I have a view set up that uses an image from a node, I have customized the views-view-field.tpl to change the url the image links to dependent on the content type. I am trying to print out the picture using field_view_field in the .tpl, but its just printing out "array". Here is what I'm working with.
<?php if ($row->node_type =='article'): ?>
<a href="http://www.someurl.com"><?php print field_view_field('node', $row, 'field_field_image');?></a>
What am I doing wrong? Thanks.
Ok, so I figured it out. First I did a node load:
<?php $node = node_load($row->nid);?>
Then I had to wrap my field_view_field in a drupal_render, finally I added my image style name in the display parameter, like this:
<?php print drupal_render(field_view_field('node', $node, 'field_image', 'image_style'));?>
Print out the field using print_r(). You will probably find that there are safe and raw values at the very least and it will be most likely the safe value you want.
EDIT:
In the comments oobie11 points to Rendering Drupal 7 fields (the right way). Bookmark this :)
Looking at the documentation for field_view_field tells me that the second argument is the actual entity. I don't think for this purpose the $row variable counts as an entity. You will need to do a node_load first getting the nid from the $row and pass the returned node to field_view_field.
