如何讓symfony 2表單從動態值返回實體

Question

好的，下面是我想要做的事情的快速概覽。我有一個與「ClientDomain」實體有關係的「客戶」實體。我需要一個表單，向我顯示給定客戶端的所有ClientDomain列表。在控制器中，我知道我需要過濾哪些客戶端，但我不確定如何將這些信息傳遞給formBuilder。

繼承人是我迄今爲止：

//src/NameSpace/ClientBundle/Entity/Client.php 
use Doctrine\ORM\Mapping as ORM; 

/** 
* @ORM\Entity 
*/ 
class Client{ 
/** 
* @ORM\Id 
* @ORM\Column(type="integer") 
* @ORM\GeneratedValue(strategy="AUTO") 
*/ 
protected $client_id; 

/** 
* @ORM\Column(type="string") 
*/ 
protected $name; 

/** 
* @ORM\OneToMany(targetEntity="ClientDomain", mappedBy="client") 
*/ 
protected $domains; 

... 
} 

而且形式：

//src/LG/ClientBundle/Form/ClientDomainSelectionForm.php 
namespace LG\ProjectBundle\Form\Projects; 

use Symfony\Component\Form\AbstractType; 
use Symfony\Component\Form\FormBuilderInterface; 

class ClientDomainSelectionForm extends AbstractType { 

    public function buildForm(FormBuilderInterface $builder, array $options) 
    { 
     $builder->add('client_domain', 'entity', array(
      'class' => 'LG\ClientBundle\Entity\ClientDomain', 
      'query_builder'=> function(EntityRepository $er) { 
       return $er->createQueryBuilder('cd') 
       /* NEEDS TO FIND DOMAINS BY CLIENT X */ 
      }, 
      'property' => 'domain', 
      'label' => 'Domain: ' 
     )); 
    } 
} 

然後終於控制器：

//src/LG/ClientBundle/Controller/DomainSelectorController.php 
namespace LG/ClientBundle/Controller; 

use Symfony\Bundle\FrameworkBundle\Controller\Controller; 
use Symfony\Component\HttpFoundation\Request; 
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Route; 
use Sensio\Bundle\FrameworkExtraBundle\Configuration\Template; 

use LG\ClientBundle\Entity\Client; 
use LG\ClientBundle\Entity\ClientDomain; 
use LG\ClientBundle\Entity\ClientActiveDomain; 
use LG\ClientBundle\Form\ClientDomainSelectionForm; 


/** 
* @Route("") 
*/ 
class DomainSelectorController extends Controller{ 

    /** 
    * @Route("/client/{client_slug}/select-domain", name="lg.client.clientdomainselection.selectclient") 
    * @Template 
    */ 
    public function selectDomainAction(Request $request, Client $client){ 
     $activeDomain = new ClientActiveDomain(); 
     $form = $this->createForm(new ClientDomainSelectionForm(), $activeDomain); 
     if ($request->isMethod('POST')) { 
      $form->bind($request); 
      if ($form->isValid()) { 
       $em = $this->getDoctrine()->getEntityManager(); 
       $em->persist($activeDomain); 
       $em->flush(); 
       return $this->redirect(/*huge long url*/); 
      } 
     } 
     return array(
      'form' => $form->createView(), 
     ); 
    } 

} 

正如你可以看到我有機會獲得控制器中的客戶端實體只是不知道如何將其提供給表單構建器，以便它僅返回th域當前客戶。

Answer 1

我已經找到了答案，你只需要一個構造函數從控制器添加到窗體，並通過在客戶端，像這樣：

//src/LG/ClientBundle/Form/ClientDomainSelectionForm.php 
namespace LG\ProjectBundle\Form\Projects; 

use Symfony\Component\Form\AbstractType; 
use Symfony\Component\Form\FormBuilderInterface; 

class ClientDomainSelectionForm extends AbstractType { 

    protected $client; 

    public function __construct(Client $client) { 
     $this->client = $client; 
    } 

    public function buildForm(FormBuilderInterface $builder, array $options) 
    { 
     $client = $this->client; 
     $builder->add('client_domain', 'entity', array(
      'class' => 'LG\ClientBundle\Entity\ClientDomain', 
      'query_builder'=> function(\Doctrine\ORM\EntityRepository $er) use ($client) { 
       return $er->createQueryBuilder('cd') 
        ->where('cd.client = :client') 
        ->orderBy('cd.domain', 'ASC') 
        ->setParameter('client',$client->getClientId()); 
      }, 
      'property' => 'domain', 
      'label' => 'Domain: ' 
     )); 
    } 
} 

然後在控制器：

//src/LG/ClientBundle/Controller/DomainSelectorController.php 

... 

public function selectDomainAction(Request $request, Client $client){ 

    ... 

    $form = $this->createForm(new ClientDomainSelectionForm($client), $activeDomain); 

    ... 
} 

...