Question

我正在Symfony 2.6上创建一个身份验证系统。注册过程有效。当我尝试登录时，我已登录并且我有“ROLE_USER”，但是探查器说我没有通过身份验证。我不明白发生了什么。这是用户实体：

<?php

namespace AppBundle\Entity;

use Doctrine\ORM\Mapping as ORM;
use Symfony\Component\Validator\Constraints as Assert;
use Symfony\Bridge\Doctrine\Validator\Constraints\UniqueEntity;
use Symfony\Component\Security\Core\User\UserInterface;

/**
* User
*
* @ORM\Table(name="baseuser")
* @ORM\Entity(repositoryClass="AppBundle\Repository\UserRepository")
* @UniqueEntity("email", message="error.user.email.taken")
*/
class User implements UserInterface, \Serializable
{
/**
 * @var integer
 *
 * @ORM\Column(name="id", type="integer")
 * @ORM\Id
 * @ORM\GeneratedValue(strategy="AUTO")
 */
private $id;

/**
 * @var string
 *
 * @ORM\Column(name="email", type="string", length=255, unique=true)
 */
private $email;

/**
 * @var string
 *
 * @ORM\Column(name="username", type="string", length=70, nullable=true)
 */
private $username;

/**
 * @var string
 *
 * @ORM\Column(name="password", type="string", length=128)
 * @Assert\Length(max = 4096)
 */
private $password;

/**
 * @var string
 *
 * @ORM\Column(name="salt", type="string", length=64)
 */
private $salt;

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

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

/**
 * @var string
 *
 * @ORM\Column(name="zipcode", type="string", length=10, nullable=true)
 */
private $zipcode;

/**
 * @var string
 *
 * @ORM\Column(name="city", type="string", length=50, nullable=true)
 */
private $city;

/**
 * @var integer
 *
 * @ORM\Column(name="country", type="integer", nullable=true)
 */
private $country;

/**
 * @var \DateTime
 *
 * @ORM\Column(name="birthdate", type="date", nullable=true)
 */
private $birthdate;

/**
 * @var string
 *
 * @ORM\Column(name="occupation", type="string", length=50, nullable=true)
 */
private $occupation;

/**
 * @var string
 *
 * @ORM\Column(name="about", type="text", nullable=true)
 */
private $about;

/**
 * @var string
 *
 * @ORM\Column(name="token", type="string", length=15, unique=true)
 */
private $token;

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

/**
 * @var boolean
 *
 * @ORM\Column(name="roles", type="integer")
 */
private $roles;

/**
 * @var \DateTime
 *
 * @ORM\Column(name="created_at", type="datetime")
 */
private $createdAt;

/**
 * @var \DateTime
 *
 * @ORM\Column(name="updated_at", type="datetime")
 */
private $updatedAt;

public function __construct()
{
    $datetime = new \DateTime();
    $this->createdAt = $datetime;
    $this->updatedAt = $datetime;
    $this->token     = base_convert(time(), 10, 36).\AppBundle\Library\StringHelper::randomString(5, "lower");
    $this->salt      = \AppBundle\Library\StringHelper::randomString();
    $this->isActive  = true;
    $this->roles     = 1;
}

/**
 * @inheritDoc
 */
public function eraseCredentials()
{
}

/**
* Return the roles
*
*/
public function getRoles() {
    switch ($this->roles) {
        case 1:
            $role =  'ROLE_USER';
            break;

        case 2:
            $role =  'ROLE_ADMIN';
            break;

        case 3:
            $role =  'ROLE_SUPER_ADMIN';
            break;

        default:
            $role = 'ROLE_USER';
            break;
    }

    return array($role);
}

/**
* Set the roles
*
* @param $roles
*/
public function setRoles($roles) {
    $this->roles = $roles;
    return $this;
}

/**
 * @see \Serializable::serialize()
 */
public function serialize()
{
    return serialize(array(
        $this->id,
        $this->email,
        $this->password
    ));
}

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

这是我的security.yml：

security:
    encoders:
         AppBundle\Entity\User:
            algorithm: bcrypt
            cost:      15

    role_hierarchy:
        ROLE_ADMIN:       ROLE_USER
        ROLE_SUPER_ADMIN: [ ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH ]

    providers:
        administrators:
            entity: { class: AppBundle:User, property: email }

    firewalls:
        main:
            pattern:    ^/
            anonymous: ~
            form_login: ~
            logout: ~

    access_control:
        - { path: ^/admin, roles: ROLE_ADMIN }

我不知道问题是什么，有什么想法吗？

此致

Answer 1

类用户必须实现EquatableInterface并实现如下方法：

public function isEqualTo(UserInterface $user)
{
    return $this->id === $user->getId();
}

Answer 2

好吧，我终于明白了。首先，您需要一个用户名，并且不能为空。所以我改变了我的方法getUsername（），所以它返回了电子邮件地址。然后，我将security.yml更改为：

security:
    encoders:
        AppBundle\Entity\User:
            algorithm: bcrypt
            cost: 13

    role_hierarchy:
        ROLE_ADMIN:       ROLE_USER
        ROLE_SUPER_ADMIN: [ ROLE_USER, ROLE_ADMIN, ROLE_ALLOWED_TO_SWITCH ]

    providers:
        users:
            entity: { class: AppBundle:User, property: email }

    firewalls:
        dev:
            pattern:  ^/(_(profiler|wdt)|css|images|js)/
            security: false

        default:
            pattern:    ^/
            anonymous: ~
            form_login: ~
            logout: ~

    access_control:
        - { path: ^/admin, roles: ROLE_ADMIN }
        - { path: ^/, roles: IS_AUTHENTICATED_ANONYMOUSLY }

Answer 3

我有同样的症状，这是因为我正在尝试没有角色的用户。

如果用户至少有一个角色，那么探查器似乎会考虑用户进行身份验证。

以下是分析器详细信息中的内容：

用户未经过身份验证，可能是因为他们没有角色。

即使他有IS_AUTHENTICATED_REMEMBERED和/或ImportError: /usr/local/lib/python2.7/dist-packages/tensorflow/python/_pywrap_tensorflow.so: invalid ELF header伪角色。

Answer 4

是的，您从dev

中删除了security.yml防火墙

dev:
    pattern: ^/(_(profiler|wdt|error)|css|images|js)/
    security: false

确保在任何其他防火墙之前插入...

查看原始security.yml

Symfony2已登录但未在Profiler中进行身份验证

4 个答案: