Answers.Body: <p>This is the solution:</p>
  
  <pre><code>{% set category = craft.categories.slug(craft.request.getParam('category')).first() %}
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2018-03-17T06:39:26.483Z
Answers.Text: This seems to only take the first category page. How would I get the category that is related to the current page I'm on, like, If this page has a category related to it, display that category title and link? Thanks @BradBell
Answers.UserDisplayName: mateostabio
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-15T09:48:57.800Z
Answers.LastActivityDate: 2017-03-15T22:07:33.620Z
Answers.LastEditDate: 2017-03-15T22:07:33.620Z
Answers.UserDisplayName: MGX
Answers.Body: <p>It depends how you're fetching the category. But if you've defined the template when setting up the categories it would be called be within a var called 'category', which would be a category object. You should then be able to access it using:</p>
  
  <pre><code>{{ category.title }}
  </code></pre>
  
  <p>If, on the other hand, you were getting the category via a slug in the URL, you would do something like:</p>
  
  <pre><code>{% set category = craft.categories.group('productType').slug(craft.request.lastSegment).first %}
  {{ category.title }}
  </code></pre>
  
  <p>UPDATE: Following on from your updated question, you've already got the category as an object so there is no need to get it again. Try replacing:</p>
  
  <pre><code>// THIS IS YOUR LOOP, BTW
  {% for category in craft.categories.group('productType').find() %}       
      &lt;header class="flush"&gt;&lt;h1&gt;{{ category.title }}&lt;/h1&gt;&lt;/header&gt; 
  {% endfor %}
  </code></pre>
  
  <p>with:</p>
  
  <pre><code>{{ category.title }}
  </code></pre>
  
  <p>But you might need to add a conditional in case the page is loaded without a category. Something like:</p>
  
  <pre><code>{% if category | length %}
      {{ category.title }}
  {% endif %}
  </code></pre>
  
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-14T09:58:45.250Z
Answers.Text: Many thanks for your response Clive - much appreciated.
  Still struggling to get this to work....have amended code in original question (sorry - should have included this in my original post) :(
Answers.UserDisplayName: MGX
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-14T16:20:08.410Z
Answers.Text: Many thanks for your response Clive - much appreciated.
  I tried both those alternatives - but unfortunately, they just output nothing/blank for the Category Title.
Answers.UserDisplayName: MGX
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-15T10:01:54.360Z
Answers.Text: What does `{{ craft.request.getParam('category') }}` output? And didn't you have `.first` on the end of your initial `craft.categories` query before?
Answers.UserDisplayName: Clive Portman
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-15T13:25:38.380Z
Answers.Text: I added that extra bit early today as it's the solution - apologies if I didn't make that clear!
Answers.UserDisplayName: MGX
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-16T08:59:31.800Z
Answers.Text: Cool. You should accept MGX's answer below, as it better answers the question in it's current form.
Answers.UserDisplayName: Clive Portman
Answers.ContentLicense: CC BY-SA 3.0
Answers.CreationDate: 2017-03-13T17:46:53.963Z
Answers.LastActivityDate: 2017-03-14T14:36:55.473Z
Answers.LastEditDate: 2017-03-14T14:36:55.473Z
Answers.UserDisplayName: Clive Portman
Body: <p>I need to be able to display the Category Title of a single category (as a h1 tag for a page displaying a list of the entries for that category).  I can successfully display the relevant entries for the category on the page, but when I have tried to display the category title it displays all of the categories titles (in a loop), not just the single category title.</p>
  
  <p>My code:</p>
  
  <pre><code>{% if craft.request.getParam('category') is not null %}    
  
          {% set category = craft.categories.slug(craft.request.getParam('category')) %}
  
          {% set products =  craft.entries.section('products').relatedTo(category).find() %}
  
  
          {% else %}
                 {% set products = craft.entries.section('products').find() %}
          {% endif %}
  
  
  
  &lt;div class="top-title"&gt; 
  
       {% for category in craft.categories.group('productType').find() %}       
          &lt;header class="flush"&gt;&lt;h1&gt;{{ category.title }}&lt;/h1&gt;&lt;/header&gt; 
       {% endfor %}
  </code></pre>
  
  <p>Any help would be very much appreciated!
  Many thanks!</p>
  
Tags: <category>
Title: How do I display the Category Title?
ThreadText: # How do I display the Category Title?
  
  (Asked by: username_0 on 2017-03-13)
  
  I need to be able to display the Category Title of a single category (as a h1 tag for a page displaying a list of the entries for that category). I can successfully display the relevant entries for the category on the page, but when I have tried to display the category title it displays all of the categories titles (in a loop), not just the single category title.
  
  My code:
  
  ```
  {% if craft.request.getParam('category') is not null %}    
  
          {% set category = craft.categories.slug(craft.request.getParam('category')) %}
  
          {% set products =  craft.entries.section('products').relatedTo(category).find() %}
  
          {% else %}
                 {% set products = craft.entries.section('products').find() %}
          {% endif %}
  
  <div class="top-title"> 
  
       {% for category in craft.categories.group('productType').find() %}       
          <header class="flush"><h1>{{ category.title }}</h1></header> 
       {% endfor %}
  
  ```
  
  Any help would be very much appreciated!
  Many thanks!
  
  **Tags:** category
  
  
  ---
  
  ## 2 answers
  
  ### Answer by username_0 on 2017-03-15. Score: 3 | Accepted answer
  
  This is the solution:
  
  ```
  {% set category = craft.categories.slug(craft.request.getParam('category')).first() %}
  
  ```
  
  > **Comments:**
  > **username_1**: This seems to only take the first category page. How would I get the category that is related to the current page I'm on, like, If this page has a category related to it, display that category title and link? Thanks @BradBell  
  
  ---
  
  ### Answer by username_2 on 2017-03-13. Score: 4
  
  It depends how you're fetching the category. But if you've defined the template when setting up the categories it would be called be within a var called 'category', which would be a category object. You should then be able to access it using:
  
  ```
  {{ category.title }}
  
  ```
  
  If, on the other hand, you were getting the category via a slug in the URL, you would do something like:
  
  ```
  {% set category = craft.categories.group('productType').slug(craft.request.lastSegment).first %}
  {{ category.title }}
  
  ```
  
  UPDATE: Following on from your updated question, you've already got the category as an object so there is no need to get it again. Try replacing:
  
  ```
  // THIS IS YOUR LOOP, BTW
  {% for category in craft.categories.group('productType').find() %}       
      <header class="flush"><h1>{{ category.title }}</h1></header> 
  {% endfor %}
  
  ```
  
  with:
  
  ```
  {{ category.title }}
  
  ```
  
  But you might need to add a conditional in case the page is loaded without a category. Something like:
  
  ```
  {% if category | length %}
      {{ category.title }}
  {% endif %}
  
  ```
  
  > **Comments:**
  > **username_0**: Many thanks for your response username_2 - much appreciated. Still struggling to get this to work....have amended code in original question (sorry - should have included this in my original post) :(  
  > **username_0**: Many thanks for your response username_2 - much appreciated. I tried both those alternatives - but unfortunately, they just output nothing/blank for the Category Title.  
  > **username_2**: What does `{{ craft.request.getParam('category') }}` output? And didn't you have `.first` on the end of your initial `craft.categories` query before?  
  > **username_0**: I added that extra bit early today as it's the solution - apologies if I didn't make that clear!  
  > **username_2**: Cool. You should accept MGX's answer below, as it better answers the question in it's current form.  
  
  ---
  
  