Symfony4 Forms - 带有两个choice_label的EntityType

问题描述：

I'm pretty new to symfony and symfony forms.

I have a form with an EntityType, that looks like this:

->add('customer', EntityType::class, [
    'label' => 'Kunde: ',
    'class' => Customer::class,
    'choice_label' => 'Name',
    'query_builder' => function(EntityRepository $er) {
        return $er->createQueryBuilder('c')
            ->select('CONCAT(c.firstname, " ", c.surname) AS Name');
    }
])

But I now get an Error/Warning:

Warning: spl_object_hash() expects parameter 1 to be object, string given

Customer Entity:

/**
 * @ORM\Entity(repositoryClass="App\Repository\CustomerRepository")
 */
class Customer
{
    /**
     * @ORM\Id
     * @ORM\GeneratedValue
     * @ORM\Column(type="integer")
     */
     private $id;

    /**
     * @var string
     *
     * @ORM\Column(name="firstname", type="string", length=50, nullable=false)
     */
    private $firstname;

    /**
     * @var string
     *
     * @ORM\Column(name="surname", type="string", length=50, nullable=false)
     */
    private $surname;

    ...

Thank you very much for your time and help.

我是symfony和symfony形式的新手。 p>

我有具有EntityType的表单，如下所示： p>

   - ＆gt; add（'customer'，EntityType :: class，[
'label'=＆gt;'Kunde  ：'，
'class'=＆gt; Customer :: class，
'select_label'=＆gt;'Name'，
'query_builder'=＆gt; function（EntityRepository $ er）{
 return $ er-＆gt  ; createQueryBuilder（'c'）
  - ＆gt; select（'CONCAT（c.firstname，“”，c.surname）AS Name'）; 
} 
 
]）
  code>  pre>  
 
 但我现在得到一个错误/警告： p> 
 
 
 警告：spl_object_hash（）期望参数1为对象，给定字符串 code>   p> 
 
 
    p> 
 
 

 
客户实体 strong>  p  > 
 
 
  / ** 
 * @ORM \ Entity（repositoryClass =“App \ Repository \ CustomerRepository”）
 * / 
class Customer 
 {
  / ** 
 * @ORM \ Id 
 * @ORM \ GeneratedValue 
 * @ORM \ Column（type =“integer”）
 * / 
 private $ id; 
 
 / ** 
  * @var string 
 * 
 * @ORM \ Column（name =“firstname”，type =“string”，length = 50，nullable = false）
 * / 
 private $ firstname; 
 
 / n  ** 
 * @var string 
 * 
 * @ORM \ Column（name =“surname”，type =“string”，length = 50，nullable = false）
 * / 
 private $ surname; \  n 
 ... 
  code>  pre> 
 
 非常感谢您的时间和帮助。 p> 
  div>

答

You could also simply use a callback for the choice_label

E.g.:

->add('customer', EntityType::class, [
    'label' => 'Kunde: ',
    'class' => Customer::class,
    'choice_label' => function (Customer $customer) {
        return $customer->getFirstname() . ' ' . $customer->getSurname();

        // or better, move this logic to Customer, and return:
        // return $customer->getFullname();
    },
])

答

Have you created a __toString() method in your customer entity.

/**
 * toString
 *
 * @return string
 */
public function __toString() {
    return $this->getFirstname().' '.$this->getSurname();
}

Then, something like this should be enough :

->add('customer')

If customer is related to your entityType, this should be enough

Symfony4 Forms - 带有两个choice_label的EntityType

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