Question

我正在使用Doctrine 2.4和Symfony 2.8，而我正试图建立一个友谊系统。

我有我的MateRelationship实体：

class MateRelationship
{
    /**
     * @var integer
     *
     * @ORM\Column(type="integer", name="id")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @ORM\ManyToOne(targetEntity="Acme\UserBundle\Entity\User")
     * @ORM\JoinColumn(name="sender", referencedColumnName="id", nullable=false)
     *
     */
    private $sender;

    /**
     * @ORM\ManyToOne(targetEntity="Acme\UserBundle\Entity\User")
     * @ORM\JoinColumn(name="receiver", referencedColumnName="id", nullable=false)
     *
     */
    private $receiver;

    /**
     * @var \DateTime
     *
     * @ORM\Column(type="datetime", nullable=false)
     */
    private $date;

    /**
     * @var boolean
     *
     * @ORM\Column(type="boolean", nullable=false)
     */
    private $invitationAccepted;

我的目标是进行DQL查询以获得给定用户的所有已接受的朋友关系（无论用户是发件人还是收件人）。

我设法通过2个查询和合并结果来实现，但它没有进行优化，而且以这种方式实现$limit更难。

public function getMates($user, $limit = 0){

    $query1 = $this->_em->createQuery('SELECT m, u.nickname, u.username FROM AcmeUserBundle:MateRelationship m JOIN m.receiver u WHERE m.invitationAccepted = 1 AND m.sender = :user')
            ->setParameter('user', $user);

    try{
        $result1 = $query1->getResult();
    }
    catch(\Doctrine\ORM\NoResultException $e){
        $result1 = [];
    }

    $query2 = $this->_em->createQuery('SELECT m, u.nickname, u.username FROM AcmeUserBundle:MateRelationship m JOIN m.sender u WHERE m.invitationAccepted = 1 AND m.receiver = :user')
            ->setParameter('user', $user);

    try{
        $result2 = $query2->getResult();
    }
    catch(\Doctrine\ORM\NoResultException $e){
        $result2 = [];
    }

    return array_merge($result1, $result2);
}

我想要一个查询，所以我可以使用->setMaxResults($limit)，但我不知道如何在1中合并这两个查询。谢谢你的帮助:)

Answer 1

您可以使用if(Pattern.matches("[a-zA-Z]:", path)) path = path.substring(2);和LEFT JOIN来获取有效的用户信息：

CASE

如果public function getMates($user, $limit = 0) { $dql = <<<DQL SELECT m, CASE usen.nickname IS NULL WHEN TRUE THEN urec.nickname ELSE usen.nickname END AS nickname, CASE usen.username IS NULL WHEN TRUE THEN urec.username ELSE usen.username END AS username, CASE usen.nickname IS NULL WHEN TRUE THEN 'sender' ELSE 'receiver' END AS relationship FROM AcmeUserBundle:MateRelationship m LEFT JOIN m.sender usen LEFT JOIN m.receiver urec WHERE m.invitationAccepted = 1 AND (m.receiver = :user OR m.sender= :user) DQL; return $this->_em->createQuery($dql) ->setParameter('user', $user) ->setMaxResults($limit) ->getResult(); }为空，则它已经是一个空数组$result。

注意： []例外仅针对调用它们的NoResultException，getOneOrNullResult或getSingleResult方法抛出。

Answer 2

以下是使用doctrine查询构建器

的提案

public function getMates($user, $limit = 0)
{
    // Create query builder
    $queryBuilder = $this->_em->getRepository("AcmeUserBundle:MateRelationship")->createQueryBuilder('m');

    // Create 'or' expression
    $or = $queryBuilder->expr()->orX();
    $or
        ->add('m.sender = :user')
        ->add('m.receiver = :user');

    // Create the query
    $queryBuilder
        ->andWhere('m.invitationAccepted = 1')
        ->andWhere($or)
        ->setMaxResults($limit)
        ->setParameter('user', $user);

    // Return the result
    return $queryBuilder->getQuery()->getResult();
}

Answer 3

您只需在匹配OR时添加:user即可。这可以是receiver或sender。

SELECT m, usen.nickname, usen.username, urec.nickname, urec.username
FROM AcmeUserBundle:MateRelationship m 
JOIN m.sender usen 
JOIN m.receiver urec 
WHERE m.invitationAccepted = 1 AND 
(m.receiver = :user OR m.sender= :user)

合并2个Doctrine DQL

3 个答案: