Symfony 4,如何从自定义服务类获取services.yaml参数? (在构造函数中不注入ParameterBagInterface)

时间:2018-10-17 22:47:31

标签: php yaml symfony4

从定制服务类中,可以通过将服务ParameterBagInterface注入服务类的构造函数中来轻松获得在services.yaml文件中定义的参数,如下所示:

#from the services.yaml
...
services:
...
    App\Utils\myCustomService:
        arguments:
            $faa: 'some value'
...

//from my custom service Class
...
public function __construct(ParameterBagInterface $faa){
    dump($faa); //print 'some value';
}

但是由于某种原因,我想在服务的构造函数中使用“正常”参数,如下所示:

//from my custom service Class
...
public function __construct($aNormalParameter){
    // do some stuff with $aNormalParameter
    // but I also want to work with the $faa value located in the services.yaml file, how can I do that ? 
}

//from a custom Controller
...
public function home(){
    $myCustomService = new App\Utils\myCustomService("blabla");
}

所以我的问题是,如何在不将ParameterBagInterface服务注入构造函数的情况下访问在services.yaml文件中定义的参数?

谢谢:)

0 个答案:

没有答案