加入实体结果的问题:未定义的索引:Doctrine/ORM/UnitOfWork.php中的id

Question

我有以下代码:

        $rsm = new ResultSetMapping();
        $rsm->addEntityResult('App\MainBundle\Entity\InstagramShopPicture', 'p');
        $rsm->addFieldResult('p', 'id', 'id');
        $rsm->addFieldResult('p','lowresimageurl','lowresimageurl');
        $rsm->addFieldResult('p','medresimageurl','medresimageurl');
        $rsm->addFieldResult('p','highresimageurl','highresimageurl');
        $rsm->addFieldResult('p','caption','caption');
        $rsm->addFieldResult('p','numberoflikes','numberoflikes');
        $rsm->addFieldResult('p','numberofdislikes','numberofdislikes');
        $rsm->addJoinedEntityResult('App\MainBundle\Entity\InstagramShop', 's', 'p', 'shop');
        $rsm->addFieldResult('s', 'id', 'id');
        $rsm->addFieldResult('s', 'username', 'username');

        $query = $em->createNativeQuery('SELECT picture.id, picture.lowresimageurl, picture.medresimageurl, picture.highresimageurl, picture.caption, picture.numberoflikes, picture.numberofdislikes, shop.id AS shop_id , shop.username
                                        FROM App_instagram_picture_category category
                                        INNER JOIN App_instagram_shop_picture picture ON category.picture_id = picture.id
                                        INNER JOIN App_instagram_shop shop ON shop.id = picture.shop_id
                                        WHERE category.first_level_category_id = ?
                                        AND picture.deletedAt IS NULL
                                        AND shop.deletedAt IS NULL
                                        AND shop.isLocked = 0
                                        AND shop.expirydate IS NOT NULL 
                                        AND shop.expirydate >  ?
                                        AND shop.owner_id IS NOT NULL 
                                        GROUP BY shop.id
                                        LIMIT ?'

                                        , $rsm);

        $query->setParameter(1, 10);
        $query->setParameter(2, '2014-05-20');
        $query->setParameter(3, 10);
        $itemsFromDifferentShops = $query->getResult();

但是我经常收到以下错误/警告:

Notice: Undefined index: id in /Users/Alex/Sites/App/vendor/doctrine/orm/lib/Doctrine/ORM/UnitOfWork.php on line 2433

这是我的实体的样子:

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

     /**
     *
     * @var string
     * @ORM\Column(name="username", type="string", nullable=true)
     */
    private $username;


    /**
    * @Exclude()
    * @ORM\OneToMany(targetEntity="InstagramShopPicture", mappedBy="shop", cascade=  
     {"persist"})
    * @ORM\OrderBy({"createdtimestamp" = "DESC"})
    */
    protected $userPictures;

}

class InstagramShopPicture
{

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

    /**
     * @Exclude()
     * @ORM\ManyToOne(targetEntity="InstagramShop", inversedBy="userPictures")
     * @ORM\JoinColumn(name="shop_id", referencedColumnName="id", nullable=false, onDelete="CASCADE")
     */
    protected $shop;
}

为什么是这样？我如何解决它？我的怀疑是因为有两个身份证.一个是产品ID,另一个是商店ID,两者都具有相同的参考.但我尝试改变它,它仍然给我警告.

Answer 1

几周前我遇到了类似的问题,我可以解决如下问题:

    $em = $this->getEntityManager();

    $rsm = new \Doctrine\ORM\Query\ResultSetMapping();
    $rsm->addEntityResult('MyBundle:Actor', 'a');
    $rsm->addFieldResult('a', 'id', 'id');
    $rsm->addFieldResult('a', 'name', 'name');
    $rsm->addFieldResult('a', 'surname', 'surname');
    $rsm->addMetaResult('a', 'presentation_id', 'presentation_id');

    $query = $em->createNativeQuery(
            'SELECT a.id, a.name, a.surname, a.presentation_id
             FROM actors AS a 
             INNER JOIN presentations AS p 
             WHERE p.id = a.presentation_id 
             AND p.finished = 0
             AND p.id IN (?)', $rsm);
    $query->setParameter(1, $presentations);

    $actors = $query->getResult();

与您的代码存在一些差异的地方,但也许您可以调整它以获得您想要的.

请注意addMetaResult()函数,该函数用于外键或鉴别器列.您可以在官方Doctrine文档http://doctrine-orm.readthedocs.org/en/latest/reference/native-sql.html的第17.3.5节中阅读此内容.在此函数中,我将presentation_id作为参数传递,该参数是actors表中Presentation的外键字段.请注意,我不会为演示文稿请​​求信息,但是,Doctrine会对对象进行水合,我可以通过这种方式从结果的Actor实体访问演示文稿的所有信息:

    foreach($actors as $a){
        //
        // Some code here
        //

        $actorId        = $a->getId();
        $actorName      = $a->getName();
        $actorSurname   = $a->getSurname();
        $preTitle       = $a->getPresentation()->getTitle();
        $preDesc        = $a->getPresentation()->getDescription();
        $preDirector    = $a->getPresentation()->getDirector()->getFullName();

        //
        // Some code here
        //   
    }

我认为你也可以用这种方式解决你的问题.也许,你可以这样做:

    $rsm = new ResultSetMapping();
    $rsm->addEntityResult('App\MainBundle\Entity\InstagramShopPicture', 'p');
    $rsm->addFieldResult('p', 'id', 'id');
    $rsm->addFieldResult('p','lowresimageurl','lowresimageurl');
    $rsm->addFieldResult('p','medresimageurl','medresimageurl');
    $rsm->addFieldResult('p','highresimageurl','highresimageurl');
    $rsm->addFieldResult('p','caption','caption');
    $rsm->addFieldResult('p','numberoflikes','numberoflikes');
    $rsm->addFieldResult('p','numberofdislikes','numberofdislikes');
    $rsm->addMetaResult('p', 'shop_id', 'shop_id');

    $query = $em->createNativeQuery('SELECT picture.id, picture.lowresimageurl, picture.medresimageurl, picture.highresimageurl, picture.caption, picture.numberoflikes, picture.numberofdislikes
            FROM App_instagram_picture_category AS category
            INNER JOIN App_instagram_shop_picture AS p ON category.picture_id = p.id
            INNER JOIN App_instagram_shop AS shop ON shop.id = p.shop_id
            WHERE category.first_level_category_id = ?
            AND p.deletedAt IS NULL
            AND shop.deletedAt IS NULL
            AND shop.isLocked = 0
            AND shop.expirydate IS NOT NULL 
            AND shop.expirydate >  ?
            AND shop.owner_id IS NOT NULL 
            GROUP BY shop.id
            LIMIT ?'

            , $rsm);

    $query->setParameter(1, 10);
    $query->setParameter(2, '2014-05-20');
    $query->setParameter(3, 10);
    $itemsFromDifferentShops = $query->getResult();

因此,您可以获得图片,通过这些,您可以获得所需商店的信息.