How to output siblings including the current entry?

Question

I'm trying to output all the siblings of the current entry in a multi level structure like this:

{% for sibling in entry.siblings() %}
    <li>{{ sibling.link }}</li>
{% endfor %}

This works well, but how can I make this output all the siblings and the current entry?

I know I could add:

    <li>{{ entry.link }}</li>

before or after the sibling.link, but that would mess up the sort order.

If this can't be done this way, maybe I could somehow output the current branch/depth using a different approach? Any suggestions?

Answer 1

What about doing it like so:

{% set parent = entry.getParent() %}
{% set siblingsPlusMe = parent.getDescendants(1) %}

{% for entry in siblingsPlusMe %}
    <li>{{ entry.link }}</li>
{% endfor %}

Answer 2

There are two possibilities. If the entry has a parent, then the entry+siblings will be all the children of that parent. If the entry does not have a parent (because it is a top level entry), then entry+siblings will be all the top level entries of the section.

{# define siblingsPlusMe so it is in scope outside of the if #}
{% set siblingsPlusMe = [] %}

{% if entry.level == 1 %} {# no parent #}
  {% set siblingsPlusMe = craft.entries.section(entry.section).level(1) %}

{% else %} {# parent #}
  {% set siblingsPlusMe = entry.parent.children %}

{% endif %}

{# now that it is set, we can use it #}
{% for sibling in siblingsPlusMe %}
  ...
{% end for %}

Answer 3

This will also work:

{% for sibling in entry.parent().children() %}
    <li>{{ sibling.link }}</li>
{% endfor %}

The code above uses deprecated Craft properties, use this instead:

{% for sibling in entry.getParent().getDescendants(1) %}
    <li>{{ sibling.link }}</li>
{% endfor %}

Answer 4

I ran into a problem trying to use the above solution, as getParent() wasn't returning anything (i think this may be because the entry url's i was trying to output where not children of another entry within the section?). Instead i did the following:

<ul>
  {% set sectionHandle = entry.getSection().handle %}
  {% set pages = craft.entries.section(sectionHandle).level(1) %}
    {% for page in pages %}
      <li>{{ page.getLink() }}</li>
    {% endfor %}
</ul>

I was following this tutorial: http://buildwithcraft.com/help/structure-nav but it was requiring me to specify a handle name for which section to use. I was trying to create a generic template for multiple sections, so wanted to get the section of the current entry dynamically.