Question

我有一个名为“发票”的实体，其中包含许多“订阅”，以便“发票”和“订阅”处于多对多的关系中。

class Invoice
{
/**
 * @ORM\Column(type="integer")
 */
protected $a;

/**
 * @ORM\Column(type="integer")
 */
protected $b;

/**
 * @ORM\ManyToMany(targetEntity="Subscription", inversedBy="invoices")
 */
protected $subscriptions;

public function __construct()
{
    $this->subscriptions = new ArrayCollection();
}

//typical setters and getters
}

使用querybuilder，如何使用subcription上的where子句获取一组匹配的实体？

$subscription = ;//something pulled from the db and is a "subscription" object
$em->createQueryBuilder()
            ->select('p')
            ->from('Invoice', 'p')
            ->where('p.a = :a')
            ->andWhere('p.b = :b')
            ->andWhere('p.subscriptions has :subscription')   //this line here I need help
            ->setParameter('a', 1)
            ->setParameter('b', 2)
            ->setParameter('subscription', $subscription)
            ->getQuery()
            ->getResult();

Answer 1

为此目的，学说有MEMBER OF特别声明。另一种解决方案是制作子查询。

->andWhere(':subscription MEMBER OF p.subscriptions')

您可以在official doctrine documentation找到示例。

Answer 2

我不完全确定，但也许你应该加入订阅。

$subscription = ;//something pulled from the db and is a "subscription" object
$em->createQueryBuilder()
        ->select('p')
        ->from('Invoice', 'p')
        ->join('p.subscriptions', 'subscription');
        ->where('p.a = :a')
        ->andWhere('p.b = :b')
        ->andWhere('subscription = :subscription')   //this line here I need help
        ->setParameter('a', 1)
        ->setParameter('b', 2)
        ->setParameter('subscription', $subscription)
        ->getQuery()
        ->getResult();

或者，如果订阅具有id，则可以使用该ID而不是对象。

带有数组的Doctrine where子句

2 个答案: