I am creating a frontpage image scroller. I have created a new content type with a title, blurb and image field, and created a view block and set it up on the frontpage. All good so far; this works.

I obviously want to use some javascript and custom HTML to create the image scroller so I have overwritten the views 'row style output' template (views-view-fields--content-type-for-scroller--block.tpl.php). This template is getting picked up correctly and I can customize my 3 fields.

The problem is that the output from the image field:


includes markup for the image, similar to:

<div class="..."><a href="..."><img src="..." />

whereby I only want it to output the Image src so that I can use the image as a background to another element.

I'm not too familiar with Drupal 7 but I image somewhere 'further up' I need to overwrite what sort of output I want for that particular field using some preprocessor but I'm not sure how to do this? Or is there a better way in general to be doing this?

1 Answer 1


here two alternatives to get the image address: (to both examples the machine name of the image field is *image_test_field*).

1) this example is using the $row offer by view template (in this case the name of the field was obtained using dsm($row) or print_r($row)):

  $uriImagefromView = $row->field_field_image_test_field[0]['rendered']['#item']['uri'];
  $urlImagefromView = file_create_url($uriImagefromView);
  // $urlImagefromView variable now has the path to the image

2) this example is loading the data from entity_load (in this case the name of the field was obtained using dsm($entity) or print_r($entity)):

  $nid = $row->nid;
  $entity = entity_load('node', array($nid));
  $uriImagefromNode = $entity[$nid]->field_image_test_field['und'][0]['uri'];
  $urlImagefromNode = file_create_url($uriImagefromNode);
  // $urlImagefromNode variable now has the path to the image

I hope this information be useful.

  • The first option worked perfectly. Thanks very much. Sep 19, 2011 at 14:25

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