Question

我有一个用户编辑表单，我想管理分配给用户的角色。

目前我有一个多选列表，但我没办法用security.yml中定义的角色层次填充它。

我是否有某种方法可以将此信息提供给FormType类中的表单生成器？

$builder->add('roles', 'choice', array(
                'required' => true,
                'multiple' => true,
                'choices' => array(),
            ));

环顾I found我可以从控制器中的容器中获取角色：

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

我还发现我可以将它设置为在services.xml中注入FormType类的依赖项：

<parameters>
    <parameter key="security.role_heirarchy.roles">ROLE_GUEST</parameter>
</parameters>
<services>
    <service id="base.user.form.type.user_form" class="Base\UserBundle\Form\UserType" public="false">
        <tag name="form.type" />
        <call method="setRoles">
            <argument>%security.role_heirarchy.roles%</argument>
        </call>
    </service>
</services>

然而，这不起作用，似乎没有调用setRoles方法。

那我怎么能让它发挥作用呢？

Answer 1

在您的控制器中

$editForm = $this->createForm(new UserType(), $entity, array('roles' => $this->container->getParameter('security.role_hierarchy.roles')));

在UserType中：

$builder->add('roles', 'choice', array(
    'required' => true,
    'multiple' => true,
    'choices' => $this->refactorRoles($options['roles'])
))

[...]

public function getDefaultOptions()
{
    return array(
        'roles' => null
    );
}

private function refactorRoles($originRoles)
{
    $roles = array();
    $rolesAdded = array();

    // Add herited roles
    foreach ($originRoles as $roleParent => $rolesHerit) {
        $tmpRoles = array_values($rolesHerit);
        $rolesAdded = array_merge($rolesAdded, $tmpRoles);
        $roles[$roleParent] = array_combine($tmpRoles, $tmpRoles);
    }
    // Add missing superparent roles
    $rolesParent = array_keys($originRoles);
    foreach ($rolesParent as $roleParent) {
        if (!in_array($roleParent, $rolesAdded)) {
            $roles['-----'][$roleParent] = $roleParent;
        }
    }

    return $roles;
}

Answer 2

webda2l提供的带有$ options数组的解决方案不适用于Symfony 2.3。它给了我一个错误：

“角色”选项不存在。

我发现我们可以将参数传递给表单类型构造函数。

在您的控制器中：

$roles_choices = array();

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

# set roles array, displaying inherited roles between parentheses
foreach ($roles as $role => $inherited_roles)
{
    foreach ($inherited_roles as $id => $inherited_role)
    {
        if (! array_key_exists($inherited_role, $roles_choices))
        {
            $roles_choices[$inherited_role] = $inherited_role;
        }
    }

    if (! array_key_exists($role, $roles_choices))
    {
        $roles_choices[$role] = $role.' ('.
            implode(', ', $inherited_roles).')';
    }
}

# todo: set $role as the current role of the user

$form = $this->createForm(
    new UserType(array(
        # pass $roles to the constructor
        'roles' => $roles_choices,
        'role' => $role
    )), $user);

在UserType.php中：

class UserType extends AbstractType
{
    private $roles;
    private $role;

    public function __construct($options = array())
    {
        # store roles
        $this->roles = $options['roles'];
        $this->role = $options['role'];
    }
    public function buildForm(FormBuilderInterface $builder,
        array $options)
    {
        // ...
        # use roles
        $builder->add('roles', 'choice', array(
            'choices' => $this->roles,
            'data' => $this->role,
        ));
        // ...
    }
}

我从Marcello Voc接受了这个想法，谢谢他！

Answer 3

您可以创建自己的类型，然后传递服务容器，然后您可以从中检索角色层次结构。

首先，创建自己的类型：

class PermissionType extends AbstractType 
{
    private $roles;

    public function __construct(ContainerInterface $container)
    {
        $this->roles = $container->getParameter('security.role_hierarchy.roles');
    }
    public function getDefaultOptions(array $options)
    {
        return array(
              'choices' => $this->roles,
        );
    );

    public function getParent(array $options)
    {
        return 'choice';
    }

    public function getName()
    {
        return 'permission_choice';
    }
}

然后您需要将您的类型注册为服务并设置参数：

services:
      form.type.permission:
          class: MyNamespace\MyBundle\Form\Type\PermissionType
          arguments:
            - "@service_container"
          tags:
            - { name: form.type, alias: permission_choice }

然后在创建表单期间，只需添加* permission_choice * field：

public function buildForm(FormBuilder $builder, array $options)
{
    $builder->add('roles', 'permission_choice');
}

如果您只想获得没有层次结构的单个角色列表，那么您需要以某种方式平面层次结构。可能的解决方案之一是：

class PermissionType extends AbstractType 
{
    private $roles;

    public function __construct(ContainerInterface $container)
    {
        $roles = $container->getParameter('security.role_hierarchy.roles');
        $this->roles = $this->flatArray($roles);
    }

    private function flatArray(array $data)
    {
        $result = array();
        foreach ($data as $key => $value) {
            if (substr($key, 0, 4) === 'ROLE') {
                $result[$key] = $key;
            }
            if (is_array($value)) {
                $tmpresult = $this->flatArray($value);
                if (count($tmpresult) > 0) {
                    $result = array_merge($result, $tmpresult);
                }
            } else {
                $result[$value] = $value;
            }
        }
        return array_unique($result);
    }
    ...
}

Answer 4

对于symfony 2.3：

<?php

namespace Labone\Bundle\UserBundle\Form\Type;

use Symfony\Component\Form\AbstractType;
use Symfony\Component\Form\FormBuilderInterface;
use Symfony\Component\OptionsResolver\OptionsResolverInterface;
use Symfony\Component\DependencyInjection\ContainerInterface as Container;

class PermissionType extends AbstractType 
{
    private $roles;

    public function __construct(Container $container)
    {
        $roles = $container->getParameter('security.role_hierarchy.roles');
        $this->roles = $this->flatArray($roles);
    }

    public function setDefaultOptions(OptionsResolverInterface $resolver)
    {
        $resolver->setDefaults(array(
            'choices' => $this->roles
        ));
    }

    public function getParent()
    {
        return 'choice';
    }

    public function getName()
    {
        return 'permission_choice';
    }

    private function flatArray(array $data)
    {
        $result = array();
        foreach ($data as $key => $value) {
            if (substr($key, 0, 4) === 'ROLE') {
                $result[$key] = $key;
            }
            if (is_array($value)) {
                $tmpresult = $this->flatArray($value);
                if (count($tmpresult) > 0) {
                    $result = array_merge($result, $tmpresult);
                }
            } else {
                $result[$value] = $value;
            }
        }
        return array_unique($result);
    }
}

如何将完整的安全角色列表/层次结构传递给Symfony2中的FormType类？

4 个答案: