By whimsy on

I have a cck multi-value field (field_images) and I'm trying to use the Computed Field module to calculate and display a count of images in a view. I know this is probably insanely simple to do, but I'm really not familiar with php and I've been tearing my hair out for two days.

I located the following snippet on the documentation for the Computed Field module:

$node_field[0]['value'] = count($node->field_images);

This works great if the node has one or more images. Nodes without any images still display a value of 1.

I would weep with gratitude if someone can help me figure this out :)

Comments

gpk’s picture

gpk commented

Enable/install devel module and use it to examine (e.g. using the dev load tab) $node->field_images for the cases of 0 and 1 images. There should be a difference that you can then detect and respond to in your code snippet.

gpk
----
www.alexoria.co.uk

whimsy’s picture

whimsy commented

Thank you, gpk! The devel module did the trick. In case anyone else wants to know how to do this:

To count the number of values in a cck multi value field. In my case the number of images in a multi-value image field. When using the Computed Field module, put this code in the Computed Code field:

if (is_null($node->field_images[0]['fid']))
{
$node_field[0]['value'] = 0;
}
else
{
$node_field[0]['value'] = count($node->field_images);
}

Put this in the Display Format field:

$display = $node_field_item['value'];

It seems that when I add a node that has the field_images field, a row is created in the content_field_images table even if the node has no images. If there are no images, the field_images_fid field will be NULL. If you only count $node->field_images, the code will always return a minimum value of 1 because of that row in the table, even though it has a null value. So I had to first test for nodes with NULL in the field_images_fid field and return 0 for them. Hope that makes sense.

drupal a11y’s picture

drupal a11y commented

Does this also work in D7?