Symfony：当用户具有 x 角色时装饰服务

Question

我一直在寻找这个问题的答案，但似乎在任何地方都找不到。

我目前已经定义了一个装饰器服务来装饰翻译器服务。但是，我只想在用户具有特定角色时装饰翻译服务。

服务.yml

services:
    app.my_translator_decorator:
        class: MyBundle\MyTranslatorDecorator
        decorates: translator
        arguments: ['@app.my_translator_decorator.inner']
        public:    false

MyTranslatorDecorator.php

class MyTranslatorDecorator {

    /**
     * @var TranslatorInterface
     */
    private $translator;

    /**
     * @param TranslatorInterface $translator
     */
    public function __construct(TranslatorInterface $translator)
    {
        $this->translator = $translator;
    }

    // more code...

}

Answer 1

容器在运行时之前“编译”。您不能根据上下文装饰服务，它总是会被装饰。

但是，在您的装饰器中，您可以添加一个保护子句，以便在没有必要时不执行您的自定义代码。

服务定义：

services:
    app.my_translator_decorator:
        class:     AppBundle\MyTranslatorDecorator
        decorates: translator
        arguments: ['@app.my_translator_decorator.inner', '@security.authorization_checker']
        public:    false

装饰器：

<?php

namespace AppBundle;

use Symfony\Component\Security\Core\Authorization\AuthorizationCheckerInterface;
use Symfony\Component\Translation\TranslatorInterface;

class MyTranslatorDecorator implements TranslatorInterface
{
    private $translator;
    private $authorizationChecker;

    public function __construct(TranslatorInterface $translator, AuthorizationCheckerInterface $authorizationChecker)
    {
        $this->translator = $translator;
        $this->authorizationChecker = $authorizationChecker;
    }

    public function trans($id, array $parameters = [], $domain = null, $locale = null)
    {
        if (!$this->authorizationChecker->isGranted('ROLE_ADMIN')) {
            return $this->translator->trans($id, $parameters, $domain, $locale);
        }

        // return custom translation here
    }

    // implement other methods
}