EasyAdmin 3.1 CrudControllers Symfony

Question

我在设置 Crud Controller 的关联字段时遇到问题。我只想在klient_id_klienta字段中看到某个ROLE_的用户，不知道怎么设置。

这是我的 CrudController：

class AdresKlientaCrudController extends AbstractCrudController
{
    public static function getEntityFqcn(): string
    {
        return AdresKlienta::class;
    }

    /*
    public function configureFields(string $pageName): iterable
    {
        return [
            IdField::new('id'),
            TextField::new('title'),
            TextEditorField::new('description'),
        ];
    }
    */

//    public function configureFields(string $pageName): iterable
//    {
//        return [
//            'id',
//            'klient_id_klienta',
//            'miejscowosc',
//            'ulica',
//            'nr_domu',
//            'nr_lokalu',
//            'kod_pocztowy'
//        ];
//    }
    public function configureFields(string $pageName): iterable
    {

        //moje
//        $qb = new QueryBuilder($this->getDoctrine()->getManager());
//        $qb->select('u')->from('User','u')->where('u.roles = ?ROLE_USER');
//
//
//        dump(EntityFilter::new('klient_id_klienta')->apply($qb));

        //koniec moje

        $foreignKey = AssociationField::new('klient_id_klienta'); //here is my problem as it shows every user
        return [
//            IdField::new('id'),
            TextField::new('miejscowosc'),
            TextField::new('ulica'),
            TextField::new('nr_domu'),
            TextField::new('nr_lokalu'),
            TextField::new('kod_pocztowy'),
            //AssociationField::new('klient_id_klienta')
            $foreignKey

        ];
    }
}

这里是用户实体

<?php

namespace App\Entity;

use App\Repository\UserRepository;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Security\Core\User\UserInterface;

/**
 * @ORM\Entity(repositoryClass=UserRepository::class)
 */
class User implements UserInterface
{
    /**
     * @ORM\Id()
     * @ORM\GeneratedValue()
     * @ORM\Column(type="integer")
     */
    private $id;

    /**
     * @ORM\Column(type="string", length=180, unique=true)
     */
    private $email;



    /**
     * @ORM\Column(type="json")
     */
    private $roles = [];



    /**
     * @var string The hashed password
     * @ORM\Column(type="string")
     */
    private $password;

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

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

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


    public function getId(): ?int
    {
        return $this->id;
    }

    public function getEmail(): ?string
    {
        return $this->email;
    }

    public function setEmail(string $email): self
    {
        $this->email = $email;

        return $this;
    }

    /**
     * A visual identifier that represents this user.
     *
     * @see UserInterface
     */
    public function getUsername(): string
    {
        return (string) $this->email;
    }

    /**
     * @see UserInterface
     */
    public function getRoles(): array
    {
        $roles = $this->roles;
        // guarantee every user at least has ROLE_USER


        return array_unique($roles);
    }

    public function setRoles(array $roles): self
    {
        $this->roles = $roles;

        return $this;
    }

    /**
     * @see UserInterface
     */
    public function getPassword(): string
    {
        return (string) $this->password;
    }

    public function setPassword(string $password): self
    {
        $this->password = $password;

        return $this;
    }

    /**
     * @see UserInterface
     */
    public function getSalt()
    {
        // not needed when using the "bcrypt" algorithm in security.yaml
    }

    /**
     * @see UserInterface
     */
    public function eraseCredentials()
    {
        // If you store any temporary, sensitive data on the user, clear it here
        // $this->plainPassword = null;
    }

    public function getName(): ?string
    {
        return $this->name;
    }

    public function setName(string $name): self
    {
        $this->name = $name;

        return $this;
    }

    public function getSurname(): ?string
    {
        return $this->surname;
    }

    public function setSurname(string $surname): self
    {
        $this->surname = $surname;

        return $this;
    }

    public function getTel(): ?string
    {
        return $this->tel;
    }

    public function setTel(string $tel): self
    {
        $this->tel = $tel;

        return $this;
    }


    //moje funkcje
    public function __toString()
    {
        // TODO: Implement __toString() method.
        $userAndRole = implode($this->roles);
        return $this->email.'-'.$userAndRole;
    }
}

I only want to see users who have ROLE_USER

我尝试使用筛选器，但从我在 Easyadmin 文档中看到的内容来看，筛选器允许我根据他们得到的内容设置选项，所以这对我不起作用。我还尝试使用 QueryBuilder 获取具有特定 ROLE_ 的用户，但也失败了。

Answer 1

试试这个：

public function configureFields(string $pageName): iterable
{
    $users = $this->entityManager->getRepository(User::class)->findBy([
            'roles' => 'ROLE_USER']);
    yield AssociationField::new('klient_id_klienta')->onlyOnForms()->setFormTypeOptions(["choices" => $users->toArray()]);
      
}

Answer 2

我弄明白了，感谢您的回答。我正在 post 制定我的解决方案，因为我不想成为那些说“我想通了”而不知道 post 他们实际上是如何想通的人。

public function configureFields(string $pageName): iterable
    {
        //utworzenie wyświetlania tylko tych użytkowników, którzy maja role ROLE_USER
        $association = AssociationField::new('klient_id_klienta', 'Email klienta')
            ->setFormTypeOption(

                    'query_builder', function (UserRepository  $userRepository){
                        return $userRepository->createQueryBuilder('u')
                            ->andWhere('u.roles LIKE :role')->setParameter('role', '%"ROLE_USER"%');
                    }
            );
        return [
//            IdField::new('id'),
            TextField::new('miejscowosc', 'Miejscowość'),
            TextField::new('ulica', 'Ulica'),
            TextField::new('nr_domu', 'Numer domu'),
            TextField::new('nr_lokalu', 'Numer Lokalu'),
            TextField::new('kod_pocztowy', 'Kod pocztowy'),
            $association,//wywołanie klucza obcego który odfiltrowuje użytkowników

        ];
    }

如您所见，我通过使用查询生成器获得了具有我想要获得的特定角色的用户。神奇的工具，通过那个查询构建器，我几乎可以从我的数据库中得到我想要的任何东西，并将它放在我的 Crud 控制器中。我希望有一天能对某人有所帮助。

EasyAdmin 3.1 CrudControllers Symfony

EasyAdmin 3.1 CrudControllers Symfony

mysql

controller

crud

symfony

easyadmin