Question

在ManyToMany关系中使用Artist和Show的实体，尝试构建一个列出现有艺术家尚未在给定节目中的表单会导致错误

错误：无效的PathExpression。必须是 CollectionValuedAssociationField

编辑：

在Doctrine's documentation处有notIn($x, $y)表达式，但它完全没有明确（即尝试失败）如何在表单类中使用它。

编辑2：

可以在Artist Repository中获得预期结果，其功能如下所示。但是它在表单类中不可用，因为它的结果不是QueryBuilder的实例！ [感谢xurshid29＆＃39; s SO answer。]

public function notInShow($show)
{
    $qb = $this->getEntityManager()->createQueryBuilder();

    $ids = $qb
        ->select('a.id')
        ->from('AppBundle:Artist', 'a', 'a.id')
        ->leftJoin('a.shows', 's')
        ->where('s.show = ?1')
        ->setParameter(1, $show->getShow())
        ->getQuery()
        ->getArrayResult();

    $ids = array_keys($ids);

    $qbA = $this->getEntityManager()->createQuery(
        'SELECT a FROM AppBundle:Artist a '
        . 'WHERE a.id NOT IN (:ids)')
        ->setParameter(':ids', $ids)
        ->getResult();

    return $qbA;
}

构建字段的最新尝试（很多）如下所示：

表单类（代码段）：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $show = $options['show'];
    $builder->add('artists', EntityType::class,
        [
            'class' => Artist::class,
            'choice_label' => function($artist, $key, $index) {
                return $artist->getLastName() . ', ' . $artist->getFirstName();
            },
            'query_builder' => function (EntityRepository $er) use($show) {
                return $er->createQueryBuilder('a')
                        ->join('a.shows', 's')
                        ->where('a NOT MEMBER of s')
                        ->andWhere('IDENTITY(s) = :show')
                        ->setParameter(':show', $show->getId())
                        ->orderBy('a.firstName', 'ASC')
                        ->orderBy('a.lastName', 'ASC')
                ;
            },
            'expanded' => true,
            'multiple' => true,
    ]);
}

实体摘录：

艺术家：

/**
 * @var \Doctrine\Common\Collections\Collection
 *
 * @ORM\ManyToMany(targetEntity="Show", mappedBy="artists")
 */
protected $shows;

显示：

/**
 * @var \Doctrine\Common\Collections\Collection
 *
 * @ORM\ManyToMany(targetEntity="Artist", inversedBy="shows", cascade={"persist"})
 * @ORM\JoinTable(name="participation",
 *      joinColumns={@ORM\JoinColumn(name="show_id", referencedColumnName="id")},
 *      inverseJoinColumns={@ORM\JoinColumn(name="artist_id", referencedColumnName="id")}
 *      ))
 */
protected $artists;

Answer 1

最后！最后的一些调整，这是有效的：

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $show = $options['show'];
    $builder->add('artists', EntityType::class,
            [
                'class' => Artist::class,
                'choice_label' => function($artist, $key, $index) {
                    return $artist->getLastName() . ', ' . $artist->getFirstName();
                },
                'query_builder' => function (EntityRepository $er) use($show) {
                    $qb = $er->createQueryBuilder('a');
                    $ids = $qb
                        ->select('a.id')
                        ->leftJoin('a.shows', 's')
                        ->where('s.show = ?1')
                        ->setParameter(1, $show->getShow())
                        ->getQuery()
                        ->getResult();
                    if (empty($ids)) {
                        return $er->createQueryBuilder('a');
                    } else {
                        return $er->createQueryBuilder('a')
                            ->where($er->createQueryBuilder('a')->expr()->notIn('a.id', ':ids'))
                            ->setParameter(':ids', $ids);
                    }
                },
                'expanded' => true,
                'multiple' => true,
        ])
        ->add('save', SubmitType::class,
            array(
                'label' => 'Add artist(s)',
                'label_format' => ['class' => 'text-bold']
    ));
}

必须是一个CollectionValuedAssociationField，试图找不到多对多

编辑：

编辑2：

1 个答案: