我使用的是来自Fresh的doctrineenumundle，我定义了这个类型:
namespace CommonBundle'DBAL'Types;
use Doctrine'DBAL'Platforms'AbstractPlatform;
use Fresh'Bundle'DoctrineEnumBundle'DBAL'Types'AbstractEnumType;
class AdminRoleType extends AbstractEnumType
{
const ROLE_ADMIN = "ROLE_ADMIN";
const ROLE_GERENTE = "ROLE_GERENTE";
const ROLE_FISCAL = "ROLE_FISCAL";
const ROLE_CENTRO_HIPICO = "ROLE_CENTRO_HIPICO";
const ROLE_OPERADOR = "ROLE_OPERADOR";
/**
* @var string Name of this type
*/
protected $name = 'AdminRoleType';
/**
* @var array Readable choices
* @static
*/
protected static $choices = [
self::ROLE_ADMIN => 'Administrador',
self::ROLE_GERENTE => 'Gerente',
self::ROLE_FISCAL => 'Fiscal',
self::ROLE_CENTRO_HIPICO => 'Centro Hípico',
self::ROLE_OPERADOR => 'Operadora'
];
}
然后在我的表单类型中，我有一个像这样的字段:
....
->add('roleType', 'choice', array(
'choices' => AdminRoleType::getChoices(),
'required' => true,
'label' => "User Type",
'trim' => true
))
....
在一个视图中，我只需要渲染ROLE_ADMIN和ROLE_OPERADOR，但在另一个视图中，我需要渲染所有的，我怎么做?
我找到了我的问题的解决方案，只需要添加尽可能多的方法，因为我想返回我需要的，见下面的例子:
class RoleType extends AbstractEnumType
{
const ROLE_ADMIN = "ROLE_ADMIN";
const ROLE_GERENTE = "ROLE_GERENTE";
const ROLE_FISCAL = "ROLE_FISCAL";
const ROLE_CENTRO_HIPICO = "ROLE_CENTRO_HIPICO";
const ROLE_OPERADOR = "ROLE_OPERADOR";
/**
* @var string Name of this type
*/
protected $name = 'RoleType';
/**
* @var array Readable choices
* @static
*/
protected static $choices = [
self::ROLE_ADMIN => 'Administrador',
self::ROLE_GERENTE => 'Gerente',
self::ROLE_FISCAL => 'Fiscal',
self::ROLE_CENTRO_HIPICO => 'Centro Hípico',
self::ROLE_OPERADOR => 'Operadora'
];
public static function getFrontChoices()
{
return [
self::ROLE_CENTRO_HIPICO => 'Centro Hípico',
self::ROLE_OPERADOR => 'Operadora'
];
}
}
如果我只想显示ROLE_CENTRO_HIPICO和ROLE_OPERADOR而不是调用getChoices()，我调用getFrontChoices():
....
->add('roleType', 'choice', array(
'choices' => AdminRoleType::getFrontChoices(),
'required' => true,
'label' => "User Type",
'trim' => true
))
....