Symfony2 Doctrine2 UniqueEntity验证指定值

Question

Symfony2 Doctrine2 UniqueEntity验证指定值

3

我已经以这种方式使用UniqueEntity：

/**
 * @ORM\Entity()
 * @ORM\Table(name="benefit_group_category")
 * @UniqueEntity(fields={"name", "project"}, ignoreNull=false, message="Duplicated group category for this project")
 */

我已经知道如何使用它。

在另一种情况下，我需要做以下操作：

 * @UniqueEntity(fields={"entityId", "status"}, message="There is already a Quality Review Pending for this entity")

但需要指出，对于相同的entityId如果status == 1，则不允许有重复项，但如果status == 0，则应该通过验证。

基本上，例如，我可以有任意数量的记录，其中entityId = 37并且status = 0，但只有一个status = 1。

这是可能的吗？

编辑：请在此处找到更新后的版本，以进行二次检查。在我的原始问题中，我忘记提及另一个参数（entityName），但这并不改变其本质。

存储库类

<?php

namespace AppBundle\Entity;

/**
 * QualityReviewRequestRepository
 *
 * This class was generated by the Doctrine ORM. Add your own custom
 * repository methods below.
 */
class QualityReviewRequestRepository extends \Doctrine\ORM\EntityRepository
{
/**
 * @param array $criteria
 * @return array
 */
public function getByEntityIdAndEntityNameAndStatusCriteria(array $criteria)
{
    // Todo remove this first return, it's done on purpose to force the system to throw an error, but it doesn't.
            return $this->createQueryBuilder('e') 
            ->andWhere('e.entityId = :entityId')
            ->setParameter('entityId', $criteria['entityId'])
            ->getQuery()
            ->getResult()
        ;

    if (QualityReviewRequest::STATUS_NEW == $criteria['status']) {
        return $this->createQueryBuilder('e')
            ->andWhere('e.entityId = :entityId')
            ->setParameter('entityId', $criteria['entityId'])
            ->andWhere('e.entityName = :entityName')
            ->setParameter('entityName', $criteria['entityName'])
            ->andWhere('e.status = :status')
            ->setParameter('status', $criteria['status'])
            ->getQuery()
            ->getResult()
        ;
    }

    return [];
}
}

主类：

<?php

namespace AppBundle\Entity;

use Doctrine\ORM\Mapping as ORM;
use Gedmo\Timestampable\Traits\TimestampableEntity;
use Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity;

/**
 * QualityReviewRequest
 *
 * @ORM\Table(name="quality_review_requests")
 * @ORM\Entity(repositoryClass="AppBundle\Entity\QualityReviewRequestRepository")
 * @UniqueEntity(fields={"entityId","entityName","status"}, repositoryMethod="getByEntityIdAndEntityNameAndStatusCriteria", message="There is already a Quality Review Pending for this entity")
 */
class QualityReviewRequest
{
    const STATUS_NEW = 0;
    const STATUS_DONE = 1;

    use TimestampableEntity;

    // removed useless stuff

    /**
     * @var string
     *
     * @ORM\Column(name="entityName", type="string", length=255)
     */
    private $entityName;


    /**
     * @var integer
     *
     * @ORM\Column(name="entityId", type="integer")
     */
    private $entityId;

    // More stuff
}

- Sergio Negri

1个回答

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- Mikhail Prosalov · Accepted Answer

您可以将repositoryMethod选项添加到UniqueEntity约束注释中，具体请参考文档。

* @UniqueEntity(fields={"entityId", "status"}, repositoryMethod="getEntityByIdAndStatusCriteria", message="There is already a Quality Review Pending for this entity")

并实现存储库方法以按定义的条件查找实体。

/**
 * @param array $criteria
 * @return array
 */
public function getEntityByIdAndStatusCriteria(array $criteria)
{
    if (1 == $criteria['status']) {
        return $this->createQueryBuilder('e')
            ->andWhere('e.entityId = :entityId')
            ->setParameter('entityId', $criteria['entityId'])
            ->andWhere('e.status = :status')
            ->setParameter('status', $criteria['status'])
            ->getQuery()
            ->getResult()
        ;
    }

    return [];
}