I'm kind of new with CodeIgniter and I'm still learning (a lot).
So I have a view and when I submit a form I 'call' the controller by surfing to the right URL dynamically e.g. site/delete
class Site extends Controller {
function index(){$this->load->view('...')}
function delete() {
$this->site_model->delete_row();
$this->index();
}
}
Now when that action is done (deleted the row) I'm calling $this->index(); to redirect to my initial page (which is good) but my url stays: site/delete . I want my URL to be ../site/index (or without the /index)
Any help would be appreciated :-) .
So far I found something to solve this:
instead of:
$this->index();
I'm using:
redirect('site');
Does anyone know this is a good practice?
Redirect is what you should use.
In the user guide:
http://codeigniter.com/user_guide/helpers/url_helper.html
they use it after checking if a user is logged in. Depending on if they are or not, they redirect to a different place.
Also, note that any code after the redirect won't run. Make sure and redirect after you've done everything you need to.
My preferred method is to have actions like that handled by the same method that will be seen by the user afterwards.
What if you go to /site/delete afterwards, as a user? It will either have to detect and throw a error (show a message) or redirect to an appropriate page. /site/delete has no meaning.
For example, if a user would normally see an overview after deleting, then my form will be posted to /site/index; with index quickly checking for the condition and calling _delete() in the same controller, before doing its normal work.
That way, if the user refreshes the page, or presses 'back', things should look consistent to them.
Another example would be that /settings/edit would post to itself - this means that it can act on the post and show any output (e.g. validation errors). It means there's no /settings/do_edit location on my site, and also means that the user can go back to /settings/edit safely, and see a form for editing their settings.
I suppose this is a subjective take on a perhaps objective question, and I would encourage feedback on my view, but it's my way of avoiding the problem you have asked about.
$this->index();
Call of function in a function simply execute the functionality within that function.
And url never changed.
for changing the url you should use.
redirect ( base_url().'site');
but you should load url helper in constructor.
Related
This is a little bit hard to understand even the title I put. Sorry about that I just do not know how to clearly explain this, but I will try...
So for example I have a controller and inside that controller I have a function which return the data in the table of my database. then in the last column of every row, I make view,add,edit,delete options as links. When a user clicks on the add for example, they will redirect to an add page. After they submit the form of the add page. they should be redirected to the first page that return the data from the table. but the problem is, the variables of foreach loop in the first page do not got recognized by laravel anymore. because they do not got processed since the route does not goes to the controller and to the function that return the data instead it goes to add function.
So I want to know is there anyway to solve this? If you could provide sample code, I would appreciate a lot thanks
From your explanation, I believe the code to go back to the original page after adding, editing etc is simply return redirect()->back(). This will take the user back to the previous page. As for data continuity, one approach would be considering using sessions. They save data globally and can be accessed from any controller once saved. To do this, simply save the data with session(['session_name' => $data]) and to retrieve the data use session('session_name'). Let me know if this helps!
If you want ti redirect after something like a login or an activation process you can do it with something like this:
return redirect()->to('login')
You can specify the path from your web.php file as you can see in my example in 'myPath'
As #DerickMasai correctly pointed out it is better to use named routes instead of hard coding the path itself.
Naming a route can work like so:
Route::get('/login', [LoginController::class, 'index'])->name('login');
I have the typical scenario of an online newspaper where you have the article and below a form to comment, but where you need to login to comment. So on clicking it takes you to the log page and it should return you to that page once you have authenticated.
So, because the form is part of a page that the user can see without being logged in, I cannot write the middleware for that page created by a PostController, (get route).
The CommentController only has one method, which is the store one for the Form. so, of course, if I placed a middleware for that controller, it would fail because although it would indeed take you to the login page on clicking the submit button, the Intended URL saved would be that Post for the form, so on returning after authenticating, it would take you to a non existent URL page under the name of the Post route for that Form.
I have read about Auth::guards and the like but could not come clear with it,
This
Laravel 5 - After login redirect back to previous page
asks exactly the same question that I do, it is the same scenario, but I dont see how his answer works, because defining a protected variable (and I have that already) like protected $redirectTo in the Auth controller only tells where to go after authenticating, and it is only a fixed route, in my case it takes you to the dashboard. But I dont want to be taken to the dashboard, it has to return to the page where the article and the comment form are.
I found the solution at Laracasts. I must say I really dont understand it, but it works. It adds two functions to the AuthController.
https://laracasts.com/discuss/channels/laravel/redirect-to-page-where-login-button-was-clicked
public function showLoginForm()
{
if(!session()->has('from')){
session()->put('from', url()->previous());
}
return view('auth.login');
}
public function authenticated($request,$user)
{
return redirect(session()->pull('from',$this->redirectTo));
}
It's been a while since i've done this, but this should work normally.
You could add a GET param in your link from the comment section to login page.
http://....com/login?intended=http://yourredirectlink.com/#form
Put the intended url in in the session variable url.intended and after login you redirect like so
Redirect::intended('/');
This will redirect back to the the url '/' if the session variable is not available.
I have the need to pass a URL to Magento, where it should redirect the User after completing the logout. To store it, I do the following:
$BackTo = Mage::app()->getRequest()->getParam('backto');
if(!empty($BackTo)) {
Mage::getSingleton('core/session')->setBackTo($BackTo);
}
When needed, I retrieve the URL using Mage::getSingleton('core/session')->getBackTo(). The issue is that, while this works well on login, it doesn't work on logout (where it's most needed). I can store the session variable, I can also immediately retrieve it, but, when I am in logout.phtml, where the redirect JavaScript is located, such variable is set to null.
I suspect that the redirect performed by Magento upon logout has something to do with this "disappearing" session variable, but I can't say for sure.
For completeness, here is the relevant code (there's more code than this, but they are mainly auxiliary functions, which don't get called on logout).
Account Controller
class MyPackage_Redirectplugin_AccountController extends Mage_Customer_AccountController {
/**
* #see AccountController:logoutAction()
*/
public function logoutAction() {
$this->_getSession()
->logout()
->setBeforeAuthUrl(Mage::getUrl());
// Store the "back to" URL in a session variable
$this->StoreBackToURL();
$this->_redirect('*/*/logoutSuccess');
}
protected function StoreBackToURL() {
// Store the value of the "backto" argument, if it was passed
$BackTo = Mage::app()->getRequest()->getParam('backto');
if(!empty($BackTo)) {
Mage::getSingleton('core/session')->setBackTo($BackTo);
// At this point I can see the correct value stored in the session variable
}
}
}
Logout.phtml
// The following command returns null
$redirectURL = Mage::getSingleton('core/session')->getBackTo();
Thanks in advance for the help.
Update 12/09/25 - Found a workaround
Since I couldn't find a way to pass a session variable to the logout page opened by redirect, I chose an alternative way: I'm passing it via the URL. Specifically, I implemented a logoutAction() which calls $this->_redirect('*/*/logoutSuccess', array('myvar' => $MyValue));.
In the template, where I have to do the redirect, I simply read such parameter using Mage::helper('core')->urlDecode(Mage::app()->getRequest()->getParam('myvar'));.
I'm aware that there might have been better ways to implement the whole thing, but I needed a bug fix solution and this does the job. Thanks to all people who answered.
The cleanest option which I see (no controller rewrite necessary!) is to observe the dynamically-dispatched controller_action_postdispatch_customer_account_logout event - see the relevant line fromMage_Core_Controller_Varien_Action::postDispatch(). The postDispatch() method is called after the controller action completes.
Example observer method:
public function logoutRedirect($obs)
{
$redirectUrl = Mage::getUrl(/* url args */);
$obs->getControllerAction()->getResponse()->setRedirect($redirectUrl);
}
Doing this will redirect the user to the desired URL directly upon logout meaning that the logoutSuccess page with the JS redirect will not be accessed.
If the desire is to have the logoutSuccess page render as normal, but redirect to a different URL, this can be achieved a couple of ways:
1. By creating a custom template
2. By creating a custom template block class, overriding the getUrl() method to retrieve the URL of your choice, and assigning that as block to render in the content area (by removing or displacing the customer_logout block) in a custom layout XML update file.
just observe this event customer_logout, and when event call method you save the session key on database.
Learn more: http://www.magentocommerce.com/wiki/5_-_modules_and_development/reference/events
http://www.magentocommerce.com/wiki/5_-_modules_and_development/0_-_module_development_in_magento/customizing_magento_using_event-observer_method
GL.
Update 12/09/25 - Found a workaround
This workaround has also been posted in the question itself.
Since I couldn't find a way to pass a session variable to the logout page opened by redirect, I chose an alternative way: I'm passing it via the URL. Specifically, I implemented a logoutAction() which calls $this->_redirect('*/*/logoutSuccess', array('myvar' => $MyValue));.
In the template, where I have to do the redirect, I simply read such parameter using Mage::helper('core')->urlDecode(Mage::app()->getRequest()->getParam('myvar'));.
I'm aware that there might have been better ways to implement the whole thing, but I needed a bug fix solution and this does the job. Thanks to all people who answered.
I know that we can design the layout in *.xml then in the action just invoke loadLayout, and renderLayout to render the blocks/views.
But, I have a question is:
- How can I load the layout at runtime?
If we have an action which does not really design its layout and will be decided how to render at runtime.
You can please consider the answer from the question for more clear.
Writing a new answer because it seems that you actually DO still want to render, you just want to render a different route's layout XML updates. I believe the _forward() method from Mage_Core_Controller_Varien_Action will allow you to do what you are describing with the least amount of pain.
You should add your action controller directory ahead of the catalog directory, create a ProductController with a viewAction, and check customer is not logged in - in this check you would call $this->_forward('customer','account','login');.
This approach though is going to require more effort in order to be usable, as I imagine that you want the user to be sent to the product page upon login. Have you seen Vinai Kopp's Login Only Catalog module? It should do this for you.
loadLayout() and renderLayout() just execute block output method toHtml() (usually) and take the resulting strings and apply them to the response object via appendBody(). In an action controller you can just call $this->getResponse()->setBody('response string'). How you build the string is up to you.
You can also use Mage_Core_Block_Flush to immediately send output to the browser without using the response object.
I have footer view that's included on all my pages which contains a form. I would like to be able to make use of CI's form validation library to validate the form. Is that possible?
Currently the form posts back to the current page using the PHP_SELF environment variable. I don't want to get it to post to a controller because when validation fails it loads the controller name in the address bar, which is not the desired behaviour.
Any suggestions gratefully received.
Thanks,
Gaz
One way, whilst far from ideal, would be to create a "contact" function in every controller. This could be in the form of a library/helper.
CI doesn't natively let you call one controller from another, although I believe there are extensions that enable this.
Another option would be an AJAX call instead, which would allow you to post to a generic controller, validate etc whilst remaining on the current page.
In this use case, I would definitely go for an AJAX call to a generic controller. This allows you to show errors even before submitting in the origin page.
Another way (slightly more complex), involves posting your form data to a generic controller method, passing it a hidden input containing the current URL.
The generic controller method handling your form can then redirect to the page on which the user submitted the form, passing it the validation errors or a success message using flash session variables: $this->session->set_flashdata('errors',validation_errors()) might do the trick (untested)
The good thing about this is that you can use the generic form-handling method for both the ajax case (suppressing the redirect) and the non-ajax case
AJAX would be best, just like everyone else says.
I would redirect the form to one function in one controller, you could make a controller just for the form itself. Then have a hidden value with the return URL. As far as errors go you could send them back with flashdata.
Just remember to never copy paste code, it a bad practice and guarantees bugs.
//make sure you load the proper model
if ($this->form_validation->run() == FALSE){
// invalid
$redirect = $this->input->post('url');
$this->session->set_flashdata('errors',validation_errors());
redirect($redirect);
} else {
/*
success, do what you want here
*/
redirect('send them where ever');
}