在 URL 中发送视图字段 - Drupal

Question

I have a search page created with embedded filters in Drupal 8 View. In the list of fields I have a fields - [Field_program_id]. I also have a button 'Learn more' when clicking this button I want to pass the [field_program_id] and show the details of the Programs in that page. Not sure where to start. Any guidance?!

My view looks as below:

The '968' is the Program ID. Upon clicking the learn more, I want my url to be https://myproj/search?nodeid=[field_program_id]

and in this page I want to show the details again. Any help?!

Answer 1

I did something similar to this.

Have the view render fields, and add a field type of "Custom text".

Under "configure field", expand 'Rewrite results', and check, "Output this field as a custom link".

You can use replacement patterns to get the ID you need into the url.

For example, set the link path to "/search?nodeid={{nid}}"

That should render the link you need.