Is there a Controller property that will allow me to get just /controller/action from the URL without any additional parameters there might be?
At the moment I am having to join $this->name . '/' . $this->action.
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BadHorsie
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What do you need it for? – deceze Jul 27 '11 at 12:27
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To pass to my login page as a redirect – BadHorsie Jul 27 '11 at 12:28
3 Answers
17
You don't want to construct the string /users/login, you want the URL that corresponds to the login action of your users controller (for example). That is not necessarily the same as /users/login, and you should not hardcode it!
To get a URL that will lead to a controller action, use reverse routing:
Router::url(array('controller' => 'users', 'action' => 'login'));
//or
Router::url(array('controller' => $this->name, 'action' => $this->action));
Yes, that's even longer, but it's the correct way to do it. If one day you decide you want the login URL to be /login or /members/entrance instead of /users/login, you only need to define an appropriate route in routes.php without rewriting all your hardcoded links.
deceze
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4
$this->here
Available in view and controller. Minor note: It's getting removed in 2.0.
Dunhamzzz
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@BadHorsie Which as you want to use it for a login redirect, I would say are *quite* relevant. If you just want the controller and action then just join the controller and action like you have been! – Dunhamzzz Jul 27 '11 at 12:37
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No, I don't want the extra parameters. Thanks, I'll just carry on with joining the controller/action. – BadHorsie Jul 27 '11 at 12:40
2
It is also possible to use HtmlHelper::url method in 2.x.
$this->Html->url(array(
"controller" => "controller",
"action" => "action",
"parameter"
));
For CakePHP 3.x, UrlHelper is a good choice:
$this->Url->build([
"controller" => "controller",
"action" => "action",
"parameter"
]);
Both examples produce
/controller/action/parameter
betatester07
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