I have created a new content type and show a list of all content of this type (list page). I have used the page URL article.

The detail page should be a child of article f.e. article/1. I configured the second view with path article/%.

How can I configure the detail view correctly and change the link in the list?

2 Answers 2


You don't need a URL argument for this. You can make the View path 'articles' and give it the following:

  • Filter by content type = article
  • Filter by status = published
  • Sort by created date, descending
  • Enable pager

That will give you a basic View listing articles, paged out, in descending order, on the url "articles".

  • Thanks, how do I add then the detail page of a spefic article with corresponding path like articles/1 for the first article.
    – CSchulz
    Mar 23, 2017 at 20:25
  • I don't understand what you mean. The Views will pull the articles, and you can either render the content (ex view Mode Teaser) or render fields, like Title, and check off "link to content" which will auto link it to the detail page.
    – Kevin
    Mar 23, 2017 at 20:30
  • Note: this is done via the Fields configuration area of Views.
    – Kevin
    Mar 23, 2017 at 20:38
  • I want to get some more readable URLs. I read about the autopath module. Do I need it?
    – CSchulz
    Mar 23, 2017 at 23:00
  • Pathauto? Yes. You don't need Views for that aspect of it- just the listing display.
    – Kevin
    Mar 23, 2017 at 23:01

For your list page, set the path to be "/article" in your view. For the node page (detail page) use Pathauto to configure the path for each node. Under Pathauto configuration add a pathauto pattern and choose your content type. In the Path Pattern field enter article/[node:nid]. This should give the desired url "article/1". Hope this answer your question.

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