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What I'm trying to do is to display child pages of a parent page. So far so good.

                {% if entry.parent %}
                    <div class="widget clearfix">
                        <h4 class="widgettitle"><a href="{{ entry.parent.url }}">{{ entry.parent.title }}</a></h4>
                        <ul class="clearfix">
                        {% set topic = entry.parent.id %}
                        {% for subsection in craft.entries.descendantOf(topic) if subsection.type != "state" %}
                            <li class="cat-item"><a href="{{ subsection.url }}">{{ subsection.title }}</a></li>
                        {% endfor %}
                        </ul>
                    </div>
                {% endif %}

However if the sub page has another child page everything goes wrong as I see the results of the parent page that is a child itself.

How do I ignore the hierarchy and display the results of level 1 aka grand parent page / topic?

Thanks.

Douglas McDonald · Answer 1 · 2015-02-01T08:33:27.210

3

You can use the ancestors property (which is a shortcut to getAncestors()).

{% set targetEntry = entry %}
{% if entry.level > 1 %}
     {% set targetEntry = entry.ancestors.last %}
{% endif %}

To retrieve an ancestor with a specific level:

{% set targetEntry = entry %}
{% for ancestorEntry in entry.ancestors %}
    {% if ancestorEntry.level == "2" %}
        {% set targetEntry = ancestorEntry %}
    {% endif %}
{% endfor %}

And if you want a specific distance (i.e. 2 levels above), you can use the distance property.

{% set targetEntry = entry %}
{% if entry.level > 1 %}
    {% set targetEntry = entry.ancestors(2).last %}
{% endif %}

edited Feb 01 '15 at 08:33

answered Feb 01 '15 at 01:40

Douglas McDonald

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Thanks, works great on any of the sub-pages of the main ancestor/parent page, however when I visit the parent aka level 1 page nothing shows up :( – Mark H. Feb 01 '15 at 08:08
Yes — If it's a level 1 entry then it has no ancestors. You would want to check the level or entry.parent or entry.ancestors|length first. Updated answer with example. – Douglas McDonald Feb 01 '15 at 08:34

score 0 · Answer 2 · answered Feb 01 '15 at 18:07

So, here is my final solution it looks ugly but works. I'm pretty sure that it could be done in a more effective / nicely looking way but this is my first website on Craft :)

                        {% set topic = entry.ancestors.first %}
                        {% if topic %}
                            <a href="{{ topic.url }}">{{ topic.title }}</a>
                        {% else %}
                            {{ entry.title }}
                        {% endif %}
                        </h4>
                        <ul class="clearfix">
                        {% if entry.level > 1 %}
                        {% for subsection in craft.entries.descendantOf(topic) if subsection.type != "state" %}
                            <li class="icon-list">
                                <a href="{{ subsection.url }}">{{ subsection.title }}</a>
                            {% if craft.request.segment(1) == subsection.slug %}
                                <span class="thumb-icon-list"><i class="fa fa-caret-right"></i></span>
                            {% endif %}
                            </li>
                        {% endfor %}

                        {% else %}

                        {% for subsection in craft.entries.descendantOf(entry.id) if subsection.type != "state" %}
                            <li class="icon-list">
                                <a href="{{ subsection.url }}">{{ subsection.title }}</a>
                            {% if craft.request.segment(1) == subsection.slug %}
                                <span class="thumb-icon-list"><i class="fa fa-caret-right"></i></span>
                            {% endif %}
                            </li>
                        {% endfor %}
                        {% endif %}

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