Question

我有两个实体Article和Image与ManytoMany关系相关，使用名为ArticlesHaveImages的第三个实体。我没有使用经典的ManytoMany关系，因为我在ArticlesHaveImages中有第三个字段称为序列。

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

    ...
    ...
        /**
         * One Article has Many articleHaveImage.
         * @ORM\OneToMany(targetEntity="ArticlesHaveImages", mappedBy="article")
         */
        private $images;


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

     ...
    ...

        /**
         * One Image has Many articleHaveImage.
         * @ORM\OneToMany(targetEntity="ArticlesHaveImages", mappedBy="image")
         */
        private $articles;


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

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

    /**
     * Many ArticlesHaveImages have One Article.
     * @ORM\ManyToOne(targetEntity="Article", inversedBy="images")
     * @ORM\JoinColumn(name="article_id", referencedColumnName="id")
     */
    private $article;

    /**
     * Many ArticlesHaveImages have One Image.
     * @ORM\ManyToOne(targetEntity="Image", inversedBy="articles")
     * @ORM\JoinColumn(name="image_id", referencedColumnName="id")
     */
    private $image;


    /**
     * @var int
     *
     * @ORM\Column(name="sequence", type="integer")
     */
    private $sequence;

如何创建表单？我想要一个多选项，所有图像都是名称。

我试过了：

class ArticleType extends AbstractType
{
    /**
     * {@inheritdoc}
     */
    public function buildForm(FormBuilderInterface $builder, array $options)
    {

        $images = $options['images'];

        $builder
            ->add('translations', CollectionType::class, array(
                    'entry_type' => ArticleTranslationType::class,
                    'entry_options' => array('label' => false)))
            ->add('images', ChoiceType::class, array(
                "required" => false,
                "choices" => $images,
                'multiple' => true,

但我明白了：

无法确定属性＆＃34;图像＆＃34;的访问类型在班上＆＃34;的appbundle \实体\第＆＃34;

非常感谢！

Symfony 3.3和doctrine 2

0 个答案: