it is a simple question. Probably the answer is simple too but it stole my 2 hours.

I have a contenty type: Article. When i create a content from it normally it shown a default display.

I want to create a view and control this display from it. How can i do that.. what should i write to url part in my view?


It's pretty simple.

1 - Go to admin/structure/views/add
2 - write view's name & (auto)machine name.
3 - select Content as of your content view from select list.. second select list choose YOUR content type you want to filter the result of.
4 - Select Create a page .. you'd have to write the title of the page .. then the path to this page example: my_player // example.com/my_player

other options would be up to you!

hope this helps

  • The view is for a content type, so I guess the URL cannot be static. – kiamlaluno Sep 7 '15 at 21:20
  • thanks for answer.. i think your way is for only one static page as kiamlaluno told.. – sharktale Sep 8 '15 at 7:45
  • from the question its for a content type list all the nodes in this content type .. it can be if this was the case – Essam Al-Qaie Sep 9 '15 at 9:31

I find the way after two hour waste of time;

  1. Create your view.
  2. Add node/% to url.. or for specific content type create a pattern with pathouto and write this down instead of node.
  3. Go to Contextual filters.. select "Provide default value".. and select content id from url
  4. Select fields as format and add fields wish you want

It is simple but not a clear answer in site.. I hope this not steal anyone time anymore..

  • Prepare to "waste" more time "learning" drupal. – Beebee Sep 8 '15 at 8:54

Views might not be a good match for what you want to do -- normally they are meant for lists of items, not just one item.

Have you tried Display Suite, in combination with Fences? It might be a better fit, depending on what you are trying to accomplish.

  • i does that after ask that question actually.. now, i use this method for this.. thanks anyway. – sharktale Sep 9 '15 at 11:38

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