Symfony2的形式多對一實體類型

Question

這是實體：

class MyEntity { 
    /** 
    * @var \OtherEntity 
    * 
    * @ORM\ManyToOne(targetEntity="OtherEntity") 
    * @ORM\JoinColumns({ 
    * @ORM\JoinColumn(name="otherentity_id", referencedColumnName="id") 
    * }) 
    */ 
    private $otherentity; 

    // some other fields 
} 

我控制器的行動：

someAction(Request $request) { 
    $em = $this->getDoctrine()->getEntityManager(); 
    // simplified this step here with id=5, so that all Entities of class MyEntity a link to the OtherEntity with ID=5 
    $otherEntity = $this->getDoctrine()->getRepository('MyTestBundle:OtherEntity')->find(5); 

    $myEntity = new MyEntity(); 
    $myEntity->setOtherEntity($otherEntity); 

    $form = $this->createForm(new MyEntityType(), $myEntity); 
    // do some form stuff like isValid, isMethod('POST') etc. 
} 

這是Formtype：

class MyEntityType extends AbstractType { 
    public function buildForm(FormBuilderInterface $builder, array $options) { 
     parent::buildForm($builder, $options); 
     $builder->add('name', 'text'); 
     // HOW TO ADD THE ENTITY TO JOIN THE ADDED MyEntity with the OtherEntity (with ID=5)? 
     // i tried this: 
     ->add('otherentity', 'entity', 
      array('class' => 'My\MyTestBundle\Entity\OtherEntity', 
        'read_only' => true, 
        'property' => 'id', 
        'query_builder' => function (
       \Doctrine\ORM\EntityRepository $repository) { 
       return $repository->createQueryBuilder('o') 
          ->where('o.id = ?1') 
         ->setParameter(1, 5); 
    } 
) 

） // ...一些其他領域 } // 標準formtype方法等 }

所以我的問題是，我有什麼選擇適合$ builder->添加用於添加otherEntity，所以如果我做一個$em->persist($myEntity)控制器內堅持通過形式加入myEntity所，讓我在我的數據庫這樣的記錄：

id | name | otherentity_id 
1 | 'test' | 5 

注：我不想堅持新otherEntity ，我只想創建一個新的MyEntity並添加OtherEntity的外鍵。

Answer 1

你就不能使用Entity form type這樣的：

$builder->add('otherentity', 'entity', array(
    'class' => 'MyTestBundle:OtherEntity' 
));