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Symfony4 Forms - 带有两个 choice_label 的 EntityType

[英]Symfony4 Forms - EntityType with two choice_label

I'm pretty new to symfony and symfony forms.我对 symfony 和 symfony 形式很陌生。

I have a form with an EntityType, that looks like this:我有一个带有 EntityType 的表单,如下所示:

->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:但我现在收到错误/警告:

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

在此处输入图片说明


Customer Entity:客户实体:

/**
 * @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.非常感谢您的时间和帮助。

You could also simply use a callback for the choice_label您也可以简单地为choice_label使用回调

Eg:例如:

->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();
    },
])

Have you created a __toString() method in your customer entity.您是否在客户实体中创建了__toString()方法。

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

Then, something like this should be enough :然后,这样的事情应该就足够了:

->add('customer')

If customer is related to your entityType, this should be enough如果客户与您的实体类型相关,这应该就足够了

Not sure if this works with Symfony 4 but it does work with symfony 5. This is more of an extension of Yoshis' answer than a new answer.不确定这是否适用于 Symfony 4,但它适用于 symfony 5。这更像是 Yoshis 答案的扩展,而不是新答案。

If you create a getFullname() function as suggested by Yoshi you can do the following.如果您按照 Yoshi 的建议创建getFullname()函数,您可以执行以下操作。

Customer.php:客户.php:

public function getFullname(): ?string {
    return $this->firstname . ' ' . $this->surname;
}

Then in your form:然后以您的形式:

->add('customer', EntityType::class, [
    'label' => 'Kunde: ',
    'class' => Customer::class,
    'choice_label' => 'fullname'
])

Adding the fullname function to the customer also allows you to use fullname within your twig files.向客户添加全名功能还允许您在树枝文件中使用全名。

eg.例如。

{{ customer.fullname }}

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