php - Symfony2：在 FormBuilder 中获取用户角色列表

Question

我正在制作一个用于创建用户的表单，并且我想在创建用户时为他分配一个或多个角色。

如何获取定义的角色列表security.yml？

这是我目前的表单构建器：

public function buildForm(FormBuilder $builder, array $options)
{
    parent::buildForm($builder, $options);

    // add your custom fields
    $user = new User();
    $builder->add('regionUser');
    $builder->add('roles' ,'choice' ,array('choices' => $user->getRolesNames(),
            'required'  => true,
    ));

}

并在 User.php

public function getRolesNames(){
    return array(
        "ADMIN" => "Administrateur",
        "ANIMATOR" => "Animateur",
        "USER" => "Utilisateur",        
    );
}

当然，这个解决方案是行不通的，因为roles在数据库中定义为位图，所以choices无法创建列表。

提前致谢。

score 23 · Accepted Answer

security.role_hierarchy.rolescontainer 参数将角色层次结构保存为一个数组。您可以对其进行概括以获取定义的角色列表。

score 14 · Accepted Answer

You can get a list of reachable roles from this method:

Symfony\Component\Security\Core\Role\RoleHierarchy::getReachableRoles(array $roles)

It seems to return all roles reachable from roles in array $roles parameter. It's an internal service of Symfony, whose ID is security.role_hierarchy and is not public (you must explicitely pass it as parameters, it's not acessible from Service Container).

score 9 · Accepted Answer

您可以为此提供服务并注入“security.role_hierarchy.roles”参数。

服务定义：

acme.user.roles:
   class: Acme\DemoBundle\Model\RolesHelper
   arguments: ['%security.role_hierarchy.roles%']

服务等级：

class RolesHelper
{
    private $rolesHierarchy;

    private $roles;

    public function __construct($rolesHierarchy)
    {
        $this->rolesHierarchy = $rolesHierarchy;
    }

    public function getRoles()
    {
        if($this->roles) {
            return $this->roles;
        }

        $roles = array();
        array_walk_recursive($this->rolesHierarchy, function($val) use (&$roles) {
            $roles[] = $val;
        });

        return $this->roles = array_unique($roles);
    }
}

您可以像这样在控制器中获取角色：

$roles = $this->get('acme.user.roles')->getRoles();

score 9 · Accepted Answer

为了正确表示您的角色，您需要递归。角色可以扩展其他角色。

我在 security.yml 中使用以下角色作为示例：

ROLE_SUPER_ADMIN: ROLE_ADMIN
ROLE_ADMIN:       ROLE_USER
ROLE_TEST:        ROLE_USER

您可以通过以下方式获得此角色：

$originalRoles = $this->getParameter('security.role_hierarchy.roles');

递归示例：

private function getRoles($originalRoles)
{
    $roles = array();

    /**
     * Get all unique roles
     */
    foreach ($originalRoles as $originalRole => $inheritedRoles) {
        foreach ($inheritedRoles as $inheritedRole) {
            $roles[$inheritedRole] = array();
        }

        $roles[$originalRole] = array();
    }

    /**
     * Get all inherited roles from the unique roles
     */
    foreach ($roles as $key => $role) {
        $roles[$key] = $this->getInheritedRoles($key, $originalRoles);
    }

    return $roles;
}

private function getInheritedRoles($role, $originalRoles, $roles = array())
{
    /**
     * If the role is not in the originalRoles array,
     * the role inherit no other roles.
     */
    if (!array_key_exists($role, $originalRoles)) {
        return $roles;
    }

    /**
     * Add all inherited roles to the roles array
     */
    foreach ($originalRoles[$role] as $inheritedRole) {
        $roles[$inheritedRole] = $inheritedRole;
    }

    /**
     * Check for each inhered role for other inherited roles
     */
    foreach ($originalRoles[$role] as $inheritedRole) {
        return $this->getInheritedRoles($inheritedRole, $originalRoles, $roles);
    }
}

输出：

array (
  'ROLE_USER' => array(),
  'ROLE_TEST' => array(
                        'ROLE_USER' => 'ROLE_USER',
  ),
  'ROLE_ADMIN' => array(
                        'ROLE_USER' => 'ROLE_USER',
  ),
  'ROLE_SUPER_ADMIN' => array(
                        'ROLE_ADMIN' => 'ROLE_ADMIN',
                        'ROLE_USER' => 'ROLE_USER',
  ),
)

score 4 · Accepted Answer

在 Symfony 3.3 中，您可以创建一个 RolesType.php，如下所示：

<?php

namespace AppBundle\Form\Type;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\OptionsResolver\OptionsResolver;
use Symfony\Component\Form\Extension\Core\Type\ChoiceType;
use Symfony\Component\Security\Core\Role\RoleHierarchyInterface;

/**
 * @author Echarbeto
 */
class RolesType extends AbstractType {

  private $roles = [];

  public function __construct(RoleHierarchyInterface $rolehierarchy) {
    $roles = array();
    array_walk_recursive($rolehierarchy, function($val) use (&$roles) {
      $roles[$val] = $val;
    });
    ksort($roles);
    $this->roles = array_unique($roles);
  }

  public function configureOptions(OptionsResolver $resolver) {
    $resolver->setDefaults(array(
        'choices' => $this->roles,
        'attr' => array(
            'class' => 'form-control',
            'aria-hidden' => 'true',
            'ref' => 'input',
            'multiple' => '',
            'tabindex' => '-1'
        ),
        'required' => true,
        'multiple' => true,
        'empty_data' => null,
        'label_attr' => array(
            'class' => 'control-label'
        )
    ));
  }

  public function getParent() {
    return ChoiceType::class;
  }

}

然后将其添加到表单中，如下所示：

$builder->add('roles', RolesType::class,array(
          'label' => 'Roles'
      ));

重要的是还必须包含每个角色，例如： ROLE_ADMIN: [ROLE_ADMIN, ROLE_USER]

score 3 · Accepted Answer

如果您需要获取某个角色的所有继承角色：

use Symfony\Component\Security\Core\Role\Role;
use Symfony\Component\Security\Core\Role\RoleHierarchy;

private function getRoles($role)
{
    $hierarchy = $this->container->getParameter('security.role_hierarchy.roles');
    $roleHierarchy = new RoleHierarchy($hierarchy);
    $roles = $roleHierarchy->getReachableRoles([new Role($role)]);
    return array_map(function(Role $role) { return $role->getRole(); }, $roles);
}

然后调用这个函数：$this->getRoles('ROLE_ADMIN');

score 1 · Accepted Answer

This is not exactly what you want but it makes your example working:

use Vendor\myBundle\Entity\User;

public function buildForm(FormBuilder $builder, array $options)
{
    parent::buildForm($builder, $options);

    // add your custom fields
    $user = new User();
    $builder->add('regionUser');
    $builder->add('roles' ,'choice' ,array('choices' => User::getRolesNames(),
            'required'  => true,
    ));
}

But regarding getting your Roles from an entity, maybe you can use entity repository stuff to query the database.

Here is a good example to get what to want using the queryBuilder into the entity repository:

public function buildForm(FormBuilder $builder, array $options)
{
    parent::buildForm($builder, $options);

    // add your custom fields
    $user = new User();
    $builder->add('regionUser');
    $builder->add('roles' ,'entity' array(
                 'class'=>'Vendor\MyBundle\Entity\User',
                 'property'=>'roles',
                 'query_builder' => function (\Vendor\MyBundle\Entity\UserRepository $repository)
                 {
                     return $repository->createQueryBuilder('s')
                            ->add('orderBy', 's.sort_order ASC');
                 }
                )
          );
}

http://inchoo.net/tools-frameworks/symfony2-entity-field-type/

score 1 · Accepted Answer

在 Symfony 2.7 中，在控制器中你必须使用 $this->getParameters() 来获取角色：

$roles = array();
foreach ($this->getParameter('security.role_hierarchy.roles') as $key => $value) {
    $roles[] = $key;

    foreach ($value as $value2) {
        $roles[] = $value2;
    }
}
$roles = array_unique($roles);

score -2 · Accepted Answer

这是我所做的：

表格类型：

use FTW\GuildBundle\Entity\User;

class UserType extends AbstractType
{

public function buildForm(FormBuilderInterface $builder, array $options)
{
    $builder
        ->add('username')
        ->add('email')
        ->add('enabled', null, array('required' => false))
        ->add('roles', 'choice', array(
        'choices' => User::getRoleNames(),
        'required' => false,'label'=>'Roles','multiple'=>true
    ))
        ->add('disableNotificationEmails', null, array('required' => false));
}

在实体中：

use Symfony\Component\Yaml\Parser; ...

static function getRoleNames()
{
    $pathToSecurity = __DIR__ . '/../../../..' . '/app/config/security.yml';
    $yaml = new Parser();
    $rolesArray = $yaml->parse(file_get_contents($pathToSecurity));
    $arrayKeys = array();
    foreach ($rolesArray['security']['role_hierarchy'] as $key => $value)
    {
        //never allow assigning super admin
        if ($key != 'ROLE_SUPER_ADMIN')
            $arrayKeys[$key] = User::convertRoleToLabel($key);
        //skip values that are arrays --- roles with multiple sub-roles
        if (!is_array($value))
            if ($value != 'ROLE_SUPER_ADMIN')
                $arrayKeys[$value] = User::convertRoleToLabel($value);
    }
    //sort for display purposes
    asort($arrayKeys);
    return $arrayKeys;
}

static private function convertRoleToLabel($role)
{
    $roleDisplay = str_replace('ROLE_', '', $role);
    $roleDisplay = str_replace('_', ' ', $roleDisplay);
    return ucwords(strtolower($roleDisplay));
}

请提供反馈...我使用了其他答案的一些建议，但我仍然觉得这不是最好的解决方案！

score -3 · Accepted Answer

//FormType
use Symfony\Component\Yaml\Parser;

function getRolesNames(){
        $pathToSecurity = /var/mydirectory/app/config/security.yml
        $yaml = new Parser();
        $rolesArray = $yaml->parse(file_get_contents($pathToSecurity ));

        return $rolesArray['security']['role_hierarchy']['ROLE_USER'];
}

这是迄今为止我发现从配置文件中获取或设置我想要的内容的最佳方式。

好勇气

php - Symfony2：在 FormBuilder 中获取用户角色列表

10 回答 10

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