PHP code for displaying fields conditionally [#1379680]

I have a homepage view that consists of 3 fields:

Title
Body
Image (field_image)

Both title and body field will always have data, but not for the image field. Therefore, for nodes which doesn't have image, the title and body (summary) will be displayed. For nodes which have image, only the title and image will be displayed. Can someone lend a help with the PHP code to be used? Thanks in advance!

Comments

Comment #1

johnv

Dutch

commented 24 January 2012 at 14:45

Status:

Active

» Postponed (maintainer needs more info)

So, if Image exists, the Body is hidden?
Haven't tried this, but one of these approached could work:

- create a new field 'Global PHP';
- You may need to set Image field to 'exclude from display';
- set the PHP-code to: return ($data->field_image[0]['und']['value']) ? NULL : $data->field_body[0]['und']['value'];

or:
- create a new field 'Global PHP';
- You may need to rearrage this field to be before or after Image field;
- in the PHP-code, clear $data->field_body[0]['und']['value'] when isset(Image) = TRUE;