EasyAdmin 3.1 CrudControllers Symfony
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【中文标题】EasyAdmin 3.1 CrudControllers Symfony【英文标题】: 【发布时间】:2020-12-04 06:32:22 【问题描述】: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,我不知道如何设置。
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
我尝试使用过滤器,但根据我在 Easyadmin 文档中看到的过滤器,我可以根据他们获得的内容设置选择,这样对我不起作用。我还尝试使用 QueryBuilder 来获取具有特定 ROLE_ 的用户,但也失败了。
【问题讨论】:
【参考方案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()]);
【讨论】:
您能告诉我您是如何获得“entityManage”属性的吗?我无法使用...【参考方案2】:我想通了,我想感谢你的回答。我发布我的解决方案是因为我不想成为那些说“我想通了”并且不发布他们实际上是如何想通的人之一。
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获取我想要的任何东西,并将其放入我的 Crud 控制器中。我希望有一天它对某人有所帮助。
【讨论】:
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