Drupal 7 Views: display tagged content where the url is equal to a taxonomy term

0
votes

I am having trouble wrapping my head around this one.

I have a taxonomy vocabulary called "industry" with various terms.

I also have a content type called "customers". For each customer I have tagged them with the relevant industry term.

I am trying to create a view that will check the url and if a term from the vocabulary is present, display all relevant customers.

So if the url is: categories/industry-group/insurance

Then the view should show all insurance customers.

I am guessing this has to be done by combining Contextual Filters and Relationships but I can't figure out how.

Update1 1:

Under fields i have:

- content
- Taxonomy Term: ID
- Taxonomy Term: Name

Under Contextual Filters:

- Content has taxonomy term ID (here I am using the raw url and using the 3rd parameter)

Under Relationship:

- Entity reference: industry
1
drupal-7viewstaxonomy

1 Answers

0
votes

The hardest lesson for me to learn is always the same one: patience and reading go a long way.

I figured this out.

Under contextual filters I needed "taxonomy term: name"