I'm attempting to do this: (pseudo code)
if(USER IS LOGGED IN){
Router::connect('/', array('controller' => 'films', 'action' => 'index'));
} else {
Router::connect('/', array('controller' => 'users', 'action' => 'register'));
}
Which simply redirects them to their 'dashboard' if they are logged in or asks them to register if they aren't signed up!
Is this bad practice to have this IF statement inside the routes.php of CakePHP?
it is likely to create problems because the session usually gets initiated later.
why not switching the action / view inside a so called "OverviewController" like I do?
if (UID) {
$this->_actionOne();
} else {
$this->_actionTwo();
}
which then call their own view:
$this->render('some_view');
Yes, it is. and although #mark pointed reasons from cake's view, the reason is because you are mixing modules functions.
When someone gets to the films/index and you want him to register first, you are redirecting him from the films controller. so, simply add a redirect or, a link for registration.
If you still want it to do it automatically, you probably should create a router_controller (bad idea).
The cake solution is to stay in the films controller, and in the index function..
But, in the view, don't show the default output, but some kind of "register" element..
Related
I have a custom routing setup (which might be cause of issue). As follows:
Router::connect('/o/*', array('controller' => 'open', 'action' => 'openinsert'));
Router::connect('/c/*', array('controller' => 'click', 'action' => 'clickinsert'));
Under UsersController, I have tried:
public function beforeFilter() {
parent::beforeFilter();
//$this->Auth->allow('add');
$this->Auth->allow('o/*');
$this->Auth->allow('clickinsert');
$this->Auth->allow('open');
}
clickinsert and openinsert are both apart of controller 'ClickController' and 'OpenController' respectively.
Nothing from above works. Is is possible to do what I'm wanting to do w/their default auth setup? Or will I need to make some major changes?
AuthComponent::allow
This method:
Takes a list of actions in the current controller for which authentication is not required, or no parameters to allow all actions.
It doesn't accept urls.
In particular this:
$this->Auth->allow('o/*');
Is neither the wildcard ( nothing or '*' depending on the version of CakePHP used) nor the name of an action, and this:
$this->Auth->allow('open');
Is apparently passing the name of a controller so neither of those method calls would do anything.
There's some confusion here
According to these two routes:
Router::connect('/o/*', array('controller' => 'open' <-
Router::connect('/c/*', array('controller' => 'click', <-
They point to two different controllers.
Under UsersController, I have tried:
Anything you put in your users controller is irrelevant for requests that don't use it. Either put the allow calls in the before filter for each controller - or put them in the beforeFilter for your App controller (which all controllers inherit from).
I am developing a hrm panel in cakephp. I want when employee punchin the page redirect to punchout page,but if employee logout the panel then he goes to punchin page that page must not be redirected until it checks the boolean value from the database.
I disagree with the other answer because it does not show the best way to do it in CakePHP. If you use the router the code provided by the other answer will not work with routing. Also if your app is not in the root of the domain / but for example.com/my-tool/ the string type URL won't work either and the link is wrong, it would go to example.com/punchout instead of example.com/my-tool/punchout
if ($value === true) {
$this->redirect(array('controller' => 'employees', 'action' => 'punchout'));
}
This is the correct way to do links in CakePHP for all links that point to any controller of your application. The string should be only used if it is an external URL. If you use links in your layout and have plugins and routing prefixes you also want to add the plugin and prefix key to the array and set it according to your needs.
Might it be so simple just to put your redirect code into a IF statement?
if($boolean === true) {
$this->redirect(array('controller' => 'employees', 'action' => 'punchout'));
}
//normal code here when boolean is false
I'm just getting started with cakePHP, and things aren't going so well so far.
I have a controller that handles confirming user emails. On registration the user is sent an email with a confirmcode in a link. Depending on the confirm code they give, the controller gives different text responses. One of these responses includes a hyperlink in order to log in.
I'm trying to use the Html helper, but although I've loaded it in $helpers at the top of the class, I an only make it work if I then use App::import, and then instantiate it.
It all seems overkill to simply make a hyperlink! How many times do I have to load the same class?
Wherever I look on the web it keeps telling me it's a bad idea to use a helper in a controller, but how else am I supposed to get the link made?
So I have
var $helpers = array('Html');
at the top of the controller, and:
if (isset($this->User->id)) { // Check the user's entered it right
// Do some stuff to remember the user has confirmed
// This is to load the html helper - supposedly bad form, but how else do I make the link?
App::import('Helper', 'Html');
$html = new HtmlHelper();
$this->set('message', __("Your email address has been confirmed.", TRUE)." ".$html->link(__("Please log in", TRUE), array('controller' => "users", 'action' => "login" )));
} else {
$this->set('message', __("Please check your mail for the correct URL to confirm your account", TRUE));
}
in the controller's confirm method and
<div>
<?php echo $message;?>
</div>
in the view to output the resulting message
Surely I'm going wrong somewhere - can anyone explain how?
You're not supposed to use Helpers in the Controller. As #Lincoln pointed out, you should construct the link in the View. You may construct the URL in the Controller, since a URL is basically data, but a link is a very medium-specific (HTML) implementation of a URL.
Either way, you'll need to create a full URL (including host) if you want to send it in an Email. The most universal way is to use Router::url:
$fullUrl = Router::url(array('controller' => ...), true); // 'true' for full URL
Do this in either the Controller or the View. For creating a link, use this in the View:
echo $html->link('Title', $fullUrl);
The idea is that all the data you need to render the page is sent to the view with set, then any conditional logic or formatting is done in the view with helpers, so send whole query results when appropriate (suppose you need to alter a link to include the user's screen name, you'll have it handy).
in controller action
$this->set('user', $this->User);
in view (this is slightly different depending on if your in <= 1.2 or 1.3
if ($user->id) //available because of Controller->set
{
//1.2
$link = $html->link(__("Please log in", TRUE), array('controller' => "users", 'action' => "login" ));
//1.3
$link = $this->Html->link(__("Please log in", TRUE), array('controller' => "users", 'action' => "login" ));
echo __("Your email address has been confirmed.", TRUE)." $link";
}
else
{
$this->set('message', __("Please check your mail for the correct URL to confirm your account", TRUE));
}
What you are trying to do should be done with the SessionComponent. $this->Session->setFlash('your message here');
and in your layout with the session helper put $this->Session->flash();
About your wanting urls in the controller, Router::url is correct as deceze said, but there is no use for it there as you should not be building html in a controller.
what you want to do it use the session::setFlash() method above and then redirect them using
$this->redirect(array('controller' => "users", 'action' => "login" ));
The problem is i want to call the index function, i need it to render the view and then the
afterFilter to redirect again to the index function and do the same.. Like a loop, the problem is it doesnt render it, ive tried using $this->render('index') but it doesnt work and also other things..
PS: I didnt include all the code that i have in index, because its pointless, doesnt render with or without it, just included the things i needed for the view.
function afterFilter()
{
if ($this->params['action'] == 'index')
{
sleep(3);
$this->redirect(array('action'=>'index',$id), null, true);
}
}
THE FUNCTION
function index($ido = 0)
{
$this->set('Operator', $this->Operator->read(null, $ido));
$this->set('ido', $ido);
}
THE VIEW = INDEX.CTP
<legend>Operator StandBy Window</legend>
<?php
?>
</fieldset>
<?php echo $html->link('LogIn', array('controller' => 'operators', 'action' => 'add')); ?>
<?php echo $html->link('LogOut', array('controller' => 'operators', 'action' => 'logout',$ido)); ?>
a function that constantly checks my database for a change, if a change occurs ill redirect, if not i need to have that constant loop of 'checking the database' and 'rendering the view'.
This is not possible entirely on the server with PHP, and especially not with CakePHP's template system. PHP just "makes pages" on the server and sends them to the browser and a linear fashion. If you loop on the server, the content of your page will just repeat itself:
Normal content
Normal content
Normal content
<redirect>
To redirect the client, you need to output headers. The headers need to be output before anything else. If you've already looped a few times and content has already been sent to the client, you can't redirect anymore.
There are two ways:
You just output one page at a time showing the current status, with a meta tag that'll refresh the page every x seconds. This can be rather expensive and annoying though.
Use AJAX and possibly Comet to update the information on the page dynamically in the browser.
In core.php I can define
Configure::write('Routing.admin', 'admin');
and /admin/controller/index will work.
but if I define both
Configure::write('Routing.admin', 'admin');
Configure::write('Routing.superuser', 'superuser');
and try to look at /superuser/blah/index/ instead of it saying the controller doesn't exist it says
Error: SuperuserController could not be found.
instead of saying
Error: BlahController could not be found.
When I first read the documentation I was under the impression I could run both routes, and not just one or the other. Is there something more I need to do?
I believe they are working on this for CakePHP 1.3, but for now, we have to cheat to accomplish additional routing. This is the method I've used in the past.
// SuperUser Routing
Router::connect('/superuser/:controller',
array('prefix' => 'superuser', 'superuser' => true));
Router::connect('/superuser/:controller/:action/*',
array('prefix' => 'superuser', 'superuser' => true));
There were some issues generating URLs using the array('controller' => ...) method, but I haven't touched that project in a few months, so I can't remember all the caveats with it. This should at least give you a starting point though.
The CakePHP document explains this some. The relevant section starts about halfway in talking about multiple prefixes.
If you're using Jason's trick, and having trouble with generating URLs using the array('controller' => ...) syntax, then put this in your appcontroller:
if (isset($this->params['prefix']) && $this->params['prefix'] == 'superuser') {
Configure::write('Routing.admin', 'superuser');
}
This forces the appcontroller to use the correct admin prefix, which in this case is "superuser".