Drupal 7 |迁移：如何按名称将分类标签关联到节点？答案

【问题标题】：Drupal 7 | Migrate: how to assoc taxonomy tags to node by name?Drupal 7 |迁移：如何按名称将分类标签关联到节点？
【发布时间】：2015-11-11 09:51:56
【问题描述】：

一段时间以来，我一直在尝试设置我的迁移模块，但没有成功将分类术语与节点相关联。

首先让我们说，它的不是 D2D迁移，源是旧的symfony数据库！

首先我已成功导出分类术语：

*-.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',
  ),
...

Cities.inc：

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”，其中包含城市名称，如“Paris”。

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');

【问题讨论】：

标签： php drupal-7 migrate

【解决方案1】：

首先，确保您的“field_city”字段接受“城市”条款。

然后确保 MigrateSymfonyCitiesMigration 是 MigrateSymfonySightsMigration 的依赖项。

According to the documentation，术语引用字段迁移默认具有术语名称的“source_type”。确保“ignore_case”设置正确。

如果所有其他方法都失败，则将 MigrateSymfonyCitiesMigration 映射 sourceid1 更改为“名称”而不是“id”并重试（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()
);

【讨论】：