Answers.Body: <p>Based on your example and description, it's a bit hard to tell whether you are dealing with a <code>category</code> template or an <code>entry</code> template, because in your description you say that the category handle is called 'media', but in the example you are grabbing entries in the section called 'media'.</p>
  
  <p>Assuming you are on a category template (i.e. trying to create a category index page for example), then you should be able to just grab the current category from the 'category' variable (provided by craft), entries related to that category using <code>relatedTo(category)</code>, and the next and previous categories from <code>category.nextSibling</code> and <code>category.previousSibling</code> (or <code>category.next</code> and <code>category.previous</code> depending on whether your categories are hierarchical). If so, then the following should work (not tested):</p>
  
  <pre><code>{# category template #}
  
  {% set prevCategory = category.prevSibling %}
  {% set nextCategory = category.nextSibling %}
  {% set entries = craft.entries('mySection').relatedTo(category) %}
  
  &lt;h1&gt;{{ category.title }}&lt;/h1&gt;
  &lt;section class="category-index"&gt;
      {% for entry in entries %}
          &lt;article&gt;
              &lt;h2&gt;{{ entry.title }}&lt;/h2&gt;
              &lt;p&gt;...&lt;/p&gt;
          &lt;/article&gt;
      {% endfor %}
  &lt;/section&gt;
  
  {% if prevCategory %}&lt;a href="{{ prevCategory.url }}"&gt;Previous&lt;/a&gt; {% endif %}{% if nextCategory %}&lt;a href="{{ nextCategory.url }}"&gt;Next&lt;/a&gt;{% endif %}
  </code></pre>
  
  <p>If you are dealing with an entry template (i.e. an entry detail page), then you may need to clarify the question a bit.</p>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-12-29T02:26:28.953Z
Answers.Text: This would be on an entry template. I added your code. It allowed me to go Previous and Next, but it seems like the order was based on the order ID (when I added the category), rather than in alphabetical order. So if we are on category entry "68 Color Micro Journal", the next one should be "68 Micro Journal." http://www.colorcomputergames.com/media
Answers.UserDisplayName: DevServe
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-12-29T04:56:59.310Z
Answers.Text: Categories are reorder-able (and hierarchical) — prevSibling and nextSibling will simply return the previous/next in the order that they are set in.
Answers.UserDisplayName: Douglas McDonald
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-12-29T04:58:59.203Z
Answers.Text: If this is an entry template (i.e. as in an entry in a section or channel), then wouldn't you want next/previous to go to the next entry? vs next category? You may need to clarify the original question a bit more.
Answers.UserDisplayName: Douglas McDonald
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2016-12-27T20:10:16.863Z
Answers.LastActivityDate: 2016-12-27T20:16:35.763Z
Answers.LastEditDate: 2016-12-27T20:16:35.763Z
Answers.UserDisplayName: Douglas McDonald
Body: <p>The examples I have seen to link to previous and next entry all seem to be for sections. I want to get the previous and next category.</p>
  
  <p>My category handler is called 'media'.</p>
  
  <p>So I might have entries in this category such as :
  Time
  Life
  People</p>
  
  <p>When viewing the entry 'Life', I want to be able to go back to Time (Previous) or to People (Next). Again, these entries are entries in my category.</p>
  
  <p>I tried:</p>
  
  <pre><code>{% set params = craft.entries.section('media').relatedTo(category) %}
  {% set prevEntry = entry.getPrev(params) %}
  {% set nextEntry = entry.getNext(params) %}
  {% if prevEntry %}&lt;a href="{{ prevEntry.url }}"&gt;Previous&lt;/a&gt;{% endif %}
  {% if nextEntry %}&lt;a href="{{ nextEntry.url }}"&gt;Next&lt;/a&gt;{% endif %}
  </code></pre>
  
  <p>That didn't work, so I also tried:</p>
  
  <pre><code>{% set category = entry.category.first() %}
  {% set params = craft.entries.section('media').order('title asc').relatedTo(category) %}
  {% set prevEntry = entry.getPrev(params) %}
  {% set nextEntry = entry.getNext(params) %}
  {% if nextEntry %}
      {% set nextEntryUrl = nextEntry.url %}
  {% else %}
      {% set firstEntry = craft.entries.section('media').first() %}
      {% set nextEntryUrl = firstEntry.url %}
  {% endif %}
  {% if prevEntry %}
      {% set prevEntryUrl = prevEntry.url %}
  {% else %}
      {% set lastEntry = craft.entries.section('media').last() %}
      {% set prevEntryUrl = lastEntry.url %}
  {% endif %}
  </code></pre>
  
  <p>But also no good.</p>
  
Comments.ContentLicense: CC BY-SA 3.0
Comments.CreationDate: 2016-12-26T19:35:32.120Z
Comments.Text: Assuming `category` is a valid CategoryModel, and `entry` belongs to a section with the handle `"media"`, your first example looks legit. In other words, this is pretty much impossible to answer. Maybe share your entire template (as a Github gist, if its big)?
Comments.UserDisplayName: Mats Mikkel Rummelhoff
Tags: <category><params><previous-next>
Title: Display the next and previous category
ThreadText: # Display the next and previous category
  
  (Asked by: username_0 on 2016-12-26)
  
  The examples I have seen to link to previous and next entry all seem to be for sections. I want to get the previous and next category.
  
  My category handler is called 'media'.
  
  So I might have entries in this category such as :
  Time
  Life
  People
  
  When viewing the entry 'Life', I want to be able to go back to Time (Previous) or to People (Next). Again, these entries are entries in my category.
  
  I tried:
  
  ```
  {% set params = craft.entries.section('media').relatedTo(category) %}
  {% set prevEntry = entry.getPrev(params) %}
  {% set nextEntry = entry.getNext(params) %}
  {% if prevEntry %}<a href="{{ prevEntry.url }}">Previous</a>{% endif %}
  {% if nextEntry %}<a href="{{ nextEntry.url }}">Next</a>{% endif %}
  
  ```
  
  That didn't work, so I also tried:
  
  ```
  {% set category = entry.category.first() %}
  {% set params = craft.entries.section('media').order('title asc').relatedTo(category) %}
  {% set prevEntry = entry.getPrev(params) %}
  {% set nextEntry = entry.getNext(params) %}
  {% if nextEntry %}
      {% set nextEntryUrl = nextEntry.url %}
  {% else %}
      {% set firstEntry = craft.entries.section('media').first() %}
      {% set nextEntryUrl = firstEntry.url %}
  {% endif %}
  {% if prevEntry %}
      {% set prevEntryUrl = prevEntry.url %}
  {% else %}
      {% set lastEntry = craft.entries.section('media').last() %}
      {% set prevEntryUrl = lastEntry.url %}
  {% endif %}
  
  ```
  
  But also no good.
  
  **Tags:** category, params, previous-next
  
  > **Comments:**
  > **username_1**: Assuming `category` is a valid CategoryModel, and `entry` belongs to a section with the handle `"media"`, your first example looks legit. In other words, this is pretty much impossible to answer. Maybe share your entire template (as a Github gist, if its big)?  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_2 on 2016-12-27. Score: 2
  
  Based on your example and description, it's a bit hard to tell whether you are dealing with a `category` template or an `entry` template, because in your description you say that the category handle is called 'media', but in the example you are grabbing entries in the section called 'media'.
  
  Assuming you are on a category template (i.e. trying to create a category index page for example), then you should be able to just grab the current category from the 'category' variable (provided by craft), entries related to that category using `relatedTo(category)`, and the next and previous categories from `category.nextSibling` and `category.previousSibling` (or `category.next` and `category.previous` depending on whether your categories are hierarchical). If so, then the following should work (not tested):
  
  ```
  {# category template #}
  
  {% set prevCategory = category.prevSibling %}
  {% set nextCategory = category.nextSibling %}
  {% set entries = craft.entries('mySection').relatedTo(category) %}
  
  <h1>{{ category.title }}</h1>
  <section class="category-index">
      {% for entry in entries %}
          <article>
              <h2>{{ entry.title }}</h2>
              <p>...</p>
          </article>
      {% endfor %}
  </section>
  
  {% if prevCategory %}<a href="{{ prevCategory.url }}">Previous</a> {% endif %}{% if nextCategory %}<a href="{{ nextCategory.url }}">Next</a>{% endif %}
  
  ```
  
  If you are dealing with an entry template (i.e. an entry detail page), then you may need to clarify the question a bit.
  
  > **Comments:**
  > **username_0**: This would be on an entry template. I added your code. It allowed me to go Previous and Next, but it seems like the order was based on the order ID (when I added the category), rather than in alphabetical order. So if we are on category entry "68 Color Micro Journal", the next one should be "68 Micro Journal." http://www.colorcomputergames.com/media  
  > **username_2**: Categories are reorder-able (and hierarchical) — prevSibling and nextSibling will simply return the previous/next in the order that they are set in.  
  > **username_2**: If this is an entry template (i.e. as in an entry in a section or channel), then wouldn't you want next/previous to go to the next entry? vs next category? You may need to clarify the original question a bit more.  
  
  ---
  
  