[英]Symfony2 FOS User password check callback constraint
我在創建的多站點平台上使用FOS用戶捆綁包。 我需要根據用戶角色設置密碼檢查的自定義約束(對管理員用戶進行更強的正則表達式檢查)。
我該如何以及在何處安全地執行此操作? 斷言/約束似乎在用戶實體上不可用,例如此答案 ,不允許對角色進行不同的檢查。
感謝您的回答,
薩科
您可以創建一個自定義驗證約束並將其附加到類而不是屬性。
ACME \\ UserBundle \\驗證\\約束\\ PasswordForRoleRegex
namespace Acme\UserBundle\Validator\Constraints;
/**
* @Annotation
* @Target({"CLASS", "ANNOTATION"})
*/
class PasswordForRoleRegex extends Constraint
{
/**
* {@inheritdoc}
*/
public function getTargets()
{
return self::CLASS_CONSTRAINT;
}
}
ACME \\ UserBundle \\驗證\\約束\\ PasswordForRoleRegexValidator
namespace Acme\UserBundle\Validator\Constraints;
class PasswordForRoleRegexValidator extends ConstraintValidator
{
const REGEX_SUPER_ADMIN = '/..../';
// or an actual message if you don't use translations
const MESSAGE_SUPER_ADMIN = 'acme.user.password.regex.super_admin';
const REGEX_ADMIN = '/..../';
const MESSAGE_ADMIN = 'acme.user.password.regex.admin';
/// and so on
const REGEX_NORMAL_USER = '/..../';
const MESSAGE_NORMAL_USER = 'acme.user.password.regex.normal_user';
public function validate($user, Constraint $constraint)
{
if (!$constraint instanceof PasswordForRoleRegex) {
throw new UnexpectedTypeException($constraint, PasswordForRoleRegex::class);
}
if (!$user instanceof UserInterface) {
throw new UnexpectedTypeException($user, UserInterface::class);
}
if (null === $password = $user->getPlainPassword()) {
return;
}
if (preg_match($this->getPasswordRegexForUserRole($user), $password) {
return;
}
$this->context->buildViolation($this->getErrorMessageForUserRole($user))
->atPath('plainPassword')
->addViolation();
}
/**
* @param UserInterface $user
* @return string
*/
private function getPasswordRegexForUserRole(UserInterface $user)
{
if ($user->hasRole('ROLE_SUPER_ADMIN')) {
return self::REGEX_SUPER_ADMIN;
}
if ($user->hasRole('ROLE_ADMIN')) {
return self::REGEX_ADMIN;
}
// and so on
return self::REGEX_NORMAL_USER;
}
/**
* @param UserInterface $user
* @return string
*/
private function getErrorMessageForUserRole(UserInterface $user)
{
if ($user->hasRole('ROLE_SUPER_ADMIN')) {
return self::MESSAGE_SUPER_ADMIN;
}
if ($user->hasRole('ROLE_ADMIN')) {
return self::MESSAGE_ADMIN;
}
// and so on
return self::MESSAGE_NORMAL_USER;
}
}
然后可以將其用於驗證中,例如...
ACME \\ UserBundle \\型號\\用戶
namespace Acme\UserBundle\Model;
use Acme\UserBundle\Validator\Constraints\PasswordForRoleRegex;
use FOS\UserBundle\Model\User as BaseUser;
/**
* @PasswordForRoleRegex(groups={"Registration", "ChangePassword", ....})
*/
class User extends BaseUser
{
//...
}
要么..
@ AcmeUserBundle /資源/配置/ validation.yml
Acme\UserBundle\Model\User:
properties:
Acme\UserBundle\Validator\Constraints\PasswordForRoleRegex:
groups: ["Registration", "ChangePassword", ....]
或我不願意做的XML。
我很確定這是可行的,但尚未經過測試,因此可能不是100%。
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