I have an content type called Image that has two fields: Title and Image
I have a Views block created so that it displays the images I want based on my criteria.
What I need is to have the raw image URL available.
I found this article Rendering Drupal 7 fields (the right way) and quite frankly, I am at a loss.
From the Drupal API:
field_view_field($entity_type, $entity, $field_name, $display = array(), $langcode = NULL)
I know I have to code the views-view-fields--my_custom_view.tpl.php, but everything I do results in an error.
What needs to go in this file so that I can simply get the URL from the image that I am looking for?
Also point in the right direction to a tutorial would be great. Everything I have found assumes you are an expert in Drupal and at this point, everything feels circular.