okay, so I am trying to load a user's information into form fields after the user has finished creating an account. This is sort of a "re-check your personal info" but here his previously entered values are already set. How should I do so using ajax. I want to use the prototype ajax to avoid obtrusive coding.
Again,these are the steps
1.User creates an account on signUp.php
2.He is redirected to edit.php where he checks his personal info which is set by default.
if I do this
function setFormData()
{
new Ajax.requestHTML("edit.php")
{ onSuccess:someFunction()
});
}
then how can I set the form values, like
function someFuncton(ajax)
{
$("firstname").value=??
$("lastname").value=??
$("country").value=??
}
you can in the edit.php output an xml, json or string with the values, parse it and use in your someFunction
Related
I'm using zf2 form object on the server and ajax code on the client to implement my registration form.
I post the form values in the ajax request, no problem, and the form gets them fine with
$form->setData($request->getPost());
After I validate the form and perform the registration on the server, I want to send the form back to the client, especially if there are errors, so I can show them to the user.
I'm looking for a standard way using zend or any plugin to serialise the form object into JSON format, so I can send it in the response to the AJAX call.
Any idea?
Well what you can do is run the validation on your form and after that you will return your form within a new JsonModel.
Here is a little example of how to handle your controller:
class RegistrationController extends AbstractActionController
{
public function RegisterAction()
{
$form = new RegisterForm();
$form->setInputFilter(new RegisterInputFilter());
if ($this->getRequest()->isPost()) {
$form->setData($this->getRequest()->getPost());
if($form->isValid()) {
// Handle your registration as the form is valid!
// return to some path after registration is complete.
// Show user he registered succesfully, etc. ;)
}
// Checks if the request is from JavaScript
if($this->getRequest()->isXmlHttpRequest()) {
return new JsonModel(array('registerForm' => $form));
}
}
return new ViewModel(array('registerForm' => $form));
}
}
Notice that the form object is holding all the invalid inputs including its message after validation.
I would take another approach just to completely render the ViewModel again so you can display the validation message much easier. On the side you could add Client Side (Javascript) validation as it's much more user-friendly, but that is just some fancy shizzle I would do ;) In case of rendering the ViewModel:
use Zend\View\Renderer\PhpRenderer;
if($this->getRequest()->isXmlHttpRequest()) {
$renderer = new PhpRenderer;
$registerViewModel = new ViewMOdel();
$registerViewModel->setTemplate('view/register.phtml');
return new JsonModel(array('registerViewModel' => $renderer->render($registerViewModel));
}
Note that not setting a template to your viewModel will result in ZF2 getting the default of the action (view/moduleName/registration/register.phtml) you are in! So in your case you don't need to use PhpRenderrer::setTemplate(). But I just hand it to you so you can change it if you are using any other file.
So now you will receive Json from our controller, in your javascript. Retrieve the new ViewModel from Json and remove the old ViewModel and replace it with the new. By removing the old, you also remove any Javascript that is bound to any element within the viewModel, so you might set the events on your body within your javascript or have it on your attributes in Form/RegistrationForm.
Hope this pushes you in the right direction.
This a screen shot of what I get when I call my ajax request:
How do I run only the task, without printing the whole page? This is my ajax call:
$.ajax
({
type: "POST",
url: "index.php?option=com_similar&task=abc",
data: {
id: id,
name: name,
similar_id: similar_id,
},
cache: false,
success: function(html)
{
$("#flash").fadeOut("slow");
$("#content"+similar_id).html(html);
}
});
});
$(".close").click(function()
{
$("#votebox").slideUp("slow");
});
});
Don't go with exit or die, Joomla! has it's nice way of dealing with this stuff.
The answers below are tested in Joomla! 2.5 & 3 (for 1.5. may work as well).
General
Your URL for the task needs to look like this:
index.php?option=com_similar&task=abc&format=raw
You than create the controller which will use the view, let's say Abc, which will contain the file view.raw.html (identical to a normal view file).
Below you have the code for generate a raw HTML response:
/controller.php
public function abc()
{
// Set view
JRequest::setVar('view', 'Abc');
parent::display();
}
/views/abc/view.raw.php
<?php
defined('_JEXEC') or die;
jimport('joomla.application.component.view');
class SimilarViewAbc extends JView
{
function display($tpl = null)
{
parent::display($tpl);
}
}
/views/abc/tmpl/default.php
<?php
echo "Hello World from /views/abc/tmpl/default.php";
Note: This is the solution I would use if I had to return HTML (it's cleaner and follows Joomla logic). For returning simple JSON data, see below how to put everything in the controller.
If you make your Ajax request to a subcontroller, like:
index.php?option=com_similar&controller=abc&format=raw
Than your subcontroller name (for the raw view) needs to be abc.raw.php.
This means also that you will / may have 2 subcontrollers named Abc.
If you return JSON, it may make sense to use format=json and abc.json.php. In Joomla 2.5. I had some issues getting this option to work (somehow the output was corrupted), so I used raw.
If you need to generate a valid JSON response, check out the docs page Generating JSON output
// We assume that the whatver you do was a success.
$response = array("success" => true);
// You can also return something like:
$response = array("success" => false, "error"=> "Could not find ...");
// Get the document object.
$document = JFactory::getDocument();
// Set the MIME type for JSON output.
$document->setMimeEncoding('application/json');
// Change the suggested filename.
JResponse::setHeader('Content-Disposition','attachment;filename="result.json"');
echo json_encode($response);
You would generally put this code in the controller (you will call a model which will return the data you encode - a very common scenario). If you need to take it further, you can also create a JSON view (view.json.php), similar with the raw example.
Security
Now that the Ajax request is working, don't close the page yet. Read below.
Don't forget to check for request forgeries. JSession::checkToken() come in handy here. Read the documentation on How to add CSRF anti-spoofing to forms
Multilingual sites
It may happen that if you don't send the language name in the request, Joomla won't translate the language strings you want.
Consider appending somehow the lang param to your request (like &lang=de).
New in Joomla 3.2! - Joomla! Ajax Interface
Joomla now provides a lightweight way to handle Ajax request in a plugin or module. You may want to use the Joomla! Ajax Interface if you don't have already a component or if you need to make requests from a module your already have.
If you just want to include the response output in some HTML element, append format=raw to your URL as mentioned above. Then you could have a controller function like this:
function abc(){
//... handle the request, read variables, whatever
print "this is what I want to place in my html";
}
The AJAX response will output everything you printed / echoed in the controller.
I have a situation where I'm editing a snippet of data within a larger context. The user submits this data to a specialized action for handling and redirects back to the parent page. Because it's a redirection, validation errors aren't getting automagically set, so I'm trying to work around that.
In the event of an error, I'm writing a validation_errors key to the session with a value of $model->validationErrors. In the form, though, I'd like to tell Cake to set each error so I can leverage my existing styles and not have to make a lot of changes to my $this->Form->input() methods.
Is something like this possible? Essentially, I'm looking to manually achieve the same result you'd get if a regular form was submitted and allowed to drop through with validation errors. I was hoping I could loop over each validation error and set the field error, but that's not making any change at all.
Thanks.
This can be achieved in the controller by
$this->Model->invalidate('fieldName', __('ErrorMessage', true));
If the values are available, you can also call
$this->Model->validates();
to validate all values with the validators defined in the model.
Save the data to the session and revalidate it.
function childAction() {
if(isset($this->data)) {
$this->Session->delete('invalid_data');
if($this->Test->save($this->data)) {
// ...
} else {
$this->Session->write('invalid_data', $this->data);
}
$this->redirect(array('action'=>'parentAction'));
}
}
function parentAction() {
if($this->Session->check('invalid_data')) {
// This will cause $this->Test->validationErrors to be populated
// Assuming your parent page has the form set up properly, the
// errors will be automagically filled. ie: $form->input('Test.field1')
$this->Test->set($this->Session->read('invalid_data'));
$this->Test->validates();
}
}
If you want to do the same with CakePHP 3, use the method "errors".
Please note: I'm not trying to do this anymore, because I found an alternative, but it may be useful in the future to know the answer.
I have a form that is in a view (index.ctp) associated with the index() action on a controller. That form should post data to another action, contact(), in the same controller. This second action doesn't have a view, it's just to process the information and redirect the user according to the outcome. This action is doing the validation and redirecting the user to the referer (index in this case) in case of an error, and then the error should be displayed in index. Note that the model doesn't use a database table, but it's used only to define validation rules.
The validation is taking place correctly and reporting the expected errors. In order to retrieve the errors after the redirect, it writes the $this->ModelName->invalidFields() array to a session variable that is retrieved on the index() action after the redirection.
This array is passed on to the $errors variable to the view. Now comes the problem. The errors, although being passed correctly between redirects, aren't getting attached to the respective forms. How can I accomplish this? The form has all the conventional names, so it should be automatic, but it isn't.
Here's part of the relevant code:
Index view:
echo $this->Form->create('Contact', array('url' => '/contacts/contact'));
echo (rest of form) ...
echo $this->Form->end(__('send message', true));
Contacts controller:
function index() {
if ($this->Session->check('Contact.errors')) {
$this->set('errors', $this->Session->read('Contact.errors'));
}
}
function contact() {
if (!empty($this->data)) {
$this->Contact->set($this->data);
if ($this->Contact->validates()) {
(send the email)
}
else {
$this->Session->write('Contact.errors', $this->Contact->invalidFields());
$this->redirect($this->referer);
}
}
}
I don't think it's a good idea to write the validation errors in a session variable. I'm no CakePHP expert, but I don't think that's the way you are supposed to do it. All your forms should point to the same url you are on, so the data that the user has entered will not be lost.
Could you add some code to your question?
I need to get the quantity of items from a form and pass that to CI's paypal_lib auto_form:
This is my controller:
function auto_form()
{
$this->paypal_lib->add_field('business', 'admin_1261513315_biz#pixelcraftwebdesign.com');
$this->paypal_lib->add_field('return', site_url('home/success'));
$this->paypal_lib->add_field('cancel_return', site_url('home/cancel'));
$this->paypal_lib->add_field('notify_url', site_url('home/ipn')); // <-- IPN url
$this->paypal_lib->add_field('custom', '1234567890'); // <-- Verify return
$this->paypal_lib->add_field('item_name', 'Paypal Test Transaction');
$this->paypal_lib->add_field('item_number', '001');
$this->paypal_lib->add_field('quantity', $quant);
$this->paypal_lib->add_field('amount', '1');
$this->paypal_lib->paypal_auto_form();
}
I have a library of my own that validates the input and redirects to auto_form on validation. I just need to pass the var $quant to the controller.
How can I achieve this?!
If you're redirecting directly to the auto_form controller method you can setup an argument there to pass your data in:
auto_form($quant)
Then, depending assuming you have no routes, rewriting, or querystrings 'on' (basically a stock CI setup) to interfere, and you are using the URL helper to redirect, you would do your redirect something like this:
redirect('/index.php/your_controller/auto_form/'. $quantity_from_form);
More on passing URI segments to your functions here.
Or if you're already using CI sessions in your application you can add the quantity value to a session variable for later retrieval inside of the auto_form controller method:
// Set After Your Form Passed Validation
$this->session->set_userdata('quant', $quantity_from_form);
// Retrieve Later in Controller Method After Redirect
$this->paypal_lib->add_field('quantity', $this->session->userdata('item'));
More on CI sessions here.