I want to create a field with hook_update in my module. I have the following code that creates the configuration for a field_link field:

$module_path = drupal_get_path('module', 'my_module');
$configs['field.field.paragraph.image.field_link'] = $module_path . '/config/install/field.field.paragraph.image.field_link.yml';
$configs['field.storage.paragraph.field_link'] = $module_path . '/config/install/field.storage.paragraph.field_link.yml';

foreach ($configs as $name => $config_path) {
  $data = Yaml::parse($config_path);
  $config = \Drupal::configFactory()->getEditable($name);

As far as configuration it seems to be working correctly. When I try to create a node it all displays as it should. But if you save, you get an error that the table doesn't exist. Perhaps I have to use the ConfigInstaller class. Anyone know how to have the table get created?

4 Answers 4


Thanks @Berdir. I was able to create the field with FieldConfig & FieldStorageConfig classes.

$module_path = drupal_get_path('module', 'my_module');

$yml = Yaml::parse(file_get_contents($module_path . '/config/install/field.storage.paragraph.field_link.yml'));
if (!FieldStorageConfig::loadByName($yml['entity_type'], $yml['field_name'])) {
$yml = Yaml::parse(file_get_contents($module_path . '/config/install/field.field.paragraph.image.field_link.yml'));
if (!FieldConfig::loadByName($yml['entity_type'], $yml['bundle'], $yml['field_name'])) {

Don't use the low-level config, that doesn't trigger the necessary API calls.

Use FieldConfig::create() and FieldStorageConfig::create()

  • The thing is that my module has the config in the install directory. Not sure if FieldConfig can be used by reading from the yml config files.
    – albertski
    Feb 6, 2017 at 20:20
  • The actual logic that core uses in \Drupal\Core\Config\ConfigImporter::importInvokeOwner. It is a bit more complicated to support certain use cases like list fields that transform their structure a bit, but for most field types it should work. You can also convert it into a PHP array and then pass it to ::create().
    – Berdir
    Feb 6, 2017 at 22:51
  • @Berdir How can you transform the YAML structure for a list field? I get an error when trying to create field storage using a yaml file for a list_string field.
    – Tim
    Sep 12, 2017 at 15:14
  • @Berdir Think I answered my own question.
    – Tim
    Sep 12, 2017 at 16:00

Rather than use the Yaml class directly, it's better to use FieldStorage. Storage settings may need to be transformed, depending on the field type.

$config_path = drupal_get_path('module', 'my_module') . '/config/install';
$storage = new FileStorage($config_path);

// Note: .yml extension not needed.
$data = $source->read('field.storage.paragraph.field_link');

if (!FieldStorageConfig::loadByName($data['entity_type'], $data['field_name'])) {
  // Transform storage settings, if required.
  $field_type_manager = \Drupal::service('plugin.manager.field.field_type');
  $class = $field_type_manager->getPluginClass($data['type']);
  $data['settings'] = $class::storageSettingsFromConfigData($data['settings']);


In my case I already had the configuration in my sync folder

  $config_storage = new FileStorage('../config/sync');
  foreach ([
  ] as $config_name) {
    $entity_type = \Drupal::service('config.manager')->getEntityTypeIdByName($config_name);
    $storage = \Drupal::entityTypeManager()->getStorage($entity_type);
    $config_record = $config_storage->read($config_name);

    /** @var \Drupal\Core\Config\Entity\ConfigEntityStorageInterface $storage */
    $entity = $storage->createFromStorageRecord($config_record);
    if ($storage->load($entity->id())) {

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