Answers.Body: <p>yes you can!</p>
  
  <p>You may search for a slug <code>craft.entries.slug('slug-of-first-parent')</code>, select the first result <code>.one()</code>, and then the children by <code>.getChildren()</code>. Now you can proceed as you already did and limit your result to only elements with a level greater than 1 <code>.level('&gt;1')</code> and list.</p>
  
  <pre><code>{% set pages = craft.entries.slug('slug-of-first-parent').one().getChildren().level('&gt;1') %}
  </code></pre>
  
  <p>I am not sure whether this is the best and most effective solution, but it should definitely do the job.</p>
  
  <p><strong>Tip:</strong> You can display information about an entry and the methods and variables available by dumping them either by using <code>var_dump($variable)</code> in php code or <code>{{ d($variable) }}</code> (by using <a href="https://github.com/studioespresso/craft3-dumper" rel="nofollow noreferrer">this</a> free plugin) in your twig template. This can help a lot when dealing with entries and finding children or parents.</p>
  
Answers.ContentLicense: CC BY-SA 4.0
Answers.CreationDate: 2018-12-18T01:17:19.940Z
Answers.LastActivityDate: 2018-12-18T01:22:50.960Z
Answers.LastEditDate: 2018-12-18T01:22:50.960Z
Answers.UserDisplayName: Luke
Body: <p>I'm trying to display a list of child structure links, but only from a specific parent. The template I'm working on isn't for the parent in question so I don't think I can use 'entry.'</p>
  
  <p>My structure is set up like this</p>
  
  <ul>
  <li><strong>Parent 1</strong> </li>
  <li>--Child One</li>
  <li>--Child Two</li>
  <li>--Child Three</li>
  <li><strong>Parent 2</strong></li>
  <li>--Child One</li>
  <li>--Child Two</li>
  </ul>
  
  <p>I only want to display the children of Parent 1. What I have so far is</p>
  
  <pre><code>&lt;ul id="sub-menu"&gt;
  {% set pages = craft.entries.section('visit').level('&gt;1') %}
  {% nav entry in pages %}
  &lt;li&gt;&lt;a href="{{ entry.url }}"&gt;{{ entry.title }}&lt;/a&gt;
      {% ifchildren %}
          &lt;ul&gt;
             {% children %}
          &lt;/ul&gt;
       {% endifchildren %}
  &lt;/li&gt;
  {% endnav %}
  &lt;/ul&gt;
  </code></pre>
  
  <p>Using this outputs all of the second level children instead of just the children from Parent 1. Is there any way to specify that the only children I want to display must come from Parent 1? Can I specify a slug somehow?</p>
  
  <p>Any help much appreciated</p>
  
Tags: <craft3><structure><navigation><children><parents>
Title: How can I display second level children only from a specific parent structure item?
ThreadText: # How can I display second level children only from a specific parent structure item?
  
  (Asked by: username_0 on 2018-12-17)
  
  I'm trying to display a list of child structure links, but only from a specific parent. The template I'm working on isn't for the parent in question so I don't think I can use 'entry.'
  
  My structure is set up like this
  
  * **Parent 1**
  * --Child One
  * --Child Two
  * --Child Three
  * **Parent 2**
  * --Child One
  * --Child Two
  
  I only want to display the children of Parent 1. What I have so far is
  
  ```
  <ul id="sub-menu">
  {% set pages = craft.entries.section('visit').level('>1') %}
  {% nav entry in pages %}
  <li><a href="{{ entry.url }}">{{ entry.title }}</a>
      {% ifchildren %}
          <ul>
             {% children %}
          </ul>
       {% endifchildren %}
  </li>
  {% endnav %}
  </ul>
  
  ```
  
  Using this outputs all of the second level children instead of just the children from Parent 1. Is there any way to specify that the only children I want to display must come from Parent 1? Can I specify a slug somehow?
  
  Any help much appreciated
  
  **Tags:** craft3, structure, navigation, children, parents
  
  
  ---
  
  ## 1 answers
  
  ### Answer by username_1 on 2018-12-18. Score: 4 | Accepted answer
  
  yes you can!
  
  You may search for a slug `craft.entries.slug('slug-of-first-parent')`, select the first result `.one()`, and then the children by `.getChildren()`. Now you can proceed as you already did and limit your result to only elements with a level greater than 1 `.level('>1')` and list.
  
  ```
  {% set pages = craft.entries.slug('slug-of-first-parent').one().getChildren().level('>1') %}
  
  ```
  
  I am not sure whether this is the best and most effective solution, but it should definitely do the job.
  
  **Tip:** You can display information about an entry and the methods and variables available by dumping them either by using `var_dump($variable)` in php code or `{{ d($variable) }}` (by using [this](https://github.com/studioespresso/craft3-dumper) free plugin) in your twig template. This can help a lot when dealing with entries and finding children or parents.
  
  
  ---
  
  