Symfony2：在链配置的命名空间中找不到类'DoctrineCommonCollectionsArrayCollection'

Question

尝试保持一对多关系时出错。 特别是，一个客户可能有多个家庭成员;一个给定的家庭成员与一个且只有一个客户相关联。 [我是Symfony的新手，所以会出现错误！ 感谢您的指导。

注意：下面的备用控制器代码段通过尝试将家庭表中的客户端 ID 设置为 null 来产生完整性约束错误。

客户端实体代码段

namespace ManaAdminBundleEntity;
use DoctrineORMMapping as ORM;
use DoctrineCommonCollectionsArrayCollection;
/**
 * ManaAdminBundleEntityClients
 *
 * @ORMTable(name="clients")
 * @ORMEntity
 */
class Clients
{
    protected $members;
    public function __construct()
    {
        $this->members = new ArrayCollection();
    }
    public function getMembers()
    {
        return $this->members;
    }
    public function setMembers(ArrayCollection $members)
    {
        $this->members = $members;
    }
    /**
     * @var integer $id
     *
     * @ORMColumn(name="id", type="integer", nullable=false)
     * @ORMId
     * @ORMGeneratedValue(strategy="IDENTITY")
     * @ORMOneToMany(targetEntity="Household", mappedBy="clients")
     * @ORMManyToOne(targetEntity="EthDesc", inversedBy="clients")
     * @ORMJoinColumn(name="eid", referencedColumnName="id")
     */

家庭实体代码段

namespace ManaAdminBundleEntity;
use DoctrineORMMapping as ORM;
/**
 * ManaAdminBundleEntityHousehold
 *
 * @ORMTable(name="household")
 * @ORMEntity(repositoryClass="ManaAdminBundleEntityHouseholdRepository")
 */
class Household {
    /**
     * @var integer $hid
     *
     * @ORMColumn(name="hid", type="integer", nullable=false)
     * @ORMId
     * @ORMGeneratedValue(strategy="IDENTITY")
     * @ORMManyToOne(targetEntity="Clients", inversedBy="members")
     * @ORMJoinColumn(name="cid", referencedColumnName="id")
     * @ORMManyToOne(targetEntity="EthDesc", inversedBy="members")
     * @ORMJoinColumn(name="eid", referencedColumnName="id")
     */

客户端类型代码段

    ->add('members', 'collection', array(
        'type' => new HouseholdType(),
        'allow_add' => true, 
        'by_reference' => false,
        'cascade_validation'  => true,
        ));

控制器代码段

public function createAction(Request $request) {
    $entity = new Clients();
    $form = $this->createForm(new ClientsType(), $entity);
    $form->bind($request);
    if ($form->isValid()) {
        $em = $this->getDoctrine()->getManager();
        $em->persist($entity);
        $em->persist($entity->getMembers());
        $em->flush();
        return $this->redirect($this->generateUrl('clients_show', array('id' => $entity->getId())));
    }
}

备用控制器代码段

public function createAction(Request $request) {
    $entity = new Clients();
    $form = $this->createForm(new ClientsType(), $entity);
    $form->bind($request);
    if ($form->isValid()) {
        $em = $this->getDoctrine()->getManager();
        $em->persist($entity);
        foreach ($entity->getMembers() as $member)
        {
            $em->persist($member);
        }
        $em->flush();
        return $this->redirect($this->generateUrl('clients_show', array('id' => $entity->getId())));
    }

Answer 1

我让这个工作的时候，我没有

use DoctrineCommonCollectionsArrayCollection; 

我只是

$this->field = new DoctrineCommonCollectionsArrayCollection();

在我的构造函数中。

此外，您可能需要考虑将 Family（似乎是客户端中的成员）添加为实体窗体：

->add('members', 'entity', array(
    'type' => new HouseholdType(),
    'allow_add' => true, 
    'by_reference' => false,
    ));

Answer 2

就我而言，我（错误地）试图访问ToMany关系，因为它是ToOne关系（类似于这个答案）。