问题描述
表单的视图数据应该是类的一个实例 App\Entity\Role,但是是一个(n)字符串。您可以通过以下方式避免此错误 将“data_class”选项设置为 null 或通过添加视图 将一个(n)字符串转换为一个实例的转换器 应用\实体\角色。
控制器方法
/**
* @Route("/{id}/edit",name="admin_edit_assign_roles")
*/
public function edit(Request $request,User $user){
$form = $this->createForm(UserType::class,$user);
$form->handleRequest($request);
if ($form->isSubmitted() && $form->isValid()) {
$this->getDoctrine()->getManager()->flush();
return $this->redirectToRoute('admin_assign_roles');
}
return $this->render('admin/assign_roles/edit.html.twig',[
'user' => $user,'form' => $form->createView()
]);
}
AssignRolesType.PHP
class AssignRolesType extends AbstractType
{
public function buildForm(FormBuilderInterface $builder,array $options)
{
$builder
->add('name');
}
public function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => Role::class
]);
}
}
UserType.PHP
class UserType extends AbstractType
{
function buildForm(FormBuilderInterface $builder,array $options)
{
$builder->add('email');
$builder->add('roles',CollectionType::class,[
'entry_type' => AssignRolesType::class,'entry_options' => ['label' => false],]);
}
function configureOptions(OptionsResolver $resolver)
{
$resolver->setDefaults([
'data_class' => User::class
]);
}
}
Entity\User.PHP
namespace App\Entity;
use App\Repository\UserRepository;
use Doctrine\Common\Collections\ArrayCollection;
use Doctrine\Common\Collections\Collection;
use Doctrine\ORM\Mapping as ORM;
use Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity;
use Symfony\Component\Security\Core\User\UserInterface;
/**
* @ORM\Entity(repositoryClass=UserRepository::class)
* @ORM\Table(name="`user`")
* @UniqueEntity(fields={"email"},message="There is already an account with this email")
*/
class User implements UserInterface
{
private $prefix = "ROLE_";
/**
* @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 Collection|Role[]
* @ORM\ManyToMany(targetEntity="App\Entity\Role")
* @ORM\JoinTable(
* name="user_roles",* joinColumns={@ORM\JoinColumn(name="user_id",referencedColumnName="id")},* inverseJoinColumns={@ORM\JoinColumn(name="role_id",referencedColumnName="id")}
* )
*/
private $roles;
/**
* @var string The hashed password
* @ORM\Column(type="string")
*/
private $password;
/**
* @ORM\Column(type="boolean")
*/
private $isverified = false;
/**
* @ORM\Column(type="integer",nullable=true)
*/
private $vkontakteID;
/**
* @ORM\Column(type="string",length=255,nullable=true)
*/
private $vkontakteAccesstoken;
/**
* @ORM\Column(type="string",nullable=true)
*/
private $firstName;
/**
* @ORM\Column(type="string",nullable=true)
*/
private $lastName;
/**
* @ORM\Column(type="string",nullable=true,unique=true)
*/
private $nickName;
public function __construct()
{
$this->roles = new ArrayCollection();
}
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) $this->email;
}
/**
* @see UserInterface
*/
public function getRoles(): array
{
foreach ($this->roles->toArray() as $roleItem){
$roles[] = $this->prefix . $roleItem->getSlug();
}
// guarantee every user at least has ROLE_USER
$roles[] = 'ROLE_USER';
return array_unique($roles);
}
public function setRoles(Collection $roles): self
{
$this->roles = $roles;
return $this;
}
public function addRole(Role $role){
$this->roles->add($role);
}
public function removeRole(Role $role){
$this->roles->removeElement($role);
}
/**
* @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 isverified(): bool
{
return $this->isverified;
}
public function setIsverified(bool $isverified): self
{
$this->isverified = $isverified;
return $this;
}
public function getVkontakteID(): ?int
{
return $this->vkontakteID;
}
public function setVkontakteID(?int $vkontakteID): self
{
$this->vkontakteID = $vkontakteID;
return $this;
}
public function getFirstName(): ?string
{
return $this->firstName;
}
public function setFirstName(?string $firstName): self
{
$this->firstName = $firstName;
return $this;
}
public function getLastName(): ?string
{
return $this->lastName;
}
public function setLastName(?string $lastName): self
{
$this->lastName = $lastName;
return $this;
}
public function getNickName(): ?string
{
return $this->nickName;
}
public function setNickName(string $nickName): self
{
$this->nickName = $nickName;
return $this;
}
public function getVkontakteAccesstoken(): ?string
{
return $this->vkontakteAccesstoken;
}
public function setVkontakteAccesstoken(?string $vkontakteAccesstoken): self
{
$this->vkontakteAccesstoken = $vkontakteAccesstoken;
return $this;
}
public function getIsverified(): ?bool
{
return $this->isverified;
}
}
解决方法
错误信息的原因
您的 User.getRoles()
返回一个字符串数组。
您的 UserType
角色的 CollectionType
字段具有条目类型 AssignRoleType
,它需要 Role
实体。表单通过 getter 访问该属性(如果存在),并且该 getter (getRoles
) 仅返回一个字符串数组,CollectionType
将其提供给 AssignRoleType
,这显然是在抱怨字符串不是 Role
实体。
由于您可能正在使用 Symfony 安全角色系统,因此 getRoles
可能必须返回字符串。所以改变它基本上是不可能的。
为了盈利而更名
下面的方法不是那么优雅,但很简单。表单组件并不关心 User 类上的属性是如何调用的,它只关心访问(或)。
您可以在 getRoleObjects
上实现方法 addRoleObject
、removeRoleObject
、addRole
(后两个应该与 removeRole
和 User
相同) } 并将表单字段重命名为 role_objects
,然后你应该是金色的。 getRoleObjects
必须return $this->roles->toArray()
,之后一切正常。
评论
您可能需要 EntityType
(multiple
选项设置为 true
)而不是 CollectionType
,除非您真的想要编辑角色的名字。但只是在这里猜测。我知道什么,对吧?! ^^
具有不同名称的方法并不是特别好,还有其他选项,例如针对此特定用例使用数据转换器,这可能会很好地工作。