Filter a view by a field on a referenced entity

Question

Sorry to be yet another question about entry reference relationships in views, but I can't find the answer in pre-existing questions.

Given:

• the content type "page" has an entity reference field "artref" that references multiple nodes of the content type "article"

• the content type "article" has a boolean field "fishy"

How do I create a view that shows all pages which reference at least one article where fishy is checked.

I understand that here are 4 relevant settings:

• the basic entity bundle (pages or articles) selected for the view

• the relationship type ("content referenced from" or "content using")

• "require this relationship" setting

• Aggregation, especially if "distinct" is a possibility

All advice much appreciated

Answer 1

I figured it out. Configure in this sequence:

• Show: "content" of type "page"
• Relationships: "Content referenced from art_ref"
• Use Aggregation: Yes
• Add filter criteria: fishy, Aggregation: "group results together", relationship: "art_ref:content", operator: "is equal to" "1"

Answer 2

1. Create new content type,for example, "Article_1",having "fishy" field of boolean type.

2. Create new content type "Page_1",having entity reference field "artref" to "Article_1" content type. Values - unlimited.

3. Create new View with filter criteria Content:Type="Page_1".

4. In view "advanced" section add Relationship:"Content entity referenced(not referencing) from field_artref". "Require this relationship" you may live unchecked.

5. Add new filter criteria is based on "fishy" field = "Yes", for example.("Yes" is "On value" and "No" is a "Off value" for boolean "fishy" field).

6. Set relationship for this filter criteria to "Content entity referenced from field_artref".

7. In "advanced" section open "Query settings:settings" and set to checked "Distinct" checkbox.