我正在制作一个用于创建用户的表单,我想在创建用户时为用户提供一个或多个角色。
如何获取 security.yml
中定义的角色列表?
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,
));
}
和用户.php
public function getRolesNames(){
return array(
"ADMIN" => "Administrateur",
"ANIMATOR" => "Animateur",
"USER" => "Utilisateur",
);
}
当然,此解决方案不起作用,因为roles
在数据库中定义为位图,因此无法创建choices
列表。
提前谢谢。
security.role_hierarchy.roles
容器参数将角色层次结构保存为数组。您可以概括它以获取定义的角色列表。
您可以从此方法获取可访问角色的列表:
Symfony'Component'Security'Core'Role'RoleHierarchy::getReachableRoles(array $roles)
它似乎返回可从数组$roles
参数中的角色访问的所有角色。它是Symfony的内部服务,其ID是security.role_hierarchy
的,不是公开的(你必须明确地将其作为参数传递,它不能从服务容器中获取)。
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。
我以 security.yml 中的以下角色为例:
ROLE_SUPER_ADMIN: ROLE_ADMIN
ROLE_ADMIN: ROLE_USER
ROLE_TEST: ROLE_USER
您可以通过以下方式获取此角色:
$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',
),
)
您可以为此创建一个服务并注入"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();
在Symfony 3.3中,你可以创建一个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:
$builder->add('roles', RolesType::class,array(
'label' => 'Roles'
));
重要的是还必须包含每个角色,例如:ROLE_ADMIN: [ROLE_ADMIN, ROLE_USER]
如果需要获取特定角色的所有继承角色:
use Symfony'Component'Security'Core'Role'Role;
use Symfony'Component'Security'Core'Role'RoleHierarchy;
private function getRoles($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(); }, $roles);
}
然后调用此函子:$this->getRoles('ROLE_ADMIN');
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);
这不是你想要的,但它使你的示例有效:
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,
));
}
但是关于从实体获取角色,也许您可以使用实体存储库的内容来查询数据库。
下面是一个使用 queryBuilder 获取所需内容到实体存储库中的很好示例:
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/
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...我已经使用了其他答案中的一些建议,但我仍然觉得这不是最好的解决方案!
//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'];
}
到目前为止,这是我发现从配置文件中获取或设置我想要的内容的最佳方法。
勇气