在我的服务构造函数
中public function __construct(
EntityManager $entityManager,
SecurityContextInterface $securityContext)
{
$this->securityContext = $securityContext;
$this->entityManager = $entityManager;
我将entityManager和securityContext作为参数传递。 我的services.xml也在这里
<service id="acme.memberbundle.calendar_listener" class="Acme\MemberBundle\EventListener\CalendarEventListener">
<argument type="service" id="doctrine.orm.entity_manager" />
<argument type="service" id="security.context" />
但现在,我想在服务中使用容器,例如
$this->container->get('router')->generate('fos_user_profile_edit')
如何将容器传递给服务?
答案 0 :(得分:57)
如果服务扩展了ContainerAware,那很容易。
use \Symfony\Component\DependencyInjection\ContainerAware;
class YouService extends ContainerAware
{
public function someMethod()
{
$this->container->get('router')->generate('fos_user_profile_edit')
...
}
}
service.yml
your.service:
class: App\...\YouService
calls:
- [ setContainer,[ @service_container ] ]
答案 1 :(得分:47)
添加:
<argument type="service" id="service_container" />
在你的听众课上:
use Symfony\Component\DependencyInjection\ContainerInterface;
//...
public function __construct(ContainerInterface $container, ...) {
答案 2 :(得分:14)
2016年,您可以使用特质,它可以帮助您使用多个库扩展同一个类。
<?php
namespace iBasit\ToolsBundle\Utils\Lib;
use Doctrine\Bundle\DoctrineBundle\Registry;
use Symfony\Component\DependencyInjection\ContainerInterface;
trait Container
{
private $container;
public function setContainer (ContainerInterface $container)
{
$this->container = $container;
}
/**
* Shortcut to return the Doctrine Registry service.
*
* @return Registry
*
* @throws \LogicException If DoctrineBundle is not available
*/
protected function getDoctrine()
{
if (!$this->container->has('doctrine')) {
throw new \LogicException('The DoctrineBundle is not registered in your application.');
}
return $this->container->get('doctrine');
}
/**
* Get a user from the Security Token Storage.
*
* @return mixed
*
* @throws \LogicException If SecurityBundle is not available
*
* @see TokenInterface::getUser()
*/
protected function getUser()
{
if (!$this->container->has('security.token_storage')) {
throw new \LogicException('The SecurityBundle is not registered in your application.');
}
if (null === $token = $this->container->get('security.token_storage')->getToken()) {
return;
}
if (!is_object($user = $token->getUser())) {
// e.g. anonymous authentication
return;
}
return $user;
}
/**
* Returns true if the service id is defined.
*
* @param string $id The service id
*
* @return bool true if the service id is defined, false otherwise
*/
protected function has ($id)
{
return $this->container->has($id);
}
/**
* Gets a container service by its id.
*
* @param string $id The service id
*
* @return object The service
*/
protected function get ($id)
{
if ('request' === $id)
{
@trigger_error('The "request" service is deprecated and will be removed in 3.0. Add a typehint for Symfony\\Component\\HttpFoundation\\Request to your controller parameters to retrieve the request instead.', E_USER_DEPRECATED);
}
return $this->container->get($id);
}
/**
* Gets a container configuration parameter by its name.
*
* @param string $name The parameter name
*
* @return mixed
*/
protected function getParameter ($name)
{
return $this->container->getParameter($name);
}
}
你的对象,这将是服务。
namespace AppBundle\Utils;
use iBasit\ToolsBundle\Utils\Lib\Container;
class myObject
{
use Container;
}
您的服务设置
myObject:
class: AppBundle\Utils\myObject
calls:
- [setContainer, ["@service_container"]]
在控制器中调用您的服务
$myObject = $this->get('myObject');
答案 3 :(得分:5)
如果您的所有服务都是ContainerAware
,我建议您创建一个BaseService类,其中包含与您的其他服务相关的所有常用代码。
1)创建Base\BaseService.php
类:
<?php
namespace Fuz\GenyBundle\Base;
use Symfony\Component\DependencyInjection\ContainerAware;
abstract class BaseService extends ContainerAware
{
}
2)在services.yml
parameters:
// ...
geny.base.class: Fuz\GenyBundle\Base\BaseService
services:
// ...
geny.base:
class: %geny.base.class%
abstract: true
calls:
- [setContainer, [@service_container]]
3)现在,在您的其他服务中,扩展您的BaseService
课程,而不是ContainerAware
:
<?php
namespace Fuz\GenyBundle\Services;
use Fuz\GenyBundle\Base\BaseService;
class Loader extends BaseService
{
// ...
}
4)最后,您可以在服务声明中使用parent
选项。
geny.loader:
class: %geny.loader.class%
parent: geny.base
我更喜欢这种方式有几个原因: