EntityTypeManager database transaction

Question

I am using \Drupal::EntityTypeManager() in Drupal 8 to save data provided from a form into a content type. The following code saves the information in the content type news:

$data = [
  'type' => 'news',
  'title' => $title,
  'body' => ['value' => $content, 'format' => $format],
];

$node = \Drupal::EntityTypeManager()
  ->getStorage('node')
  ->create($data);
$node->save();

Now I need to get the nodeID of the node that was just saved. How do I do that ? Furthermore, I need to insert another set of data into a second content type that needs to store the nodeID I just mentioned.

$node2 = \Drupal::EntityTypeManager()
  ->getStorage('node')
  // $data2 contains the $node->id()
  ->create($data2);
$node2->save();

Outside of Drupal I am used to mysql transactions to make sure that both things only get stored together and get rolled back if there is an error with either of them. Is there a way to achieve this 'all-or-nothing' functionality with \Drupal::EntityTypeManager()?

Do I genrally need to put this in a try{}catch{} block in Drupal?

Answer 1

You can create the node without saving it. You wouldn't really need a try catch if you know what types of data you're saving.

    $node = \Drupal::EntityTypeManager()->getStorage('node')->create([
      'type' => 'news',
      'title' => "hello world",
      'field_some_textfield' => "Here's some text",
    ]);

    $node->set('status', 1);
    $node->set('field_site_section', [['target_id' => 6]]);
    $node->set('field_first_name', $data->firstName);
    if (!empty($data->middleName)) {
      $node->set('field_middle_name', $data->middleName);
    }
    $node->set('field_last_name', $data->lastName);

When you're satisfied that your node is set up properly, then you can call:

$node->save();

At this point, you can get the node id this way:

$nid = $node->id();