这是我的错误:

The form's view data is expected to be an instance of class My\Bundle\Entity\Tags, but is an instance of class Doctrine\Common\Collections\ArrayCollection. You can avoid this error by setting the "data_class" option to null or by adding a view transformer that transforms an instance of class Doctrine\Common\Collections\ArrayCollection to an instance of My\Bundle\Entity\Tags

这是我的表单生成器

 $builder
            ->add('name')
            ->add('tags','collection',array(
                        'data_class' => 'My\Bundle\Entity\Tags'
                        )
            )
            ->add('save','submit')
        ;

我将 data_class 更改为 null(仅此)，但出现错误:

The form's view data is expected to be of type scalar, array or an instance of \ArrayAccess, **but is an instance of class My\Bundle\Entity\Tags*. You can avoid this error by setting the "data_class" option to "My\Bundle\Entity\Tags" or by adding a view transformer that transforms an instance of class My\Bundle\Entity\Tags to scalar, array or an instance of \ArrayAccess.

我试过用变压器，所以它看起来像这样:

    $transformer = new TagTransformer($this->entityManager);
    $builder
        ->add(
            $builder->create(
                'tags','collection',array(
                    'data_class' => 'My\Bundle\Entity\Tags'
                    )
            )->addModelTransformer($transformer)
        );

和变压器:

public function transform($tag)
{
    if (null === $tag) {
        return "";
    }

    return $tag->toArray();
}

并再次将 data_class 更改为 null。我得到的:

The form's view data is expected to be of type scalar, array or an instance of \ArrayAccess, but is an instance of class My\Bundle\Entity\Tags. You can avoid this error by setting the "data_class" option to "My\Bundle\Entity\Tags" or by adding a view transformer that transforms an instance of class My\Bundle\Entity\Tags to scalar, array or an instance of \ArrayAccess.

当我将 data_class 更改为 My\Bundle\Entity\Tags

The form's view data is expected to be an instance of class My\Bundle\Entity\Tags, but is a(n) array. You can avoid this error by setting the "data_class" option to null or by adding a view transformer that transforms a(n) array to an instance of My\Bundle\Entity\Tags.

嗯..我的意思是...wtf？我究竟做错了什么？我该如何改变它？

编辑:

我的用户实体:

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

    /**
     * @var string
     *
     * @ORM\Column(name="name", type="string", length=255)
     */
    private $name;

    /**
     * @ORM\ManyToMany(targetEntity="Tags", cascade={"persist"})
     */
    protected $tags;

// methods, etc..
}

最佳答案

所以你得到错误的原因是因为你使用的集合字段类型有点不正确。首先，集合字段类型不支持data_class。当你说

->add('tags','collection',array(
    'data_class' => 'My\Bundle\Entity\Tags'
     )
)

你基本上是在说标签(根据你的声明，它是一个数组集合)实际上是一个标签。如果您查看集合类型的文档，您会注意到 data_class 甚至不是受支持的选项。 http://symfony.com/doc/current/reference/forms/types/collection.html

所以如果你想呈现一个多选标签列表，你正在寻找实体类型，但是这些是标签，如果你有任何类型的体面网站，你可能有比多选列表更多的方式将是实用的。在设计方面，您只想拥有一个自动完成器来显示您键入的文本中已经存在哪些标签，然后让用户按回车键添加标签，无论标签是否存在。然后在自动完成器上方，您将显示已添加的标签，并且旁边有 x ，他们可以按下以删除标签。

您可以通过将表单中的标签字段设置为未映射的文本类型并使用 javascript 将标签组合成表单提交时的字符串来作弊，然后在您的操作中将字符串转换为您的标签。

关于php - Symfony 不能决定哪个类需要，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22119163/