Question

So I gather I can render a specific element of the $page['content'] array like so…

<?php print render(field_view_field('node', $node, 'field_image')); ?>

Where out of the box this will render the element as expected with standard defaults. Ok, so how can I make sure that hand plucked element no longer renders in the <?php print render($page['content']) ?> call later?

Why do I want something dumb like this? Because every page WILL have a header image with a few css tricks for overlays, design and such. But not every page will have attachments, links, and so on… you know, things that are additional fields in the page. So I can’t manually print out each field since I don’t know how many or what else there is. All I know for sure is the field_image I’m printing above is wrapped in a ton of markup for styling and must be done this way. Same for a few other fields.

Basically I’m looking for a way to unset the field immediately after use.

Does anyone know how to achieve this? I’d rather not make a view or a custom block that displays for specific pages. I eventually have to hand this over to a client who will not be able to wrap their heads around a single page being administered over many places in the CMS.

Answer 1

Looks like this will work:

<?php
print render(field_view_field('node', $node, 'field_image'));
MYTHEME_remove_item($page['content'], 'field_image');
?>

and in template.php file I made this function:

function MYTHEME_remove_item(&$content, $field)
{
    foreach($content['system_main']['nodes'] AS $key => $val){
        unset($content['system_main']['nodes'][$key][$field]);
    }
}

This is ridiculous, in my opinion. I would think a system as robust as drupal would have a solution for something like this. If anyone knows the proper way to do this I will gladly mark them as correct. In the meantime, for others facing similar situations, this worked for me.