Drupal7|Migrate:如何按名称将分类法标记关联到节点


Drupal 7 | Migrate: how to assoc taxonomy tags to node by name?

我已经尝试设置迁移模块一段时间了,但没有成功地将分类术语与节点关联起来。

让我们首先说,它的不是D2D迁移,源是旧的symfony数据库!

首先,我成功地导出了分类术语:

*.migrate.inc:

...
  'symfony_db_countries' => array(
    'class_name' => 'MigrateSymfonyCountriesMigration',
    'group_name' => 'symfony_db_loc',
  ),
  'symfony_db_cities' => array(
    'class_name' => 'MigrateSymfonyCitiesMigration',
    'group_name' => 'symfony_db_loc',
    'dependencies' => array('symfony_db_countries'),
  ),    
  'symfony_db_sights' => array(
    'class_name' => 'MigrateSymfonySightsMigration',
    'group_name' => 'symfony_db_loc',
  ),
...

城市公司:

class MigrateSymfonyCitiesMigration extends MigrateSymfonyCitiesMigrationBase {
  const TABLE_NAME = 'ya_loc_city';
  public function __construct($arguments = array()) {
    parent::__construct($arguments);
    $this->description = t('Migrate cities from Symfony DB.');
    $query = Database::getConnection('symfony', 'default')->select('ya_loc_city', 'c');
    $query->leftJoin('ya_loc_city_translation', 'ct', 'ct.id = c.id');
    $query->fields('c', array());
    $query->fields('ct', array());
    $this->source = new MigrateSourceSQL($query, array(), NULL, array('map_joinable' => FALSE));
    $this->destination = new MigrateDestinationTerm('cities');
    $this->map = new MigrateSQLMap($this->machineName,
        array(
          'id' => array(
            'type' => 'int',
            'not null' => TRUE,
            'description' => 'City ID',
          ),
        ),
        MigrateDestinationTerm::getKeySchema()
    );
    $this->addFieldMapping('name', 'name');
    $this->addFieldMapping('parent', 'country_id')->sourceMigration('symfony_db_countries');
  }
} 

然后,我从另一个表中创建节点,该表具有分类术语的entityReference字段(field_city)。

<?php
class MigrateSymfonySightsMigration extends Migration {
  public function prepareRow($row) {
    parent::prepareRow($row);
    $row->point = 'point (' . $row->longitude . ' ' . $row->latitude. ')';
  }
  public function __construct($arguments = array()) {
    //entry is the name of a group.
    parent::__construct($arguments);
    $this->description = 'Migration sights';
    /*************SOURCE DATA*************/
    ... 
    $this->source = new MigrateSourceSQL($query, array(), NULL,
      array('map_joinable' => FALSE));
    $this->destination = new MigrateDestinationNode('sight');
    /*************MAPPING*************/
    $this->addFieldMapping('field_city', 'name')->sourceMigration('symfony_db_cities');
    $this->map = new MigrateSQLMap($this->machineName,
       array(
         'id' => array('type' => 'int',
                            'unsigned' => TRUE,
                            'not null' => TRUE,
                      )
       ),
       MigrateDestinationNode::getKeySchema()
    );
  }
}

symfony表中有一列"name",其中包含类似"Paris"的城市名称。

如何将此列('name')值('Paris')与分类术语相关联?以下不起作用:

$this->addFieldMapping('field_city', 'name')->sourceMigration('symfony_db_cities');

首先,确保您的"field_city"字段接受"cities"术语。

然后确保MigrateSymfonyitiesMigration是MigrateSymonySightsMigration的依赖项。

根据文档,术语引用字段迁移默认情况下具有术语名称的"source_type"。确保"ignore_case"设置正确。

如果所有其他操作都失败,则将MigrateSymfonyitiesMigration映射源id1更改为"name"而不是"id",然后重试(必须重新注册迁移):

$this->map = new MigrateSQLMap($this->machineName,
    array(
      'name' => array(
        'type' => 'varchar',
        'length' => 255,
        'not null' => TRUE,
        'description' => 'City Name',
      ),
    ),
    MigrateDestinationTerm::getKeySchema()
);