I have a form in cake php Here i have some hidden field that i want to sent when user click on submit button.
At start hidden field value is blank you can see in given form below
I set that hidden field value in playTimer() function in js file
But my prob is when i am going to submit that form it have blank value for actionId hidden field in post data .Wen i again clcik submit then i have value .
I want to set it when user first clcik the submit button so taht i am using before call back function.
echo $this->Form->create('Meditation', array('id' => 'timerform'));
echo $this->Form->hidden('actionId', array('value'=>'', 'id'=>'actionId'));
echo $this->Js->submit('Play', array(
'before' => 'playTimer();',
'update' => '#map_container',
'complete' => 'setsessionid();',
'success' => 'soundPlay();resume();',
'class' => 'btn btn-med-success playTimer',
'div' => false,
'async' => false,
'url' => array('action' => 'timerupdateDuo')
));
==========Custom.js============
function playTimer(){
$("#actionId").val('playTimer');
}
Thanks
<? echo $this->Form->submit(__('Save'), array(
'class' => 'ClassOfTheFormSubmitBTN'
)); ?>
<script type="text/javascript">
$(function(){
$('.ClassOfTheFormSubmitBTN').on('click',function(){
// implement your logic here
$('#MODELNAMEactionId').val('something');
//you can use ajax here with serialize to send the form
//to stop the form from posting return false;
});
});
</script>
Related
I am trying to create a controller from a button id in CakePHP.
Here is what I am trying to do:
<button id="initial_pay" class="btn btn-action" onclick="payWithPaystack()">Initial payment - N5,000</button>
public function initial_pay() {
$amount= 55;
$credit = $this->calculateCredit($amount, true);
$uid = $this->Auth->user('id');
$this->CreditBalances->addCredit($uid, floatval($credit));
}
How can I achieve this so that when users click on the button, the action code will be executed?
You can do it like this:
echo $this->Html->link(
'Initial payment - N5,000',
array(
'id' => 'initial_pay',
'class' => 'btn btn-action',
'controller' => 'myController',
'action' => 'initial_pay',
'full_base' => true
)
);
More info: https://book.cakephp.org/2.0/en/core-libraries/helpers/html.html
But if you want the user to click the button and stay in the same page without reloading, you'll need to call mycontroller/initial_pay with ajax.
I'm just trying to call a function from a form submision (which is also call from ajax).
I want to call the add method but it's the index which always called.
My ajax view with the ajax submission form :
<?php $count = count($files); ?>
<div id="message"></div>
<i><?= __('Nombre de fichiers liés : ') . $count ?></i>
<?= $this->Form->create('FilesManager', array('enctype' => 'multipart/form-data', 'url' => array('action' => 'add'))); ?>
<?= $this->Form->input('file', array('type' => 'file', 'label' => false, 'class' => 'form-control')); ?>
<?= $this->Js->submit('Envoyer', array('update' => '#message', 'div' => false, 'type' => 'json', 'async' => false)); ?>
<?= $this->Js->writeBuffer(); ?>
i've also tried :
<?= $this->Form->create(null, array('enctype' => 'multipart/form-data', 'url' => array('controller' => 'FilesManagers', 'action' => 'add'))); ?>
<?= $this->Form->create('FilesManager/add', array('enctype' => 'multipart/form-data')); ?>
EDIT
$this->Form->create('FilesManager', array('action' => 'add'))
Don't work for me but the form generated look like :
<form action="/agralis/files_managers/add" enctype="multipart/form-data" id="FilesManagerIndexForm" method="post" accept-charset="utf-8">...<input id="submit-838644811" type="submit" value="Envoyer"><script type="text/javascript">
//<![CDATA[
$("#submit-838644811").bind("click", function (event) {$.ajax({async:false, data:$("#submit-838644811").closest("form").serialize(), dataType:"html", success:function (data, textStatus) {$("#message").html(data);}, type:"post", url:"\/agralis\/FilesManagers"});
return false;});
//]]>
</script></form>
I can see that form action look good (when i copy/paste the action in the browser url he found the method) but the url called from the ajax submit button is wrong ! How can I change that ?
The helpers are working independently
The Js helper is not aware of the Form helper, they don't interact with each other, and so the form URL is unknown to the Js helper.
The solution for that problem is rather simple, just look at the docs:
http://book.cakephp.org/2.0/en/core-libraries/helpers/js.html#JsHelper::submit
The submit() method accepts a url option where you can specify a URL in case the default one that is based on the current request doesn't fit you.
$this->Js->submit('Envoyer', array(
'url' => array('action' => 'add')
// ...
));
File uploads are not supported
The next problem that you'll encounter is that the file upload doesn't work. That's not part of the question, but I'll broach the subject.
The Js helper doesn't support generating the required JavaScript to handle file uploads, it would need to make use of the XMLHttpRequest Level 2 (xhr2) functionalities that older browsers are lacking.
You'll have to write custom JavaScript to handle that. Explanations and tutorials regarding that can be found all over the web.
I have two form on a single page: login form and register form. When I submit the register form, it validates both: form fields that are in login and registeration. How can I handle it if both form have the same model (user model)
Register form
<?php echo $this->Form->create('User', array('url' => array('controller' => 'users', 'action' => 'add'))); ?>
<?php echo $this->Form->input('username', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('email', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('password', array('label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('confirm_password', array('type' => 'password', 'label' => false, 'div' => false, 'class' => 'reg_input'));?>
<?php echo $this->Form->submit(__('Submit', true), array ('class' => 'reg_button', 'div' => false));
echo $this->Form->end();?>
and Login form is below
<?php echo $this->Form->create('User', array('controller' => 'users', 'action' => 'login'))?>
<?php echo $this->Form->input('User.username',array('label'=>false,'div'=>false, 'class' => 'reg_input'));?>
<?php echo $this->Form->input('User.password',array('label'=>false,'div'=>false, 'class' => 'reg_input'));?>
<?php echo $this->Form->submit(__('Log in', true), array ('class' => 'reg_button', 'div' => false)); ?>
<?php echo $this->Form->end();?>
When I submit registration form it validates both forms, I want to validate only the registration form.
How can I handle that?
I've come up with a "solution" (I find the approach dirty, but it works) for a different question (very similar to this). That other question worked with elements and views, though. I'll post the entire solution here to see if it helps someone (though I rather someone else comes with a different approach).
So, first: change the creation names for the two forms.
//for the registration
<?php echo $this->Form->create('Registration',
array('url' => array('controller' => 'users', 'action' => 'add'))); ?>
//for the login
<?php echo $this->Form->create('Login',
array('controller' => 'users', 'action' => 'login'))?>
The forms should work, look and post to the same actions, so no harm done.
Second step: I don't have your action code, so I'm going to explain what needs to be done in general
public function login() {
if ($this->request->is('post')) {
//we need to change the request->data indexes to make everything work
if (isset($this->request->data['Login'] /*that's the name we gave to the form*/)) {
$this->request->data['User'] = $this->request->data['Login'];
unset($this->request->data['Login']); //clean everything up so all work as it is working now
$this->set('formName', 'Login'); //we need to pass a reference to the view for validation display
} //if there's no 'Login' index, we can assume the request came the normal way
//your code that should work normally
}
}
Same thing for the registration (only need to change 'Login' to 'Registration').
Now, the actions should behave normally, since it has no idea we changed the form names on the view (we made sure of that changing the indexes in the action). But, if there are validation errors, the view will check for them in
$this->validationErrors['Model_with_errors']
And that 'Model_with_errors' (in this case 'User') won't be displayed in the respective forms because we've changed the names. So we need to also tweak the view. Oh! I'm assuming these both forms are in a view called index.ctp, for example, but if they are on separate files (if you're using an element or similar) I recommend add the lines of code for all the files
//preferably in the first line of the view/element (index.ctp in this example)
if (!empty($this->validationErrors['User']) && isset($formName)) {
$this->validationErrors[$formName] = $this->validationErrors['User'];
}
With that, we copy the model validation of the User to the fake-named form, and only that one. Note that if you have a third form in that view for the same model, and you use the typical $this->form->create('User'), then the validation errors will show for that one too unless you change the form name for that third one.
Doing that should work and only validate the form with the correct name.
I find this a messy approach because it involves controller-view changes. I think everything should be done by the controller, and the view shouldn't even blink about validation issues... The problem with that is that the render function of Controller.php needs to be replaced... It can be done in the AppController, but for every updgrade of Cakephp, you'll have to be careful of copying the new render function of Controller.php to the one replacing it in AppController. The advantage of that approach, though, is that the "feature" would be available for every form without having to worry about changing the views.
Well, it's just not that maintainable anyway, so better to leave it alone if it's just for this one case... If anyone is interested on how to handle this just in the controller side, though, comment and I'll post it.
You can duplicate your model and change his name and define $useTable as the same table name.
Example :
class Registration extends AppModel {
public $useTable = 'users';
You define the action in form->create like Nunser for your login form
<?php
echo $this->Form->create('User',array(
'url' => array(
'controller' => 'Users',
'action' => 'login',
'user' => true
),
'inputDefaults' => array(
'div' => false,
'label' => false
),
'novalidate'=>true,
));
?>
and your registration form
<?php
echo $this->Form->create('Registration',array(
'url' => array(
'controller' => 'Users',
'action' => 'validation_registration',
'user' => false
),
'inputDefaults' => array(
'div' => false,
'label' => false
),
'novalidate'=>true,
));
?>
In your controller define a method for registration validation and the most important define the render
public function validation_registration(){
$this->loadModel('Registration');
if($this->request->is('post')){
if($this->Registration->save($this->request->data)){
--- code ---
}else{
--- code ---
}
}
$this->render('user_login');
}
Sorry for my english ! Have a nice day ! :D
The create method on your login form is missing the 'url' key for creating the action attribute. I tried to re-create this once I fixed this and could not. Maybe that will fix it?
I'm using CakePHP 2.2. I'm adapting a method of dynamically updating a selectbox which I got from: http://www.willis-owen.co.uk/2011/11/dynamic-select-box-with-cakephp-2-0/#comment-10773 which works without issue. It updates the 'hotels' selectbox contents when the users selects a 'region' from another select box.
On the same form, I want to automatically populate multiple 'team' fields with address details from the 'hotels' model when a user selects a 'hotel' from the selectbox.
The user can then modify the address ... all of this before the user clicks submit on the 'team' add view.
In Team\add.ctp view I have the following code:
echo "<div id='address'>";
echo $this->Form->input('address_1');
echo $this->Form->input('address_2');
echo $this->Form->input('address_3');
echo $this->Form->input('city');
echo $this->Form->input('postcode');
echo $this->Form->input('country');
echo "</div>";
...
$this->Js->get('#TeamHotelId')->event('change',
$this->Js->request(array(
'controller'=>'hotels',
'action'=>'getAddress'
), array(
'update'=> '#address',
'async' => true,
'method' => 'post',
'dataExpression' => true,
'data'=> $this->Js->serializeForm(array(
'isForm' => true,
'inline' => true))
)
)
);
In my HotelsController.php I have:
public function getAddress() {
$hotel_id = $this->request->data['Team']['hotel_id'];
CakeLog::write('debug', print_r($hotel_id, true));
$address = $this->Hotel->find('first', array(
'recursive' => -1,
'fields' => array('hotel.address_1', 'hotel.address_2', 'hotel.address_3', 'hotel.city', 'hotel.postcode', 'hotel.country'),
'conditions' => array('Hotel.id' => $hotel_id)
));
CakeLog::write('debug', print_r($address, true));
$this->set('hotels', $address);
$this->set(compact('address'));
$this->layout = 'ajax';
}
hotels\get_address.ctp:
<?php
echo $this->Form->input('Team.address_1', array('value'=> $address['Hotel']['address_1']));
echo $this->Form->input('Team.address_2', array('value'=> $address['Hotel']['address_2']));
echo $this->Form->input('Team.address_3', array('value'=> $address['Hotel']['address_3']));
echo $this->Form->input('Team.city', array('value'=> $address['Hotel']['city']));
echo $this->Form->input('Team.postcode', array('value'=> $address['Hotel']['postcode']));
echo $this->Form->input('Team.country', array('value'=> $address['Hotel']['country'])); ?>
This now works and the code has been updated.
You can not do in that way as you are trying to accomplish. However, there is a way in which you can update as you want by passing an id from dropdown using ajax and populating all the other fields on the basis of that id. Please make sure in 'update'=>#id you should put the div id of a div containing all the fields you want to show on the page where you want your contents to appear on ajax request.
Note: Please follow the Richard link which you have given i.e. www.willis-owen.co.uk, it will definitely help you that you are trying to do.
Quick questions
Is there Way to Append a textarea using Cakephp
view code:
<?php echo $ajax->link(
$qnote['Qnote']['subject'],
array(
'controller' => 'qnotes',
'action' => 'view', $qnote['Qnote']['id']
),
array( 'update' => 'Textarea_id')
);
?>
controller Code:
function view($id = null) {
$this->Qnote->id = $id;
$this->set('qnote', $this->Qnote->read());
}
the above code Pulls the information But Replaces the entire Text in the textarea.
Is there a way I can just append the textarea with out removing the existing text in the textarea
if possible can somebody point me in the right direction please.
You can try saving the result of your AJAX request to a hidden field and then having it execute and on page javascript function to simply slap the values from the hidden field to the visible text area.
The AJAX helper lets your specify callback functions, so something like this should work:
<?php echo $ajax->link(
$qnote['Qnote']['subject'],
array(
'controller' => 'qnotes',
'action' => 'view', $qnote['Qnote']['id']
),
array( 'update' => 'Textarea_id_hidden', "complete" => "concat_fields()" )
);
?>
and then the JavaScript in the View
<script type="text/javascript">
function concat_fields() {
$('#Textarea_id').val( $('#Textarea_id').val() . $('#Textarea_id_hidden').val() );
}
</script>
Note: My JavaScript example above assumes you're using JQuery, changes will need to be made if you're not.