如何在中间件 Laravel 中获取请求的控制器和动作的名称

声明:本页面是StackOverFlow热门问题的中英对照翻译,遵循CC BY-SA 4.0协议,如果您需要使用它,必须同样遵循CC BY-SA许可,注明原文地址和作者信息,同时你必须将它归于原作者(不是我):StackOverFlow 原文地址: http://stackoverflow.com/questions/30442746/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me): StackOverFlow

提示:将鼠标放在中文语句上可以显示对应的英文。显示中英文
时间:2020-09-14 11:37:54  来源:igfitidea点击:

How to get name of requested controller and action in middleware Laravel

phplaravellaravel-5

提问by Deejay

I am new to Laravel, and i want to get name of requested controller and action in beforefilter middelware.

我是 Laravel 的新手,我想在 beforefilter 中间件中获取请求的控制器和操作的名称。

Thanks, DJ

谢谢,DJ

回答by Limon Monte

Laravel 5.6:

Laravel 5.6:

class_basename(Route::current()->controller);

Laravel 5.5 and lower:

Laravel 5.5 及更低版本:

You can retrieve the current action name with Route::currentRouteAction(). Unfortunately, this method will return a fully namespaced class name. So you will get something like:

您可以使用 检索当前操作名称Route::currentRouteAction()。不幸的是,此方法将返回一个完全命名空间的类名。所以你会得到类似的东西:

App\Http\Controllers\FooBarController@method

Then just separate method name and controller name:

然后只需将方法名称和控制器名称分开:

$currentAction = \Route::currentRouteAction();
list($controller, $method) = explode('@', $currentAction);
// $controller now is "App\Http\Controllers\FooBarController"

$controller = preg_replace('/.*\\/', '', $controller);
// $controller now is "FooBarController"