是否可以将路由参数(或路由段)注入控制器构造函数？
你找到一些代码来澄清我的问题.
class TestController{
protected $_param;
public function __construct($paramFromRoute)
{
$this->param = $paramFromRoute;
}
public function testAction()
{
return "Hello ".$this->_param;
}
}
----------------------------------------------------
App::bind('TestController',function($app,$paramFromRoute){
$controller = new TestController($paramFromRoute);
return $controller;
});
----------------------------------------------------
// here should be some magic
Route::get('foo/{bar}','TestController');
不可能注射它们,但您可以通过以下方式访问所有这些：
class TestController{
protected $_param;
public function __construct()
{
$id = Route::current()->getParameter('id');
}
}
