在模型/控制器之外的Symfony2 getdoctrine

问题描述

我试图在控制器外面getDoctrine（）。
我创建了这个服务：

I'm trying to getDoctrine() outside of the controller.I've created this service:

config / services.yml

config/services.yml

services:
  update_command:
    class: project\projBundle\Command\Update
    arguments: ['@doctrine.orm.entity_manager']

和我的app / config / config.yml

and in my app/config/config.yml

imports:
    - { resource: "@projectprojBundle/Resources/config/services.yml" }

所以我想使用的类：

so and the class that I want to use:

namespace project\projBundle\Command;
use Doctrine\ORM\EntityManager;

class Update {
    protected $em;
    public function __construct(EntityManager $em) {
        $this->em = $em;
    }

但每次我想这样做：（这样做吧？

but every time I want to do this: (I'm doing this right?)

$up = new Update();

我收到此错误：

Catchable Fatal Error: Argument 1 passed to ...\Update::__construct() must be an instance of Doctrine\ORM\EntityManager, none given, called in .../Update.php line 7

推荐答案

strong>

Simple solution

如果您正在实现Symfony命令（可以在cron选项卡中执行），则可以从命令访问服务容器。

If you're implementing a Symfony command (that can be executed in a cron tab), you can access the service container from the command.

<?php
namespace MyProject\MyBundle\Command;

use Symfony\Bundle\FrameworkBundle\Command\ContainerAwareCommand;
use Doctrine\ORM\EntityManager;

use Symfony\Component\Console\Input\InputInterface;
use Symfony\Component\Console\Output\OutputInterface;

class UpdateCommand extends ContainerAwareCommand
{
    protected $em;

    protected function configure()
    {
        $this->setName('myproject:mybundle:update') ;
    }

    protected function execute(InputInterface $input, OutputInterface $output)
    {
        $this->em = $this->getContainer()->get('doctrine.orm.entity_manager');
    }
}

这样，您就可以从命令中获取实体管理器并且不需要将此类声明为服务。因此，您可以删除在 services.yml 文件中添加的配置。

That way, you get the entity manager from a command and don't need to declare this class as a service. You can therefore remove the configuration you added in the services.yml file.

其他解决方案更清洁）

此解决方案可以更好地分离问题，因此可以轻松地在Symfony应用程序的其他部分进行单元测试和重用（不仅仅是一个命令）

This solution allows better separation of concerns and can therefore be easily unit tested and reused in other parts of your Symfony application (not only as a command).

将update命令的所有逻辑部分移动到您将声明为服务的专用类：

Move all the logic part of your "update" command to a dedicated class that you will declare as a service:

<?php
namespace MyProject\MyBundle\Service;

use Doctrine\ORM\EntityManager;

class MyUpdater
{
    protected $em;

    public function __construct($em)
    {
        $this->em = $em;
    }

    public function runUpdate()
    {
        // All your logic code here
    }
}

在您的 services.yml 文件中声明为服务：

Declare it as a service in your services.yml file:

services:
    myproject.mybundle.myupdater:
        class: MyProject\MyBundle\Service\MyUpdater
        arguments: ['@doctrine.orm.entity_manager']

只需拨打您的服务您的命令：

Simply call your service from your command :

<?php
namespace MyProject\MyBundle\Command;

use Symfony\Bundle\FrameworkBundle\Command\ContainerAwareCommand;

use Symfony\Component\Console\Input\InputInterface;
use Symfony\Component\Console\Output\OutputInterface;

class UpdateCommand extends ContainerAwareCommand
{
    protected function configure()
    {
        $this->setName('myproject:mybundle:update') ;
    }

    protected function execute(InputInterface $input, OutputInterface $output)
    {
        $myUpdater = $this->getContainer()->get('myproject.mybundle.myupdater');
        $myUpdater->runUpdate();
    }
}

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