I know my title isn't explicit, but i will explain my problem. I created a view that displays me all documents(content) of my website, some documents are linked to a project(content) via entity reference, and in my view I display a link to a document. if there is a project the link looks like : node/100?project=5 . If there is no project the link needs to look like : node/100.

So in my view I added field excluded from display with the id of the project, and a second field which is the id of the node document and I rewrite the link to :


My problem is: This works fine for document with projects only. good url: node/100?project=5 bad url, without project: node/100?project=.

Can I check in view if the project id exists and rewrite the url to node/100?


1 Answer 1


almost there,

  1. add a link to the original content as a field, exclude it from display
  2. add a path to the original content as a field, exclude it from display fields
  3. rewrite the term reference to the full url you want to have using replacements [path]?something=[someid] reference url rewrite

  4. add a no results behavior and output the link from 1) [view_node] no results behavior

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