How do I programmatically add an image while programmatically creating nodes?

Question

I'm writing a script to add nodes programmatically, and I don't know the proper way to add/attach image. Actually I'm not that familiar with Drupal.

Here are the (sample) image objects I found while print_ring the existing $node I'm using:

field_image: Array ( [und] => Array ( [0] => Array ( [fxid] => 3089 [alt] => [title] => [width] => 95 [height] => 126 [uid] => 249 [filename] => helloworld.jpg [uri] => public://helloworld.jpg [filemime] => image/jpeg [filesize] => 3694 [status] => 1 [timestamp] => 1346748001 [type] => image [media_title] => Array ( ) [media_description] => Array ( ) [field_tags] => Array ( ) [field_license] => Array ( [und] => Array ( [0] => Array ( [value] => nothing ) ) ) [metatags] => Array ( ) [rdf_mapping] => Array ( ) ) ) )

The next one I found is the following.

field_temp_image: Array ( [und] => Array ( [0] => Array ( [value] => http://www.example.com/sample-path/helloworld.jpg [format] => [safe_value] => http://www.example.com/sample-path/helloworld.jpg ) ) ) `

How should I add an image to that node in that way?

Answer 1

Let suppose your image fields in field_body_images ,

first load your node by node_load and record your image in file tables and add it to your node image field I hope this sample help you to achieve :

$n=  node_load($nid);
$file = new stdClass();      
$file->filename =$file_name;
$file->filemime =file_get_mimetype($localimagepath.$file_name);
$file->filesize = @filesize(file_create_path($localimagepath.$file_name));
$file->uid = $user->uid;
$file->status = 1;
$file->timestamp = time();
$file->list=1;
$file->data=array('alt'=>'','title'=>$n->title);
drupal_write_record('files', $file);
$record->fid=$file->fid;
$n->field_body_images[]=(array)$file;
node_save($n);

Answer 2

Drupal 8

For Drupal 8 it would go like following. First add the file and get a file ID. Then create the node and attach the given file ID.

$data = file_get_contents(__DIR__ . '/images/my_image.jpeg');
$file = file_save_data($data, 'public://my_image.jpeg');

$node = \Drupal\node\Entity\Node::create([
  'type'             => 'page',
  'title'            => 'Foobar',
  'field_my_image' => [
    'target_id' => $file->id(),
    'alt'       => 'Lorem ipsum',
    'title'     => 'Dolor sit amet',
  ],
]);

$node->save();

Answer 3

May be not exactly what you need, but why not just add a reference to an image in a field, like your example 2 suggests ? And then in content template file render that field as an image.

In node creation stream:

$node->field_image['und'][0]['value'] = "/path_to_image/image.jpg";

In content type:

<?php
global $base_url;
$image_source_link=$base_url . $node->field_image['und'][0]['value'];
?>
<img src="<?php print($image_source_link); ?>" />

Answer 4

In a Drupal 8 form submit callback, with entityTypeManager:

public function submitForm(array &$form, FormStateInterface $form_state) {
  // Get values from a managed_file widget.
  $files = $form_state->getValue('uploaded_images');

  $storage_file = $this->entityTypeManager->getStorage('file');
  $storage_node = $this->entityTypeManager->getStorage('node');

  $field_image = [];
  
  foreach ($files as $fid) {
    /** @var \Drupal\file\Entity\File $file */
    $file = $storage_file->load($fid);
    // Set the file status to permanent.
    $file->setPermanent();
    $file->save();

    $field_image[] = [
      'target_id' => $file->id(),
      'alt'       => '',  
    ];
  }

  $values = [
    // Node bundle.
    'type'           => 'gallery',
    // Node title.
    'title'          => 'Gallery',
    // Node status: active.
    'status'         => 1,
    // Author ID.
    'uid'            => \Drupal::currentUser()->id(),
    // Uploaded images.
    'field_image'    => $field_image,
  ];

  /** @var \Drupal\node\Entity\Node $entity */
  $entity = $storage_node->create($values);
  $entity->save();
}