将参数传递到过滤器 - Laravel 4

php laravel laravel-4

2022-08-30 19:07:07

是否可以访问筛选器中的路由参数？

例如，我想访问$agencyId参数：

Route::group(array('prefix' => 'agency'), function()
{

    # Agency Dashboard
    Route::get('{agencyId}', array('as' => 'agency', 'uses' => 'Controllers\Agency\DashboardController@getIndex'));

});

我想在我的过滤器中访问此$agencyId参数：

Route::filter('agency-auth', function()
{
    // Check if the user is logged in
    if ( ! Sentry::check())
    {
        // Store the current uri in the session
        Session::put('loginRedirect', Request::url());

        // Redirect to the login page
        return Redirect::route('signin');
    }

    // this clearly does not work..?  how do i do this?
    $agencyId = Input::get('agencyId');

    $agency = Sentry::getGroupProvider()->findById($agencyId);

    // Check if the user has access to the admin page
    if ( ! Sentry::getUser()->inGroup($agency))
    {
        // Show the insufficient permissions page
        return App::abort(403);
    }
});

仅供参考，我在我的控制器中调用此过滤器：

class AgencyController extends AuthorizedController {

    /**
     * Initializer.
     *
     * @return void
     */
    public function __construct()
    {
        // Apply the admin auth filter
        $this->beforeFilter('agency-auth');
    }
...

答案 1

Input::get只能检索或（依此类推）参数。GETPOST

要获取路由参数，您必须在过滤器中抓取对象，如下所示：Route

Route::filter('agency-auth', function($route) { ... });

并获取参数（在过滤器中）：

$route->getParameter('agencyId');

（只是为了好玩）在您的路线中

Route::get('{agencyId}', array('as' => 'agency', 'uses' => 'Controllers\Agency\DashboardController@getIndex'));

您可以在参数数组中使用，而不是在构造函数中详细说明它。'before' => 'YOUR_FILTER'

答案 2

方法名称在 Laravel 4.1 中已更改为。例如，在 RESTful 控制器中：parameter

$this->beforeFilter(function($route, $request) {
    $userId = $route->parameter('users');
});

另一种选择是通过立面检索参数，当您在路径之外时，这很方便：Route

$id = Route::input('id');