How to render an image field

Question

What is the correct way to render a Drupal 7 image field and show the default image if the field is empty?

If I use field_get_items it will simply return FALSE when the field is empty.

Clive · Accepted Answer · 2012-06-04 00:01:54Z

15

I think best practice would probably be to use field_view_field():

$view = field_view_field('node', $node, 'field_image');
print render($view);

That will render the field as if it was attached to an entity view, and as such will provide the default image if none is available.

answered Jun 3, 2012 at 1:09

Clive♦

163k17 gold badges290 silver badges325 bronze badges

2

As documented, you can either use the display settings from an existing view mode or provide your own as fourth argument. Also, if you only want the field value with its label and wrapper markup, use field_view_value().
– Pierre Buyle
Jun 3, 2012 at 16:48

Add a comment |

apaderno · Accepted Answer · 2013-08-14 15:24:08Z

10

You can also image style (image cache presets in D6) like this:

render(field_view_field('user', $user, 'field_avatar', array('settings' => array('image_style' => 'avatar'))));

apaderno♦

93.8k15 gold badges158 silver badges280 bronze badges

answered Jun 11, 2012 at 14:43

Paul Hudson

1012 bronze badges

render() requires a variable passed as reference, and since field_view_field() doesn't return a reference, that code raises an error. See Clive's answer to see how render() should be called.
– apaderno ♦
Aug 14, 2013 at 15:26

Add a comment |

2 Answers 2