Maison > Article > développement back-end > Laravel中类中的构造函数传参是可以自动new一个传递进去的吗?
这是Laravel
中Auth\Guard
的构造函数:
<code> /** * Create a new authentication guard. * * @param \Illuminate\Auth\UserProviderInterface $provider * @param \Illuminate\Session\Store $session * @param \Symfony\Component\HttpFoundation\Request $request * @return void */ public function __construct(UserProviderInterface $provider, SessionStore $session, Request $request = null) { $this->session = $session; $this->request = $request; $this->provider = $provider; } </code>
其中传入了参数SessionStore $session
但是session
的构造函数是这样的:
<code>public function __construct($name, SessionHandlerInterface $handler, $id = null) { $this->setId($id); $this->name = $name; $this->handler = $handler; $this->metaBag = new MetadataBag; } </code>
这里是有参数的,为什么Guard
的构造函数可以自动生成session
?
是php
原生提供的还是Laravel
提供的?
这是Laravel
中Auth\Guard
的构造函数:
<code> /** * Create a new authentication guard. * * @param \Illuminate\Auth\UserProviderInterface $provider * @param \Illuminate\Session\Store $session * @param \Symfony\Component\HttpFoundation\Request $request * @return void */ public function __construct(UserProviderInterface $provider, SessionStore $session, Request $request = null) { $this->session = $session; $this->request = $request; $this->provider = $provider; } </code>
其中传入了参数SessionStore $session
但是session
的构造函数是这样的:
<code>public function __construct($name, SessionHandlerInterface $handler, $id = null) { $this->setId($id); $this->name = $name; $this->handler = $handler; $this->metaBag = new MetadataBag; } </code>
这里是有参数的,为什么Guard
的构造函数可以自动生成session
?
是php
原生提供的还是Laravel
提供的?
https://github.com/laravel/framework/blob/4.2/src/Illuminate/Auth/AuthManager.php#L51
<code>/** * Create an instance of the Eloquent driver. * * @return \Illuminate\Auth\Guard */ public function createEloquentDriver() { $provider = $this->createEloquentProvider(); return new Guard($provider, $this->app['session.store']); } </code>