3

I have a structure as follows:

  • Parent 1
    • Child 1
    • Child 2
    • Child 3
  • Parent 2
    • Child 1
    • Child 2
    • Child 3
  • Parent 3
    • Child 1
    • Child 2
    • Child 3 etc...

On my Child entry pages I want to pull out the other child entries under that parent to use as navigation. I think this should be fairly straight forward but I can not find any documentation on it. So far I have the below but this pulls out all level 2 child entries and not just the entries for the specified parent.

Any help much appreciated.

{% for subnav in craft.entries.section('mySection').level(2).find() %}
    <li><a href="{{ subnav.url }}">{{ subnav.title }}</a></li>
{% endfor %}
5

Ok I've worked this out from above and adding some other code and now it works as below:

{% set parent = entry.getAncestors().first() %}
{% for subnav in craft.entries.section('mySection').descendantOf(parent).descendantDist('1') %}
    <li><a href="{{ subnav.url }}">{{ subnav.title }}</a></li>
{% endfor %}

Hope this assists someone else.

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1

Sometimes you need the same parent/siblings menu from various POVs (parent or children). Here's what I came up with:

{# Parent / Children subnavigation, either from the perspective of the parent or a child #}

{% if entry.level == 1 %}
  {% set parent = entry %}
  {% set children = entry.getDescendants().all() %}
{% elseif entry.level == 2 %}
  {% set parent = entry.getAncestors().one() %}
  {% set children = entry.getSiblings().all() %}
{% endif %}

<h1><a href="{{parent.url}}">{{parent.title}}</a></h1>
<ul>
    {% for child in children %}
    <li><a href="{{child.url}}">{{child.title}}</a></li>
    {% endfor %}
</ul>

I wonder if that's a detour and there's an easier way?

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1

For this you should be able to use the getSibling() method which:

"Returns an ElementCriteriaModel object prepped to return the entry’s siblings (if it lives in a Structure section)."

https://docs.craftcms.com/v2/templating/entrymodel.html#getsiblings

{% set siblings = entry.getSiblings().all() %}

<ul>
    {% for page in siblings %}
        <li class="item-contentmenu">
            <a href="{{ page.url }}">{{ page.title }}</a>
        </li>
    {% endfor %}
</ul>
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  • I've tried this instead of the code below but with no success again the model returns Null for siblings. Any ideas why? – Lettie Jul 30 '15 at 11:13
  • could you show how you tried to implement the getSibling()? But to note it wont really improve on the answer you've given. – George D Jul 30 '15 at 16:51

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