我现在和一个客户住在Bizzaro World。
因为我们正在编写的symfony应用程序将响应注入到另一个应用程序的页面中,所以我们被迫只为该应用程序提供一个URL,但幸运的是,我们确实完好无损地获得了传递给我们的查询字符串。
因此,我需要做与symfony完全相反的事情,一种老式的MVC方法。我需要通过querystring参数进行路由,路由到正确的控制器并呈现正确的响应,而不是使用标准的、合理的路径方法。
因此url将是http://www.bizzaro.com/appname?route=%2fblog
而不是http://www.bizzaro.com/appname/blog
,依此类推。
路由部分中有一些很好的例子,说明如何确保将查询字符串作为参数传递给控制器操作,但不适用于这种相当回归的方法。
我从哪里开始
我实现了Corentin Dandoy在下面提出的解决方案2,但进行了轻微修改,以防止在查找根时出现循环。
namespace AppBundle'Controller;
use Symfony'Component'HttpFoundation'Request;
use Symfony'Component'Routing'Exception'ResourceNotFoundException;
use Symfony'Bundle'FrameworkBundle'Controller'Controller;
class FrontControllerControllerController extends Controller
{
/**
* Forward the request to the appropriate controller
* @param Request $request
* @return 'Symfony'Component'HttpFoundation'Response
*/
public function indexAction(Request $request)
{
// get the parameter that specifies the route to the 'real' homepage controller
$homeroute = $this->container->getParameter('homeroute');
$route = $request->query->get('route');
// Convert the query-string route into a valid path route
$path = '/'.$route;
// if it is the route, then use the 'real' homepage controller, otherwise you end up in a routing loop!
if ($path === '/')
{
$match = $this->get('router')->match('/' . $homeroute);
} else {
try {
$match = $this->get('router')->match($path);
} catch (ResourceNotFoundException $e) {
throw $this->createNotFoundException('The route does not exist');
}
}
$params = array(
'request' => $request,
);
return $this->forward($match['_controller'], $params);
}
}
解决方案1
一个简单的解决方案(不可扩展(是:
namespace AppBundle'Controller;
use Symfony'Component'HttpFoundation'Request;
use Symfony'Component'HttpFoundation'Response;
use Sensio'Bundle'FrameworkExtraBundle'Configuration'Route;
class MainController
{
/**
* @Route("/app", name="bizarro_app")
*/
public function mainAction(Request $request)
{
$route = $request->query->get('route');
switch ($route) {
case 'blog':
return $this->blogAction($request);
default:
throw $this->createNotFoundException('The route does not exist');
}
}
protected function blogAction(Request $request)
{
// Your blog page here
return new Response('...');
}
}
解决方案2
如果你不介意有两种可用的路线,你可以试试这个:
namespace AppBundle'Controller;
use Symfony'Component'HttpFoundation'Request;
use Symfony'Component'HttpFoundation'Response;
use Symfony'Component'Routing'Exception'ResourceNotFoundException;
use Sensio'Bundle'FrameworkExtraBundle'Configuration'Route;
class MainController
{
/**
* @Route("/app", name="bizarro_app")
*/
public function mainAction(Request $request)
{
$route = $request->query->get('route');
// Convert the query-string route into a valid path
$path = '/'.$route;
try {
$match = $this->get('router')->match($path);
} catch (ResourceNotFoundException $e) {
throw $this->createNotFoundException('The route does not exist');
}
$params = array(
'request' => $request,
);
return $this->forward($match['_controller'], $params);
}
/**
* @Route("/blog", name="bizarro_blog")
*/
public function blogAction(Request $request)
{
// Your blog page here
return new Response('...');
}
}
这样,您就可以从Sf2路由组件中获益。请注意,我自己没有测试过。