How to programmatically display an image field in Drupal 7 + default value when needed

Question

I've used field_view_field to display an image field for a specific entity in Drupal. The problem is when that entity doesn't have any value for field. I expect field_view_field to display the default image for field but it doesn't load default value of the field.

How can I do this?

Answer 1

field_view_field is a correct function to use if you want to display node image (if provided) or default (attached to field settings) if not.

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

Answer 2

In our case we have two set of images

1. Default image
2. another is node specific image

i created two block in views in the first view get the default image it will always have the default image

and in the second block render the image if the image field is empty render the whole default block there, below is the sample code

in views block 2 select the image field in the view and exclude it from display

in views global php add the code below

<?php
    $banner = isset($data->field_field_banner)? $data->field_field_banner : '';
    if(!empty($banner)){
      print drupal_render($banner);
    }else{
          print render_view_block_for_noresult('view_name','block_name');
    }
?>


/* function to render default view blocks if no banner is present */
function render_view_block_for_noresult($view_name='',$block_name=''){
        $view = views_get_view($view_name);
        if(isset($view) && !empty($view))
            return $view->preview($block_name);
}

this will help you to print the image and get the default image printed in the view.

Answer 3

By the way, I find a solution for my problem. If there aren't any value for field, it retrieves field instance information then load default image of field and display it.

       $user_loaded = user_load($uid);
if ($user_loaded->field_profile_photo[LANGUAGE_NONE][0]) {
  $user_pic = drupal_render(
    field_view_field('user', $user_loaded, 'field_profile_photo', 
      array(
        'label' => 'hidden',
        'settings' => array(
          'image_style' => '170_170'
        )
      )));
}
else {
  $instance = field_info_instance('user', 'field_profile_photo','user');
  $user_loaded->field_profile_photo[LANGUAGE_NONE][0] = (array)file_load($instance['settings']['default_image']);
  $user_pic = drupal_render(
    field_view_field('user', $user_loaded, 'field_profile_photo', 
      array(
        'label' => 'hidden',
        'settings' => array(
          'image_style' => '170_170'
        )
      )));
}

Answer 4

Here's an example of how to use a field instance's default image as a fallback when the individual image field itself is empty.

To briefly explain the concept, use field_get_items() to attempt to get the image attached to the entity (e.g. node, user, taxonomy_term, ...), then fallback to field_info_instance() if an image has not been uploaded.

Once an image is available to work with, set the $file variable to the image reference, which will get rendered after calling field_view_value().

$field_image = field_get_items('node', $node, 'field_image');

if (!empty($field_image)) {
  $file = $field_image[0];
}
else {
  $info = field_info_instance('node', 'field_image', 'blog');

  if (!empty($info) && $info['settings']['default_image'] > 0) {
    $default_image_fid = $info['settings']['default_image'];

    $file = (array) file_load($default_image_fid);
  }
}

if ($file) {
  $image = field_view_value('node', $node, 'field_image', $file, array(
    'type' => 'image',
    'settings' => array(
      'image_style' => NULL,
    ),
  ));

  print render($image);
}

All of this logic could be wrapped up into a helper function or simply used wherever needed.