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 UsersController
>> function 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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