RoleInterface抛出"调用非对象"错误

Question

我正在研究Symfony 2.0.16

我的UserProvider中有getRoles方法

public function getRoles()
{
    /**
     * @var \Doctrine\Common\Collections\ArrayCollection $rol
     */
    return $this->rol->toArray();
}

我的Rol实体有角色接口

class Rol implements \Symfony\Component\Security\Core\Role\RoleInterface
//...
public function getRole()
{
    return $this->getName();
}

但是当我尝试登录时,我收到以下错误

致命错误:在第57行的C:\ Users\julian\Code\parqueadero\vendor\symfony\src\Symfony\Bundle\SecurityBundle\DataCollector\SecurityDataCollector.php中的非对象上调用成员函数getRole()

读取类SecurityDataCollector,Closure抛出错误

array_map(function ($role){ return $role->getRole();}, $token->getRoles()

现在我把它改成了

array_map(function ($role){ var_dump($role); return $role->getRole();}, $token->getRoles()

令我惊讶的是,$role是一个对象Rol,但我无法理解为什么我得到错误.

Answer 1

我发现解决方案的问题是PHP 5.4中的一个bug(php我用的)序列化方法 github用户yoannch提出这个解决方案,是serialize/unserialize用json_encode/json_decode方法覆盖方法

class User implements \Serializable

//...

/**
 * Serializes the content of the current User object
 * @return string
 */
public function serialize()
{
    return \json_encode(
            array($this->username, $this->password, $this->salt,
                    $this->rol, $this->id));
}

/**
 * Unserializes the given string in the current User object
 * @param serialized
 */
public function unserialize($serialized)
{
    list($this->username, $this->password, $this->salt,
                    $this->rol, $this->id) = \json_decode(
            $serialized);
}

只需要更改正确的名称属性

Answer 2

我有同样的问题(Windows,PHP 5.4.5),更新到5.4.7,它仍然无法正常工作.尽管如此,我提出了一种需要较少维护的解决方法(当您按照上述文章中的描述覆盖序列化功能时,在添加/删除字段时,您必须使它们保持最新).到目前为止它对我有用,我希望我可能忘记的解决方法没有其他问题.只需改变用户的getRoles()功能:

/**
 * @inheritDoc
 */
public function getRoles()
{
    $roles = array();
    foreach ($this->userRoles as $role) {
        $roles[] = $role->getRole();
    }

    return $roles;
}

请注意,$role->getRole()将角色名称返回为String(例如ROLE_ADMIN).

Answer 3

解决这个问题很简单.与您User和Role对象的循环引用相关的所有问题.所以你不必序列化User::$roles和Role::$users字段.

看看Symfony\Component\Security\Core\Authentication\Token\AbstractToken::__construct()和Symfony\Component\Security\Core\Authentication\Token\AbstractToken::serialize().

您如何看待,Symfony通过UserInterface::getRoles()在序列化之前调用来获取用户的角色.并序列化User和Roles单独.

您必须实现\Serializable接口User和Role实体.

例:

/**
 * Acme\Bundle\UserBundle\Entity\User
 * 
 * @ORM\Table(name="`user`")
 * @ORM\Entity(repositoryClass="Acme\Bundle\UserBundle\Entity\UserRepository")
 */
class User implements AdvancedUserInterface, EquatableInterface, \Serializable
{
    /**
     * @var integer $id
     *
     * @ORM\Column(name="id", type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var string $username
     * 
     * @ORM\Column(type="string", length=30, unique=true)
     */
    private $username;

    /**
     * @var string $email
     * 
     * @ORM\Column(type="string", length=100, unique=true)
     */
    private $email;

    /**
     * @var string $salt
     *
     * @ORM\Column(type="string", length=40)
     */
    private $salt;

    /**
     * @var string $password
     *
     * @ORM\Column(type="string", length=128)
     */
    private $password;

    /**
     * @var boolean $isActive
     *
     * @ORM\Column(type="boolean")
     */
    private $isActive;

    /**
     * User's roles. (Owning Side)
     * 
     * @var ArrayCollection
     * 
     * @ORM\ManyToMany(targetEntity="Role", inversedBy="users")
     */
    private $roles;

    // .....

    /**
     * @see \Serializable::serialize()
     */
    public function serialize()
    {
        /*
         * ! Don't serialize $roles field !
         */
        return \serialize(array(
            $this->id,
            $this->username,
            $this->email,
            $this->salt,
            $this->password,
            $this->isActive
        ));
    }

    /**
     * @see \Serializable::unserialize()
     */
    public function unserialize($serialized)
    {
        list (
            $this->id,
            $this->username,
            $this->email,
            $this->salt,
            $this->password,
            $this->isActive
        ) = \unserialize($serialized);
    }

}

/**
 * Acme\Bundle\UserBundle\Entity\Role
 *
 * @ORM\Table(name="role")
 * @ORM\Entity
 * 
 */
class Role implements RoleInterface, \Serializable
{
    /**
     * @var integer $id
     *
     * @ORM\Column(type="integer")
     * @ORM\Id
     * @ORM\GeneratedValue(strategy="AUTO")
     */
    private $id;

    /**
     * @var string $role
     *
     * @ORM\Column(name="role", type="string", length=20, unique=true)
     */
    private $role;

    /**
     * Users in group (Inverse Side)
     * 
     * @var ArrayCollection
     * 
     * @ORM\ManyToMany(targetEntity="User", mappedBy="roles")
     */
    private $users;

    // .....    

    /**
     * @see \Serializable::serialize()
     */
    public function serialize()
    {
        /*
         * ! Don't serialize $users field !
         */
        return \serialize(array(
            $this->id,
            $this->role
        ));
    }

    /**
     * @see \Serializable::unserialize()
     */
    public function unserialize($serialized)
    {
        list(
            $this->id,
            $this->role
        ) = \unserialize($serialized);
    }
}

所有都将正确序列化/反序列化.

请访问https://github.com/symfony/symfony/issues/3691查看讨论

另见:http: //docs.doctrine-project.org/projects/doctrine-orm/en/latest/cookbook/entities-in-session.html#serializing-entity-into-the-session