我想使用SF 4.3上發(fā)布的新不承諾密碼：https://symfony.com/blog/new-in-symfony-4-3-compromised-password-validator我已經(jīng)在我的驗(yàn)證.yaml上設(shè)置了它，如下所示：App\Entity\User:    constraints:        - App\Validator\Constraints\ConstraintPassword: ~    properties:        plainPassword:            - Symfony\Component\Validator\Constraints\NotCompromisedPassword: ~它可以工作，但我想自定義錯(cuò)誤消息，例如，直接在我的約束密碼驗(yàn)證器上使用它.php：<?phpnamespace App\Validator\Constraints;use App\Entity\User;use Symfony\Component\Validator\Constraint;use Symfony\Component\Validator\Constraints\NotCompromisedPassword;use Symfony\Component\Validator\ConstraintValidator;class ConstraintPasswordValidator extends ConstraintValidator{    /**     * @param User $user     * @param Constraint $constraint     */    public function validate($user, Constraint $constraint)    {        if (strlen($user->getPlainPassword()) < 8 || strlen($user->getPlainPassword() < 35)) {            $this->context->buildViolation($constraint->lengthError)                ->addViolation();        }        // Doing something like that        $notCompromised = new NotCompromisedPassword();        $notCompromised->message = "My custom error message";       //Then, build the violation if password leaked    }}也許它需要在我的約束密碼中實(shí)例化和自定義.php？但我不知道如何<?phpnamespace App\Validator\Constraints;use Symfony\Component\Validator\Constraint;class ConstraintPassword extends Constraint{    public $lengthError = 'Erreur : La longueur du mot de passe doit être comprise entre 8 et 35 caractères';    public function validatedBy()    {        return \get_class($this).'Validator';    }    public function getTargets()    {        return self::CLASS_CONSTRAINT;    }}