I just came across a bit of a puzzling situation. My controller action looks like this:
public function myaction($eventId = false) {
if(!$eventId) {
//list all events
$data = foo;
} else {
if(!$this->Event->findById($eventId) )
{
$this->Session->setFlash('myerror message', 'flash_frontend_message');
$this->redirect(array('controller' => 'events', 'action' => 'myaction'));
} else {
// display event information
$data = foo;
}
}
$this->set($data);
}
If I call the function with an $eventId that is not found in the database, it outputs an error message and the user gets redirected back to the list of all events. However lets say I then select an eventId which is valid, view the relevant information and then press the browser back button, the previous flash message gets out put again, even though the URL in this case does not contain the $eventId.
I suppose what happens is that the page gets loaded from the browser cache rather then reloaded. I have tried to avoid storing the view in the cache like this:
<!--nocache-->
<?php echo $this->Session->flash(); ?>
<?php echo $this->Session->flash('auth'); ?>
<!--/nocache-->
but still, the flash message gets displayed.
Any idea how to prevent this behavior?
Is there perhaps a way to clear the flash after it has been displayed?
you are talking about the right cache but trying to fix it with the wrong cache.
<!--nocache-->
is php/html no cache statements in the file itself. it has nothing to do with the browser cache.
what you need to be doing is setting the right headers.
there is a convinience method in the request object for this exact thing:
http://book.cakephp.org/2.0/en/controllers/request-response.html#interacting-with-browser-caching
I use it in all my normal frontend actions via AppController:
public function beforeRender() {
$this->response->disableCache();
...
}
Related
I am using Yii-user extension in the main layout i have a sign up link which is common to all the Cmenu
view/main layout
echo CHtml::link('Signup','#',array('id'=>'regi'));
$("#regi").click(function(){
$.ajax({
type:'GET',
url:'<?php echo Yii::app()->request->baseUrl;?>/index.php/user/registration',
success:function(res){
$("#dispdata").show();
$("#dispdata").html(res);
}
});
});
<div id="dispdata"><div>
**yii user extension **renders this perfectly and even submit its correctly if form values a re valid.
but if the values are incorrect and blank it redirect to url .../user/registration
which is not what my need .I need guidance what do i do such that if the values are incorrect or blank it should not redirect and display the errors in model window.
I did tried but hardly could get the satisfied results
if i place the following the model window itself doesnt appear what do i do
module registrationController i placed
....//some code here (**in yiiuser register controller**)
if ($model->save()) {
echo CJSON::encode(array(
'status'=>'success',
));
}
....//some code here...
Yii::app()->clientScript->scriptMap['jquery.js'] = false;
$this->renderPartial('registration',array('model'=>$model,),false,true);
in module view registration
<?php echo CHtml::ajaxSubmitButton(Yii::t('registration'),CHtml::normalizeUrl(array('user/registration','render'=>false)),array('dataType'=>'json',
'success'=>'function(data) {
if(data != null && data.status == "success") {
$("#registration-form").append(data.data);
}
}')); ?>
can anyone please guide me am working past 10 ten days tried every hook or crook method but could not obtain the results......how can the model window with client side validation be done appear..... Please guide me or let me know something better can be done
rules in registration model
if (!(isset($_POST['ajax']) && $_POST['ajax']==='registration-form')) {
array_push($rules,array('verifyCode', 'captcha', 'allowEmpty'=>!UserModule::doCaptcha('registration')));
as well was not with attributes for reqired field
have changed to
array_push($rules,array('verifyCode', 'captcha','message' => UserModule::t("captcha cannot be blank.")));
and added the verifycode to required field
yet not working,
The simple way is using render method in your Ajax action and creating empty layout for this action. If you do so, validation scripts will be included in the server response. Also you need to exclude jquery.js and other script with Yii::app()->clientScript->scriptMap and include them in main layout always.
I have a problem repopulating my form after validation fails. Problem is my url contains an additional uri which I access by clicking a link. This is what it looks like:
http://www.example.com/admin/trivia/add/5
At first trouble was that the segment 4 of the uri completely disappeared, so even though the validation errors showed and the form was repopulated, I lost my added uri.
Then I found in another question that the solution was to set form open like this:
echo form_open(current_url());
Problem is now it isn't showing any validation errors and the form is not repopulated. Is there a way to achieve this?
This is what my controller looks like:
function add()
{
$data = array('id' => $this->uri->segment(4));
if($_POST)
{
$this->_processForm();
}
$this->load->view('admin/trivia_form', $data);
}
And inside _processForm() I got all the validation rules, error message and redirecting in case success.
[edit] Here is my _processForm() :
function _processForm()
{
$this->load->library('form_validation');
//validation rules go here
//validation error messages
$this->form_validation->set_rules($rules);
$this->form_validation->set_error_delimiters('<div style="color:red">', '</div>');
if ($this->form_validation->run())
{
//get input from form and assign it to array
//save data in DB with model
if($this->madmin->save_trivia($fields))
{
//if save is correct, then redirect
}
else
{
//if not show errors, no redirecting.
}
}//end if validation
}
To keep the same url, you can do all things in a same controller function.
In your controller
function add($id)
{
if($this->input->server('REQUEST_METHOD') === 'POST')// form submitted
{
// do form action code
// redirect if success
}
// do your actual stuff to load. you may get validation error in view file as usual if validation failed
}
to repopulate the form fields you are going to need to reset the field values when submitting it as exampled here and to meke it open the same page you can use redirect() function as bellow:
redirect('trivia/add/5','refresh');
i don't know what you are trying to do, but try this to repopulate the form with the values user entered
<?php
echo form_input('myfield',set_value('myfield'),'placeholder="xyz"');
?>
My site uses multiple languages and my users can click on flags to set their desired language. When that flag is clicked, a Session should store that information and then i want my controller to redirect the user to another page. This i do with the following code:
<?php
public function setLangAction () {
$oLanguageCookie = new Container('language');
$oLanguageCookie->lang = $this->params ('langvar');
$this->redirect()->toRoute('loadpage', array('page' => 'home'));
}
?>
However, when i print_r($_SESSION) in the indexAction (the action where loadpage routes to), $_SESSION is empty.
Can somebody help me?
Depending where you param comes from you should execute
$this->params()->fromQuery('langvar');
$this->params()->fromPost('langvar');
unless it is a route parameter then you can use either:
$this->params()->fromRoute('langvar');
$this->params('langvar');
Usually, I just set a $feedback var or array and then check for that to display in my views.
However, it occurred to me I should perhaps use flashdata instead.
The problem is sometimes - for say an edit record form, I may simply want to reload the form and display feedback - not redirect. when i use flashdata, it shows but then it shows on the next request as well.
What would be the best practice to use here?
CodeIgniter supports "flashdata", or session data that will only be available for the next server request, and are then automatically cleared.
u use hidden field for that
I would use the validation errors from the Form validation class and load those directly to the view in its 2nd argument.
$this->form_validation->set_error_delimiters('<p>', '</p>');
$content_data = array();
if (!$this->form_validation->run()) {
$content_data['errors'] = validation_errors();
}
$this->load->view('output_page', $content_data);
Then check in your view whether $errors isset.
Controller:
$data['message'] = 'some message you want to see on the form';
$this->load->view('yourView', $data);
View:
if (isset ($message)) : echo $message; endif;
...
In my CakePHP app for my login method I do some different things for when a user submits a form via AJAX calls using if ($this->request->is('ajax'))
However I also want to allow the login method to be shown in a modal for quick login which again is an ajax call. But how do I detect the difference between the AJAX GET to show the form and then AJAX POST to do the actual login?
See below I can detect native get and posts but for ajax how do I detect the difference in CakePHP??? As it seems I can only detect an ajax event and not the type :/
NATIVE:
GET = if ($this->request->is('get'))
POST = if ($this->request->is('post'))
AJAX:
GET = if ($this->request->is('ajax'))
POST = if ($this->request->is('ajax'))
Thanks
Solution:
if ($this->request->is('get'))
{
if ($this->request->is('ajax'))
{
echo json_encode('ajax get'); exit;
}
else {
echo 'Normal get'; exit;
}
}
if ($this->request->is('post'))
{
if ($this->request->is('ajax'))
{
echo json_encode('ajax post'); exit;
}
else {
echo 'Normal post'; exit;
}
}
Not sure if I understand the question, but if the problem is that the form data can come in either as a POST or GET, the solution is to check whether the POST data is there. If it is, use POST, otherwise take the data from GET. (Or other way around.)
If the function should do different things depending on whether the form has been sent as POST or GET, then simply make two different functions in the controller.