Hmm, yes. After do something like this:
-- $url = Router::parse($redirect);
$params = $url['params'];
unset($url['params'])
$redirect = Router::url(array_merge($url, $params))
On Thursday, March 19, 2015 at 6:32:07 PM UTC+1, Pgbi wrote:
On Thursday, March 19, 2015 at 6:32:07 PM UTC+1, Pgbi wrote:
The reason i need to parse and generate again an url is in the example I gave in my first message:In my App, when a user logs in, he is redirected to $this->Auth->redirectUrl() which is of the form /locale/controller/action/param . Before redirecting him, I want to replace "locale" with $user->locale. So I parse the url, replace the locale parameter, and generate the url again. Does it make sense :) ?
Le jeudi 19 mars 2015 02:09:10 UTC-7, José Lorenzo a écrit :Router::url() and Router::parse() are not symmetrical. I don't see the reason why you want to parse the url and the pass it again to the Router. Could you explain?
On Thursday, March 19, 2015 at 1:00:16 AM UTC+1, Pgbi wrote:Tell me if i'm wrong but i thought that Router::parse was the inverse of Router::url.In other words, I thought that $url == Router::url(Router::parse($url)) would always be true.Just found out this was not the case.If $url = "/users/view/123" then Router::url(Router::parse($url)) = "/users/view?pass%5B0%5D=123"This leads to the following bug in my App:// In UsersControllerfunction login(){if ($this->request->is('post')) {$user = $this->Auth->identify();if ($user) {$this->Auth->setUser($user);$url = $this->Auth->redirectUrl(); // let's say redirect url is "/en/users/view/123"$url = Router::parse($url); // now url is ['controller' => 'users', 'action' => 'view', 'locale' => 'en', 'pass' => ['123']]$url['locale'] = $user['locale']; // my user is french$url = Router::url($url); // now url is "/fr/users/view?pass%5B0%5D=123" instead of " /fr/users/view/123"return $this->redirect($url);}}}
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