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');
- }
- }
