php - Symfony 3 类的对象无法转换为字符串

Question

我有这个地址表格。一切都按预期工作，直到我为城市添加了一个下拉选择字段。当我尝试保存表单时出现此错误:

Catchable Fatal Error: Object of class AppBundle\Entity\Address could not be converted to string

文档说 EntityType 字段旨在从 Doctrine 实体加载选项。

这是表格:

 $builder
             ->add('cities', EntityType::class, array(
                'class' => \AppBundle\Entity\Cities::class,
                'choice_label' => 'cityname',
                'choice_value' => 'cityid')
            )

这是我的实体

 /**
 * @var string
 * 
 * @ORM\Column(name="cities", type="string", length=55, nullable=false)
 */
private $cities;   

和 setter/getter:

 /**
 * Set cities
 *
 * @param string $cities
 *
 * @return Address
 */
public function setCities($cities)
{
    $this->cities = $cities;

    return $this;
}

/**
 * Get cities
 *
 * @return string
 */
public function getCities()
{
    return $this->cities;
}

我还添加了这个:

public function __toString() {
  return $this->getCities();
}

得到了这样的结果:

Catchable Fatal Error: Method AppBundle\Entity\Address::__toString() must return a string value

Answer 1

你应该为你的实体实现 __toString() 方法:

__toString() allows a class to decide how it will react when it is treated like a string. For example, what echo $obj; will print. This method must return a string, as otherwise a fatal E_RECOVERABLE_ERROR level error is emitted.

例如，您可以编写如下代码来将您的类表示为字符串:

public function __toString() {
  return $this->getCities();
}

关于这个魔术方法的更多信息here .