Symfony 3.3 - Doctrine - Doctrine -持久化连接表

Question

我正在开发一个使用PHP框架Symfony 3.3和ORM Doctrine的网站。

我有用户，每个用户都可以提供服务。一个服务可以由多个用户提供。

一切都运行得很好，除了我想持久化用户的服务列表。什么都没有发生，并且我没有得到任何错误。您能看到我的代码中有什么地方错了吗？

我的User类(其中一部分)：

 class User implements AdvancedUserInterface, \Serializable
 {

     public function __construct(){
        $this->services = new ArrayCollection(); // Services offered by the user
     }

     /**
     *  
     * @var Doctrine\Common\Collections\ArrayCollection $services
     * 
     * @ORM\ManyToMany(targetEntity="AppBundle\Entity\Service", inversedBy="users")
     * @ORM\JoinTable(name="service_offer")
     * 
     */

    private $services;

    function getServices() {
        return $this->services;
    }

    function setServices(ArrayCollection $services) {
        $this->services = $services;
        return $this;
    }

}

我的Service类(部分)：

class Service
{

    /**
     * @var integer
     *
     * @ORM\Column(name="id_service", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="IDENTITY")
     */
    private $idService;

    /**
    * @var \Doctrine\Common\Collections\ArrayCollection $users
    * 
    * @ManyToMany(targetEntity="AppBundle\Entity\User", mappedBy="services")
    * 
    **/
    private $users;
}

在我的控制器中，我有：

class ServiceOfferController extends Controller {
/**
 *
 * @param EntityManagerInterface $em
 * 
 * 
 * @Route("/member/ServiceOffer", name="ServiceOffer")
 */
public function serviceOfferAction(Request $request, EntityManagerInterface $em){

    //$user = $this->getUser();
    $form = $this->createForm(ServiceOfferType::class);

    $form->handleRequest($request);

    if ($form->isSubmitted()){
        // Services ticked in the formular
        $services = $form->get('services')->getData();

        $user = $this->getUser();

        $user->setServices($services);

        print_r($user->getServices());

        // Persisting in the DB
        $em->persist($user);
        $em->flush();
    }

    return $this->render('serviceOffer.html.twig', array(
        'form' => $form->createView()
    ));
}

}

"print_r($user->getServices());“返回数据，但是表service_offer中没有持久化任何内容。有什么想法吗？

提前谢谢。

编辑：

我还尝试指定列，如下所示：

/**
 *  
 * @var Doctrine\Common\Collections\ArrayCollection $services
 * 
 * @ORM\ManyToMany(targetEntity="AppBundle\Entity\Service", inversedBy="users")
 * @ORM\JoinTable(name="service_offer",
 *   joinColumns={
 *     @ORM\JoinColumn(name="id_user", referencedColumnName="id_user", onDelete="CASCADE")
 *   },
 *   inverseJoinColumns={
 *     @ORM\JoinColumn(name="id_service", referencedColumnName="id_service", onDelete="CASCADE")
 *   }
 * )
 * 
 */
private $services;

编辑2：

我的service_offer MySQL表是：

CREATE TABLE `service_offer` (
`id_service_offer` INT(11) NOT NULL AUTO_INCREMENT,
`id_user` INT(11) NULL DEFAULT NULL,
`id_service` INT(11) NULL DEFAULT NULL,
PRIMARY KEY (`id_service_offer`),
INDEX `fk_service_idx` (`id_service`),
INDEX `fk_user_idx` (`id_user`),
CONSTRAINT `fk_service` FOREIGN KEY (`id_service`) REFERENCES `service` (`id_service`),
CONSTRAINT `fk_user` FOREIGN KEY (`id_user`) REFERENCES `user` (`id_user`)
)

Answer 1

尝试为联接表指定列参数。

/**
 *  
 * @var Doctrine\Common\Collections\ArrayCollection $services
 * 
 * @ORM\ManyToMany(targetEntity="AppBundle\Entity\Service", inversedBy="users")
 * @ORM\JoinTable(name="user_service",
 *   joinColumns={
 *     @ORM\JoinColumn(name="user_id", referencedColumnName="id", onDelete="CASCADE")
 *   },
 *   inverseJoinColumns={
 *     @ORM\JoinColumn(name="offer_id", referencedColumnName="id", onDelete="CASCADE")
 *   }
 * )
 * 
 */

private $services;