使用 Sonata Admin (Symfony 4) 创建用户时如何对密码进行编码？

Question

我正在尝试在创建用户时使用 Sonata Admin 对 plainPassword 字段进行编码。我没有像 issue 那样使用 FOSUserBundle。这是我的 UserAdmin class :

<?php
// src/Admin/UserAdmin.php
namespace App\Admin;

use Sonata\AdminBundle\Admin\AbstractAdmin;
use Sonata\AdminBundle\Datagrid\ListMapper;
use Sonata\AdminBundle\Datagrid\DatagridMapper;
use Sonata\AdminBundle\Form\FormMapper;
use Symfony\Component\Form\Extension\Core\Type\TextType;
use Symfony\Component\Form\Extension\Core\Type\PasswordType;
use Symfony\Component\Form\Extension\Core\Type\EmailType;
use Symfony\Bridge\Doctrine\Form\Type\EntityType;
use Sonata\AdminBundle\Form\Type\ModelType;
use App\Entity\Image;

class UserAdmin extends AbstractAdmin
{
    protected function configureFormFields(FormMapper $formMapper) {
        $formMapper
            ->add('username', TextType::class)
            ->add('email', EmailType::class)
            ->add('plainPassword', TextType::class)
            ->add('avatar', ModelType::class, [
                'class' => Image::class,
                'property' => 'image',
            ])
        ;
    }

    protected function configureDatagridFilters(DatagridMapper $datagridMapper) {
        $datagridMapper->add('username')
            ->add('avatar', null, [], EntityType::class, [
                'class'    => Image::class,
                'choice_label' => 'image',
            ])
        ;
    }

    protected function configureListFields(ListMapper $listMapper) {
        $listMapper
            ->addIdentifier('username')
        ;
    }

我的 services.yaml 包含以下内容：

admin.user:
        class: App\Admin\UserAdmin
        arguments: [~, App\Entity\User, ~]
        tags:
            - { name: sonata.admin, manager_type: orm, label: User }
        public: true

我不知道是否必须像 example 和 prePersist 方法中那样覆盖 preUpdate 方法。

我的 security.yaml 是这样开始的：

security:
    encoders:
        App\Entity\User:
            algorithm: bcrypt
    # https://symfony.com/doc/current/book/security.html#where-do-users-come-from-user-providers
    providers:
        #in_memory: { memory: ~ }
        my_db_provider:
            entity:
                class: App\Entity\User

Answer 1

密码编码通常在提交（并验证）表单后立即进行。也许您必须在用户控制器 class 的新方法和更新方法中为加密本身复制两行代码，但仅此而已。

Answer 2

我找到了一个解决方案：在 prePersist 方法中编写一些代码。我的 configureFormFields 方法略有改变，但没有任何影响：

protected function configureFormFields(FormMapper $formMapper) {
    $formMapper
        ->add('username', TextType::class)
        ->add('email', EmailType::class)
        ->add('plainPassword', RepeatedType::class, array(
            'type' => PasswordType::class,
            'first_options' => array('label' => 'Password'),
            'second_options' => array('label' => 'Password confirmation')
        ))
        ->add('avatar', ModelType::class, [
            'class' => Image::class,
            'property' => 'image',
        ])
    ;
}

最后是我的 prePersist 方法：

public function prePersist($object) { // $object is an instance of App\Entity\User as specified in services.yaml
    $plainPassword = $object->getPlainPassword();
    $container = $this->getConfigurationPool()->getContainer();
    $encoder = $container->get('security.password_encoder');
    $encoded = $encoder->encodePassword($object, $plainPassword);

    $object->setPassword($encoded);
}

使用 Sonata Admin (Symfony 4) 创建用户时如何对密码进行编码？

How to encode password while creating a user with Sonata Admin (Symfony 4)?

symfony

security

sonata-admin