RoleInterface 抛出“对非对象的调用"；错误

问题描述

我正在使用 Symfony 2.0.16

I'm working over Symfony 2.0.16

我的 UserProvider 中有 getRoles 方法

I have in my UserProvider the method getRoles

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

并且我的 Rol 实体具有角色接口

and my Rol entity has the role interface

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

但是当我尝试登录时出现以下错误

but when I try to login I get the following error

致命错误:在线调用 C:\Users\julian\Code\parqueadero\vendor\symfony\src\Symfony\Bundle\SecurityBundle\DataCollector\SecurityDataCollector.php 中非对象的成员函数 getRole()57

读取类SecurityDataCollector，由Closure抛出错误

Reading the class SecurityDataCollector, the error is thrown by a Closure

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

现在我把它改成

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，但我不明白为什么会出现错误.

To my surprise, $role is a object Rol but I can't understand why I get the error.

推荐答案

我发现该问题的解决方案是一个 PHP 5.4 中的错误(我正在使用的 php)序列化方法 github 用户 yoannch 提出了这个 解决方案，使用json_encode/json_decode方法覆盖serialize/unserialize方法

I found the solution the problem is a bug in PHP 5.4 (the php i'm using) serialize method the github user yoannch proposed this solution, is overwrite the serialize/unserialize methods using json_encode/json_decode methods

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

only need change the correct name properties

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