将Laravel包含在自定义PHP脚本中并路由到控制器

Question

Let's say there's a web-based PHP project which uses an independent framework and also contains a folder with Laravel 4.2:

project/
    laravel/ (Laravel install)
        app/
            routes.php
        public/
            index.php
    index.php
    test.php
    .htaccess

The .htaccess file rewrites every query to the independent framework's index.php, unless a PHP script is requested (e.g., project/test.php).

In this case, test.php contains require laravel/public/index.php to include Laravel's index.php script. The URI to access this is http://example.com/test.php.

How can I make use of Laravel's routing to pick up on this script? I've tried this:

Route::get('/test.php', function()
{
    echo 'success';exit;
});

But that doesn't work (I've tried test.php and just test). Dumping Route::getCurrentRoute() outputs this:

object(Illuminate\Routing\Route)[126]
  protected 'uri' => string '/' (length=1)
  ...
  protected 'compiled' => 
    object(Symfony\Component\Routing\CompiledRoute)[135]
      ...
      private 'staticPrefix' => string '/' (length=1)
      ...

Is it possible to access http://example.com/test.php and have Laravel's Routing see it as if /test or /test.php has been requested, without making changes to the independent framework's .htaccess file?

Answer 1

You are not forced to use Laravel routing, you can bypass it with pure PHP:

if($_SERVER['REQUEST_URI'] === '/test.php')
{
    exit('success');
}

But in your case, you probably should redirect all requests to Laravel with the .htaccess and fall back to root index.php when Laravel does not handle the page:

App::missing(function($exception)
{
    include('../../index.php');
    exit;
});