必须是一个收藏品，试图在许多人中找到很多东西

Question

与艺术家实体并在Manytomany关系中展示，试图构建一种表格，该表格列出了尚未在给定的节目中列出的现有艺术家导致错误

错误：无效的路径表达。必须是一个Collection ValuedAssociationField

编辑：

在学说的文档中有 notIn($x, $y)的表达式，但根本不清楚（即尝试失败）如何在表单类中使用。

编辑2：

可以在艺术家存储库中获得预期的结果，并显示下面的功能。不过，它在表单类中不可用，因为它的结果不是查询布构成的实例！[感谢Xurshid29的So回答。]

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;
}

（许多人）建立一个领域的最新尝试看起来像这样：

表格类（片段）：

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,
    ]);
}

实体片段：

艺术家：

/**
 * @var DoctrineCommonCollectionsCollection
 *
 * @ORMManyToMany(targetEntity="Show", mappedBy="artists")
 */
protected $shows;

显示：

/**
 * @var DoctrineCommonCollectionsCollection
 *
 * @ORMManyToMany(targetEntity="Artist", inversedBy="shows", cascade={"persist"})
 * @ORMJoinTable(name="participation",
 *      joinColumns={@ORMJoinColumn(name="show_id", referencedColumnName="id")},
 *      inverseJoinColumns={@ORMJoinColumn(name="artist_id", referencedColumnName="id")}
 *      ))
 */
protected $artists;

Answer 1

终于！一些最终的调整，这有效：

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']
    ));
}