在services.yml中定义存储库参数

Question

我试图将存储库注入服务中，为此，我以这种方式定义services.yml文件：

parameters:
    user.repository.class: App\UserBundle\Repository\UserRepository
    user_repository.factory_argument: App\UserBundle\Usuario
    user_repository.factory_argument2: Doctrine\ORM\Mapping\ClassMetadata
    user.service.class: App\UserBundle\Services\UserServiceImpl
services:
    app_user.registration.form.type:
        class: App\UserBundle\Form\Type\RegistrationFormType
        arguments: [%fos_user.model.user.class%]
        tags:
            - { name: form.type, alias: app_user_registration }
    login_listener:
        class: 'App\UserBundle\Listener\LoginListener'
        arguments: ['@security.context', '@doctrine', '@service_container']
        tags:
            - { name: 'kernel.event_listener', event: 'security.interactive_login' }
    app.user.repository:
        class: "%user.repository.class%"
        factory-service: doctrine.orm.entity_manager
        factory-method: getRepository
        arguments: [%user_repository.factory_argument%, %user_repository.factory_argument2%]
    user.service:
        class: "%user.service.class%"
        arguments: [ @app.user.repository ]

问题出在app.user.respository参数定义中： arguments: [%user_repository.factory_argument%, %user_repository.factory_argument2%]

我得到错误：

PHP可捕获的致命错误：传递给Doctrine \\ ORM \\ EntityRepository :: __ construct（）的参数2必须是Doctrine \\ ORM \\ Mapping \\ ClassMetadata的实例，给定字符串

我关注的是这篇帖子 ，它实际上以不同的方式定义了这些参数，但这对我不起作用，也许它是针对先前版本的symfony或教义完成的。

编辑

如注释中所述，“ factory_service”中有一个错字，但我仍然会绕过相同的错误，我不知道是否需要配置参数接口或实现这些接口的类

parameters:
    user.repository.class: App\UserBundle\Repository\UserRepository
    user_repository.factory_argument: App\UserBundle\Entity\Usuario
    user.service.class: App\UserBundle\Services\UserServiceImpl
    user.service.argument: App\UserBundle\Repository\UserRepository
services:
    app_user.registration.form.type:
        class: App\UserBundle\Form\Type\RegistrationFormType
        arguments: [%fos_user.model.user.class%]
        tags:
            - { name: form.type, alias: app_user_registration }
    login_listener:
        class: 'App\UserBundle\Listener\LoginListener'
        arguments: ['@security.context', '@doctrine', '@service_container']
        tags:
            - { name: 'kernel.event_listener', event: 'security.interactive_login' }
    app.user.repository:
        public: true
        class: "%user.repository.class%"
        factory_service: doctrine.orm.entity_manager
        factory_method: getRepository
        arguments: [%user_repository.factory_argument%] #, %user_repository.factory_argument2%]
    user.service:
        class: "%user.service.class%"
        arguments: [%user.service.argument%]

这些是我的课程：

UserRepositoryInterface：

<?php
namespace App\UserBundle\Repository;

use Doctrine\Common\Persistence\ObjectRepository;

interface UserRepositoryInterface extends ObjectRepository
{
    public function fetch($id_or_email);
}

UserRepository：

<?php

namespace App\UserBundle\Repository;

use Doctrine\ORM\EntityRepository;

class UserRepository extends EntityRepository
                     implements UserRepositoryInterface
{
    public function fetch($id_or_email)
    {
        if (!$id_or_email) return;

        $user = $this->find($id_or_email);

        return $user;
    }
}

UserService（接口）

<?php

    namespace App\UserBundle\Services;

    use Guzzle\Service\ClientInterface;

    interface UserService
    {
        public function getUser($id_or_email);
    }

UserServiceImpl：

<?php

namespace App\UserBundle\Services;

use Guzzle\Service\ClientInterface;
use Doctrine\ORM\EntityRepository;
use App\UserBundle\Repository;

class UserServiceImpl implements UserService {

protected $userRepository;

public function __construct(UserRepositoryInterface $fooRepository) //
{
    $this->userRepository = $fooRepository;
}

public function getUser($em, $id_or_email=null)
{
    error_log("#### getUSer {$id_or_email} ####");

    if (!$id_or_email) return;

    return $this->userRepository->fetch($id_or_email);
}

}

Answer 1

正如我在评论中提到的那样，您需要将第二个参数放在工厂中。 这是一个工作示例：

cerad_game__entity_manager__doctrine:
    alias: doctrine.orm.games_entity_manager

cerad_game__team_repository__doctrine:
    class:  Cerad\Bundle\GameBundle\Doctrine\EntityRepository\TeamRepository
    factory_service: 'cerad_game__entity_manager__doctrine'
    factory_method:  'getRepository'
    arguments:  
        - 'Cerad\Bundle\GameBundle\Doctrine\Entity\Team'

您的服务应如下所示：

app.user.repository:
    class: App\UserBundle\Repository\UserRepository
    factory_service: doctrine.orm.entity_manager
    factory_method: getRepository
    arguments:
        - App\UserBundle\Usuario