I'm sure this must be simple but I've tried multiple options with little success. What is the best way to build a query to get the children of more than one entry via an entry field?

For one parent would be:

{% set parent = craft.entries.id(parent_id).one() %}
{% for child in parent.children.all() %}
{% endfor %}

But I cant work out how to build the initial query to include more than one parent entry. The output I am looking for is a query that allows me to loop through all of the children from both(or more) entries selected. eg:

{% set query = craft.entries.id([4,35]).all().children.all() %}

Any help much appreciated!

2 Answers 2


There's a method available to all element queries called getDescendants() (or like any method in Yii you can omit the get prefix and use descendants()):

{% for child in craft.entries.id([4,35]).descendants().all() %}
    <pre>{{ dump(child.title) }}</pre>
{% endfor %}

You can also pass in a distance integer if you want to limit to direct children/grandchildren.


  • This is exactly what I was looking for! Thank you :) Jun 5, 2023 at 9:10
  • Although saying that having just tried this it only returns the parent entry...no child entries are output using the above... Jun 5, 2023 at 15:26
  • hm, I tested this and it worked ok for me. What version of Craft are you using? Jun 5, 2023 at 15:34
  • Craft version: Craft Pro 4.4.13 Have tried a variety of things and can only return the parent entry. Interestingly looking through the docs and trying getAncestors() returns an error: Calling unknown method: craft\elements\db\EntryQuery::getAncestors() Jun 7, 2023 at 8:54
  • Are all of the entries you're querying part of a Craft Structure? Those methods are only available to such elements. Jun 7, 2023 at 10:25

Why don't you get the children from each entry individually and then merge them together like so:

{% set childrenOne = craft.entries.id(entryID).one().children.all() %}
{% set childrenTwo = craft.entries.id(entryID).one().children.all() %}

{% set allChildren = childrenOne|merge(childrenTwo) %}
  • Beware that pretty much every time you write .one() or .all(), you are executing DB queries... and fairly heavy ones in the case of element queries. So I wouldn't recommend this way. You could mitigate slightly by eager loading the children: craft.entries.id(entryID).with('children').one(), but still not great. Jun 3, 2023 at 8:38

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