Multiple user classes - symfony

Question

I have 4 different user types in a system (on top of Symfony 2). Each type have some specific fields and behaviour, but all of them have a common base. So it seems it would be good idea to implement single class for each user extending the same superclass.
How can it be achieved? All I have found on the topic is some RollerworksMultiUserBundle.

Answer 1

Using table inheritance in the ORM level and OOP inheritance. Go for Single Table Inheritance if performance is critical (no JOINs) or Class Table Inheritance if you are a purist.
E.g.
Common base class:
use Symfony\Component\Security\Core\User\AdvancedUserInterface;
/**
* #ORM\Entity(repositoryClass="Some\Bundle\Repository\UserRepository")
* #ORM\InheritanceType("SINGLE_TABLE")
* #ORM\DiscriminatorColumn(name="userType", type="string")
* #ORM\DiscriminatorMap({
* "userType1" = "UserType1",
* "userType2" = "UserType2",
* "userType3" = "UserType3",
* "userType4" = "UserType4"
* })
*/
abstract class User implements AdvancedUserInterface
{
/**
* #ORM\Id()
* #ORM\Column(type="integer")
* #ORM\GeneratedValue(strategy="AUTO")
*/
protected $id;
/**
* #ORM\Column(type="string", length=250, unique=true)
*/
protected $email;
/**
* #ORM\Column(type="string", length=128, nullable=true)
*/
protected $password;
// other fields
public function getSalt()
{
return "some salt number";
}
public function getUsername()
{
return $this->email;
}
public function getPassword()
{
return $this->password;
}
public function getRoles()
{
return array('ROLE_USER');
}
public function eraseCredentials() {}
public function isCredentialsNonExpired()
{
return true;
}
public function isAccountNonLocked()
{
return true;
}
public function isAccountNonExpired()
{
return true;
}
public function isEnabled()
{
return true;
}
public function equals(UserInterface $user)
{
return $user->getUsername() === $this->getUsername() || $user->getEmail() === $this->getEmail();
}
}
The children classes are straightforward (below an example for class UserType1 only):
/**
* #ORM\Entity
*/
class UserType1 extends User
{
// fields of UserType1 class go here
public function getRoles()
{
return array('ROLE_USER_TYPE_1', 'ROLE_USER');
}
}
The rest is pretty much like in the examples. In security.yml:
security:
encoders:
Some\Bundle\Repository\User:
algorithm: sha512
encode_as_base64: false
iterations: 1000
providers:
provider1:
entity: { class: "SomeBundle:User" }
role_hierarchy:
ROLE_USER_TYPE_1: ROLE_USER
ROLE_USER_TYPE_2: ROLE_USER
ROLE_USER_TYPE_3: ROLE_USER
ROLE_USER_TYPE_4: ROLE_USER
firewalls:
firewall1:
pattern: ^/
provider: provider1
form_login:
login_path: /login
check_path: /auth
post_only: true
username_parameter: email
password_parameter: password
always_use_default_target_path: true
default_target_path: /
logout:
path: /logout
target: /login
anonymous: ~
The repository class:
use Doctrine\ORM\EntityRepository;
use Symfony\Component\Security\Core\User\UserInterface;
class UserRepository extends EntityRepository implements UserProviderInterface
{
public function loadUserByUsername($username)
{
$qb = $this->createQueryBuilder('u');
$query = $qb->where('LOWER(u.email) = :email')
->setParameter('email', strtolower($username))
->getQuery();
try {
$user = $query->getSingleResult();
}
catch (NoResultException $e) {
throw new UsernameNotFoundException('User not found.', null, $e);
}
return $user;
}
public function supportsClass($class)
{
return $this->getEntityName() === $class ||
is_subclass_of($class, $this->getEntityName());
}
public function refreshUser(UserInterface $user)
{
$class = get_class($user);
if (!$this->supportsClass($class)) {
throw new UnsupportedUserException(sprintf('Instances of "%s" are not supported.', $class));
}
return $this->find($user->getId());
}
}