如何解决EasyAdmin 3.1 CrudControllers Symfony
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,却不知道如何设置它。
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,但是从Easyadmin文档中看到的内容来看,过滤器使我可以根据选择内容进行设置,以免对我起作用。我还尝试使用QueryBuilder获取具有某些ROLE_的用户,但这也失败了。
解决方法
我知道了,感谢您的回答。我之所以发布解决方案,是因为我不想成为那些说“我明白了”并且不发表他们实际上如何解决的人。
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,通过使用查询生成器,我获得了具有特定角色的用户。惊人的工具,通过该查询构建器,我几乎可以从数据库中获取所需的任何内容,并将其放入Crud Controllers中。我希望有一天能对某人有所帮助。
,尝试一下:
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()]);
}
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