Filter by page in field, query language

Hello, jhow to do something like this with query language?

query: site.children.listed.filterBy('artist', '==', page);

…meaning: return all listed children who has this page in their ‘artist’ field (assuming that is a page selection field).

This probably boils down to if filterBy can deal with a $page in its comparison.

Thank you

No, you would have to use filterBy with a callable and that’s not possible in query language. So you need a custom method that returns the desired value and then use that method in your query.

Thank you,

What is wrong with this code ?

	'allTheirProducts' => function () {
		return $site->children()->listed()->filterBy('intendedtemplate','product')->filter(function ($p) {
			return $p->artist()->toPage()->is($this);

I use it in a pagesdisplay field like this:

query: page.allTheirProducts;

…but I am getting the error:

The section “products” could not be loaded: Invalid query result - Result must be of type Kirby\Cms\Pages, Kirby\Cms\Field given.


   'allTheirProducts' => function () {
      return site()->children()->listed()->filterBy('intendedtemplate','product')->filterBy(function($p) {
        return $p->artist()->toPage() === $this;

Ah, I see.

Thank you

Actually @texnixe , it still does not work.

I am getting the same error.

If I do this on home template:

dump(site()->children()->listed()->filterBy('intendedtemplate','product')->filter(function ($p) use ($page) {
	return $p->artist()->toPage() === $page;

… I get an empty Pages object as expected:

<pre>Kirby\Cms\Pages Object

Why would pagesdisplay say a field is given instead?

Perhaps @rasteiner can help?

Thank you

Ok , I got it working.

Possibly some typo, although I can’t see what was it.

Thank you.