视图：排除具有与当前节点相同的分类术语ID的节点

Question

Drupal 8

I have a content type with a field that holds a Taxonomy Term Reference . I want a View that shows other nodes of the same content type that do not have that field with the same Taxonomy Term ID as the currently viewed node.

Example:

URL: / story/storyA

Currently viewing a node of Content Type Story that has a Taxonomy Term Reference to value B . I need the view to show other stories that do not have that same Taxonomy Term Reference to value B .

I thought I could achieve that by

• Set up a relationship to the field that holds the Taxonomy Term
• Add a Filter Criteria of Term ID associated to above relationship and with operator is not equal to . PROBLEM: how do I get the token for Value?

I also tried to go about adding a Contextual Filter associated to the created relationship and selecting Exlude but no dice.

Answer 1

I finally figured it out, I was missing the adequate additional Contextual Filter :

• Term ID , Provide default value with Taxonomy term ID from URL and the options Load default filter from node page and limit terms by vocab checked. And the last bit, in More , I have Exclude checked. This contextual filter is associated to the created relationship mentioned in the question.