[英]How to inject a repository into a service in Symfony?
我需要将两个对象注入ImageService
。 其中之一是Repository/ImageRepository
的实例,我得到这样的:
$image_repository = $container->get('doctrine.odm.mongodb')
->getRepository('MycompanyMainBundle:Image');
那么我如何在我的 services.yml 中声明呢? 这是服务:
namespace Mycompany\MainBundle\Service\Image;
use Doctrine\ODM\MongoDB\DocumentRepository;
class ImageManager {
private $manipulator;
private $repository;
public function __construct(ImageManipulatorInterface $manipulator, DocumentRepository $repository) {
$this->manipulator = $manipulator;
$this->repository = $repository;
}
public function findAll() {
return $this->repository->findAll();
}
public function createThumbnail(ImageInterface $image) {
return $this->manipulator->resize($image->source(), 300, 200);
}
}
对于像我这样来自谷歌的人来说,这是一个清理过的解决方案:
更新:这是 Symfony 2.6(及更高版本)解决方案:
services:
myrepository:
class: Doctrine\ORM\EntityRepository
factory: ["@doctrine.orm.entity_manager", getRepository]
arguments:
- MyBundle\Entity\MyClass
myservice:
class: MyBundle\Service\MyService
arguments:
- "@myrepository"
已弃用的解决方案(Symfony 2.5 及更低版本):
services:
myrepository:
class: Doctrine\ORM\EntityRepository
factory_service: doctrine.orm.entity_manager
factory_method: getRepository
arguments:
- MyBundle\Entity\MyClass
myservice:
class: MyBundle\Service\MyService
arguments:
- "@myrepository"
我找到了这个链接,这对我有用:
parameters:
image_repository.class: Mycompany\MainBundle\Repository\ImageRepository
image_repository.factory_argument: 'MycompanyMainBundle:Image'
image_manager.class: Mycompany\MainBundle\Service\Image\ImageManager
image_manipulator.class: Mycompany\MainBundle\Service\Image\ImageManipulator
services:
image_manager:
class: %image_manager.class%
arguments:
- @image_manipulator
- @image_repository
image_repository:
class: %image_repository.class%
factory_service: doctrine.odm.mongodb
factory_method: getRepository
arguments:
- %image_repository.factory_argument%
image_manipulator:
class: %image_manipulator.class%
如果不想将每个存储库定义为服务,从2.4
版开始,您可以执行以下操作( default
为实体管理器的名称):
@=service('doctrine.orm.default_entity_manager').getRepository('MycompanyMainBundle:Image')
查看我的文章How to use Repository with Doctrine as Service in Symfony以获得更一般的描述。
对于您的代码,您需要做的就是使用组合而不是继承- SOLID 模式之一。
<?php
namespace MycompanyMainBundle\Repository;
use Doctrine\ORM\EntityManagerInterface;
use MycompanyMainBundle\Entity\Image;
class ImageRepository
{
private $repository;
public function __construct(EntityManagerInterface $entityManager)
{
$this->repository = $entityManager->getRepository(Image::class);
}
// add desired methods here
public function findAll()
{
return $this->repository->findAll();
}
}
# app/config/services.yml
services:
_defaults:
autowire: true
MycompanyMainBundle\:
resource: ../../src/MycompanyMainBundle
use MycompanyMainBundle\Repository\ImageRepository;
class ImageService
{
public function __construct(ImageRepository $imageRepository)
{
$this->imageRepository = $imageRepository;
}
}
就我而言,基于@Tomáš Votruba 的回答和这个问题,我提出了以下方法:
创建一个通用的适配器类:
namespace AppBundle\\Services; use Doctrine\\ORM\\EntityManagerInterface; class RepositoryServiceAdapter { private $repository=null; /** * @param EntityManagerInterface the Doctrine entity Manager * @param String $entityName The name of the entity that we will retrieve the repository */ public function __construct(EntityManagerInterface $entityManager,$entityName) { $this->repository=$entityManager->getRepository($entityName) } public function __call($name,$arguments) { if(empty($arrguments)){ //No arguments has been passed $this->repository->$name(); } else { //@todo: figure out how to pass the parameters $this->repository->$name(...$argument); } } }
然后foreach实体定义一个服务,例如在我的例子中定义一个(我用php来定义symfony服务):
$container->register('ellakcy.db.contact_email',AppBundle\\Services\\Adapters\\RepositoryServiceAdapter::class) ->serArguments([new Reference('doctrine'),AppBundle\\Entity\\ContactEmail::class]);
与上述相同的步骤 1
扩展RepositoryServiceAdapter
类,例如:
namespace AppBundle\\Service\\Adapters; use Doctrine\\ORM\\EntityManagerInterface; use AppBundle\\Entity\\ContactEmail; class ContactEmailRepositoryServiceAdapter extends RepositoryServiceAdapter { public function __construct(EntityManagerInterface $entityManager) { parent::__construct($entityManager,ContactEmail::class); } }
注册服务:
$container->register('ellakcy.db.contact_email',AppBundle\\Services\\Adapters\\RepositoryServiceAdapter::class) ->serArguments([new Reference('doctrine')]);
如果您有一个很好的可测试方法来测试您的数据库行为,它也可以帮助您进行模拟,以防您想对您的服务进行单元测试,而无需过多担心如何去做。 例如,假设我们有以下服务:
//Namespace definitions etc etc
class MyDummyService
{
public function __construct(RepositoryServiceAdapter $adapter)
{
//Do stuff
}
}
并且 RepositoryServiceAdapter 适应以下存储库:
//Namespace definitions etc etc
class SomeRepository extends \Doctrine\ORM\EntityRepository
{
public function search($params)
{
//Search Logic
}
}
因此,您可以通过模拟非继承方法中的RepositoryServiceAdapter
或继承方法中的ContactEmailRepositoryServiceAdapter
来轻松模拟/硬编码/模拟SomeRepository
定义的方法search
的行为。
或者,您可以定义以下工厂:
namespace AppBundle\ServiceFactories;
use Doctrine\ORM\EntityManagerInterface;
class RepositoryFactory
{
/**
* @param EntityManagerInterface $entityManager The doctrine entity Manager
* @param String $entityName The name of the entity
* @return Class
*/
public static function repositoryAsAService(EntityManagerInterface $entityManager,$entityName)
{
return $entityManager->getRepository($entityName);
}
}
然后通过执行以下操作切换到 php 服务注释:
将它放到文件./app/config/services.php
(为symfony1.2 V3.4, .
假设你ptoject根)
use Symfony\Component\DependencyInjection\Definition;
use Symfony\Component\DependencyInjection\Reference;
$definition = new Definition();
$definition->setAutowired(true)->setAutoconfigured(true)->setPublic(false);
// $this is a reference to the current loader
$this->registerClasses($definition, 'AppBundle\\', '../../src/AppBundle/*', '../../src/AppBundle/{Entity,Repository,Tests,Interfaces,Services/Adapters/RepositoryServiceAdapter.php}');
$definition->addTag('controller.service_arguments');
$this->registerClasses($definition, 'AppBundle\\Controller\\', '../../src/AppBundle/Controller/*');
并更改./app/config/config.yml
( .
假定您的 ptoject 的根)
imports:
- { resource: parameters.yml }
- { resource: security.yml }
#Replace services.yml to services.php
- { resource: services.php }
#Other Configuration
然后,您可以按如下方式设置服务(从我的示例中使用,其中我使用了一个名为Item
的虚拟实体):
$container->register(ItemRepository::class,ItemRepository::class)
->setFactory([new Reference(RepositoryFactory::class),'repositoryAsAService'])
->setArguments(['$entityManager'=>new Reference('doctrine.orm.entity_manager'),'$entityName'=>Item::class]);
同样作为一个通用提示,切换到php
服务注释可以让您无障碍地进行更高级的服务配置。 对于代码片段,请使用我使用factory
方法制作的特殊存储库。
对于 Symfony 5,它真的很简单,不需要 services.yml 来注入依赖:
private $em; public function __construct(EntityManagerInterface $em) { $this->em = $em; }
$this->em->getRepository(ClassName::class)
通过将 ClassName 替换为您的实体名称。
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