Question

I have created a simple class inside my bundle in Symfony 2:

class MyTest {
    public function myFunction() {
        $logger = $this->get( logger );
        $logger->err( testing out );
    }
}

如何进入集装箱?

Answer 1

你们需要注入服务集装箱。你的班子将研究:

use SymfonyComponentDependencyInjectionContainerInterface;

class MyTest
{
    private $container;

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

    public function myFunction()
    {
        $logger = $this->container->get( logger );
        $logger->err( testing out );
    }
}

然后在控制器或集装箱内意识到:

$myinstance = new MyTest($this->container);

如果需要更多的解释:。

Answer 2

在大多数情况下,对整个集装箱进行拦截是坏的。单独拒绝所需服务。

namespace Vendor;

use SymfonyComponentHttpKernelLogLoggerInterface;

class MyTest 
{
    private $logger;

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

    public function myFunction() 
    {
        $logger->err( testing out );
    }
}

登记<代码>服务:yml:

services:
    my_test:
        class: VendorMyTest
        arguments: [@logger]

Answer 3

为什么不添加<条码>@logger 只是服务? e.g

arguments: [@logger]

如果你想添加集装箱(WHICH IS NOT RecommendationsED),那么你可以在<代码>@service_container上添加<条码>。

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