I have several fields in my view:

  1. Path (excluded from display)
  2. Image
  3. Body
  4. Button (rewritten to use [path])

I need to output the image like so (this is a requirement):

<a href='[path]' style='background: url([image]);'>TITLE</a>

I cannot do this using rewrite rules because of https://www.drupal.org/node/417956 which explains that views runs all rewrites through filter_xss_admin(). This function strips out the style attribute on the a tag.

I am attempting to implement this in a view template, specifically the Field Content: Image (ID: Image) template:


 * @file
 * This template is used to print a single field in a view.
 * It is not actually used in default Views, as this is registered as a theme
 * function which has better performance. For single overrides, the template is
 * perfectly okay.
 * Variables available:
 * - $view: The view object
 * - $field: The field handler object that can process the input
 * - $row: The raw SQL result that can be used
 * - $output: The processed output that will normally be used.
 * When fetching output from the $row, this construct should be used:
 * $data = $row->{$field->field_alias}
 * The above will guarantee that you'll always get the correct data,
 * regardless of any changes in the aliasing that might happen if
 * the view is modified.
<a href='/' style='background-image: url(<?php print $output; ?>);'><?php print $row->node_title; ?></a>

I have tried this, [path] remains unchanged:

<a href='[path]' style='background-image: url(<?php print $output; ?>);'><?php print $row->node_title; ?></a>

and this, [path] remains unchanged:

<a href='<?php print token_replace("[path]", array('node' => $row->_field_data['nid']['entity'])); ?>' style='background-image: url(<?php print $output; ?>);'><?php print $row->node_title; ?></a>

I cannot find the value of the path anywhere in any of the variables. How can I output the value of [path] in my view template?

  • Which template views template file are you overriding? field or fields?
    – Collins
    Jun 17, 2016 at 14:40
  • I think it's because you excluded it, try un-excluding your path from display.
    – No Sssweat
    Jun 18, 2016 at 3:51
  • including the path (not excluding it) has no effect Jun 22, 2016 at 17:18

1 Answer 1


Can you get the node id you are aiming for? If so, try using drupal_get_path_alias()

$linknid = $fields['field_link_to_content']->content;

$alias = drupal_get_path_alias('node/' . $linknid);

<a href='<?php print $alias; ?>' style='background-image: url(<?php print $output; ?>);'><?php print $row->node_title; ?></a>

UPDATE - this is how I did it, based on this answer:

<a href='<?php print drupal_get_path_alias('node/' . $row->nid); ?>' style='background-image: url(<?php print $output; ?>);'><?php print $row->node_title; ?></a>
  • I'm sure I could. I was hoping to avoid another db call...but I will give it a shot Jun 17, 2016 at 16:13

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