.categories.group('products').slug(entry.someCustomField) %}`. Don't echo the variable inside the `.slug()` parameter, i.e. don't use `{{ }}`.
Comments.UserDisplayName: Mats Mikkel Rummelhoff
Tags: <category>
Title: Category listing.
ThreadText: # Category listing.
  
  (Asked by: username_0 on 2016-08-04)
  
  So I have a categories with sub categories.
  
  ```
  Anakena (Wine Brand Name)
  -Ona (Wine line)
  
  --product one
  --product two
  
  -Tama (Wine Line)
  
  --product one
  --product two
  
  ```
  
  Im trying to make links to the specific lines. So I want to add Anakena under a field, and have it display an ul of Ona, Tama, etc. all the subcategories within Anakena. Once I click on this link, will take me to the category page with all of Ona's produtcs, etc..
  
  ```
  {% nav category in craft.categories.group('products').slug('anakena') %}
     <a href="{{ category.url }}">{{ category.title }}<a>
  {% endnav %}
  
  ```
  
  With this it gives me the main category, Anakena, how can I display the child categories assigned to Anakena.?
  
  Thanks
  
  **Tags:** category
  
  > **Comments:**
  > **username_0**: Great,!  Thanks username_1.. One more thing now, Is there a way to replace the .slug name with a user defined something.? each entry will have a different slug, can I have the name filled by an entry.?   `{% set category = craft.categories.group('products').slug('{{ user defines entry?? }}').first() %}`  Thanks  
  > **username_1**: Sure, just do `{% set category = craft.categories.group('products').slug(entry.someCustomField) %}`. Don't echo the variable inside the `.slug()` parameter, i.e. don't use `{{ }}`.  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2016-08-05. Score: 1
  
  By far, the easiest way to do this is to use the [`{% children %}`](https://craftcms.com/docs/templating/nav#showing-children) tag which is available inside a `{% nav %}{% endnav %}` tag pair, i.e. something like this:
  
  ```
  <ul>
  {% nav category in craft.categories.group('products') %}
      <li>
        <a href="{{ category.url }}">{{ category.title }}<a>
        {% ifchildren %}
          <ul>
              {% children %}
          </ul>
        {% endifchildren %}
      </li>
  {% endnav %}
  </ul>
  
  ```
  
  Note: The above will print a recursive, unordered list of all your top level `product` categories (i.e. "Akana", "Tama" etc) and their children.
  
  In your current code, you won't actually be able to use the `{% children %}`, because you have a `slug('akana')` parameter on your `craft.categories` query for some reason. Using the `slug()` parameter means that Craft will *only pull categories with that particular slug* – in other words, the child category "Ona" won't be returned by your query and the `{% children %}` tag will output nothing.
  
  If you have to use the `slug` parameter, there's no real use to the `{% nav %} tag (it's designed for recursive, hierarchical lists).
  
  For a more "manual" approach, you can also get to a categories children by doing something like this:
  
  ```
  {% set category = craft.categories.group('products').slug('akana').first() %}
  <a href="{{ category.url }}">{{ category.title }}</a>
  {% set children = category.children %}
  {% for child in children %}
      <a href="{{ category.url }}">{{ category.title }}</a>
  {% endfor %}
  
  ```
  
  One gotcha to note is that `category.children` will pull all your category's children, from all levels. To get children for different levels, you can use the `level` parameter:
  
  ```
  {% set childrenA = category.children.level(2) %}
  {% set childrenB = category.children.level(3) %)    
  
  ```
  
  
  ---
  
  