Zend Framework fancybox confirmation dialog with ajax and posted values - ajax

I created a confirmation page for deleting an item. This works perfectly.
Now I want to appear the confirmation page in fancybox, still passing all the variables and delete the data or close the dialog.
It's important that the process still works as it does now (delete->confirmationpage->redirect/deletion) so that users with javascript disabled can perform these actions without a problem.
Now have I been reading about zend framework,ajax, json and more, but the more I read, the less I understand.
So in a nutshell, my question:
I want to pass a variable to the fancybox and if 'yes' perform the delete action or on 'no' close the dialog. This within the zend framework and jquery.
Any advice or tips are greatly appreciated!

You need to use content switching in your ajax which will then render your action appropriately, for example:
function confirmDeleteAction() {
if($this->_request->isXmlHttpRequest()) {
//This is ajax so we want to disable the layout
$this->_helper->layout->disableLayout();
//Think about whether you want to use a different view here using: viewRenderer('yourview')
}
//The normal code can go here
}
function deleteAction() {
//You probably don't want to show anything here, just do the delete logic and redirect therefore you don't need to worry where it's coming from
}
And in your fancybox, have a view that has a form and two buttons, the form should point to your delete action with the id of whatever your deleting as a get param. Set some javascript up that says something like (and this is mootools code, but you can convert it easily):
$('confirmButton').addEvent('click', function(e) {
e.preventDefault();
this.getParent('form').submit();
}
$('cancelButton').addEvent('click', function(e) {
e.preventDefault();
$('fancyBox').destroy(); //I'm not sure it has an id of fancyBox, never used it
}

Came across the question today and figured I could give it a fresh look. For Zend Framework the solution is really simple:
This is how the action looks:
public function deleteAction()
{
$this->_helper->layout()->disableLayout();
$this->view->news = $this->newsService->GetNews($this->_getParam('id'));
if($this->getRequest()->isPost())
{
if($this->getRequest()->getPost('delete') == 'Yes')
{
$this->newsService->DeleteNews($this->_getParam('id'), $this->view->user->username, $this->view->translate('deleted: ').'<strong>'.$this->view->pages[0]['title'].'</strong>');
$this->_helper->flashMessenger(array('message' => $this->view->translate('The page is deleted'), 'status' => 'success'));
$this->_helper->redirectToIndex();
}
elseif($this->getRequest()->getPost('delete') == 'No')
{
$this->_helper->flashMessenger(array('message' => $this->view->translate('The page is <u>not</u> deleted'), 'status' => 'notice'));
$this->_helper->redirectToIndex();
}
}
}
The delete.phtml
<div>
<h2><?php echo $this->translate('Delete news'); ?></h2>
<?php
foreach($this->news as $news)
{
?>
<p>
<?php echo $this->translate('You are now deleting <strong>\'').$news['title'].$this->translate('\'</strong>. Are you sure about this?'); ?>
</p>
<p>
<?php echo $this->translate('<strong>Note! </strong>This action is inreversable, even for us!'); ?>
</p>
<form action="<?php echo $this->baseUrl(false).'/news/index/delete/'.$news['id']; ?>" method="post">
<?php
}
?>
<input class="submit deleteYes" type="submit" name="delete" value="<?php echo $this->translate('Yes'); ?>" />
<input class="submit deleteNo" type="submit" name="delete" value="<?php echo $this->translate('No'); ?>" />
</form>
And this is how the link to delete a file looks (within a foreach loop with my database results)
<a class="deleteConfirmation" href="<?php echo $this->baseUrl(false).'/news/index/delete/'.$news->id; ?>">delete</a>
This works like you would expect; when you click on delete the user goes to the delete confirmation page and redirects the user back to the index after the form is submitted. But I wanted the confirmation in a dialog (in my case I use fancybox). To achieve this, add the following jquery to your index:
$('.deleteConfirmation').fancybox({
// Normal fancybox parameters
ajax : {
type : "POST"
}
});

Related

Showing notification with AJAX

I have a navbar on my users' panel. A part of the navbar indicates if the user has a new unread message. In this case a badge will appear next to an icon. I've simplified the codes here to make them easier to understand and read.
So this is the simplified HTML code:
<div class="btn-group msg-box">
<i class="fa fa-envelope"></i>
// this is the default state, no badge is shown
</div>
Here is the AJAX request which calls a custom function every 10 seconds:
<script type='text/javascript'>
$(document).ready(function(){
setInterval(checkMsg,10000);
});
function checkMsg(){
$.get('ajax.php',{user_id : <?php echo $user_id; ?>},function(data){
$('.msg-box').html(data);
});
}
</script>
And this is the ajax.php file content:
if(isset($_GET['user_id']){
// a few lines of code here to check if that particular user has any unread message.
// In such case a variable name $newMessage is set to 1. Now ... :
if($newMessage>0){
$data='
<i class="fa fa-envelope"></i>
<span class="badge"><i class="fa fa-info"></i></span>
';
}else{
$data='
<i class="fa fa-envelope"></i>
';
}
echo $data;
}
First of all, I know the way I've written this AJAX request is very rookie, but it works fine anyway, up to one point!
In case the user has a new message, and if they stay on a page, the code runs perfectly and shows the badge. But when the user refreshes the page or goes to another page, even-though they have a new message, that default state is shown again where there's no badge. And I know it's of course because I have specified a default state via HTML codes.
I need to know how I can keep the result of the AJAX request regardless of how many times the user refreshes the page or goes to another page.
UPDATE
I tried storing the query result in a SESSION in my ajax.php file. So instead of $data I wrote $_SESSION['data'].
Back on my HTML I made the following change:
<div class="btn-group msg-box">
<?php
if(!isset($_SESSION['data'])){
?>
<i class="fa fa-envelope"></i>
<?php
}else{
echo $_SESSION['data'];
}
?>
</div>
I made this change because I considered the fact that SESSIONS, by definition, are created and accessed globally within the domain. So once it's set, it can be checked and used on all other pages.
So that only if that SESSION isn't set, the default state should be displayed. But that as well doesn't seem to have my desired result. Still the same thing happens.
Ok, answering my own question now. :)
My UPDATE seemed to be a good idea which I tried.
The problem there was that I had written session_start(); on my main PHP file which was included in all other PHP files of the project.
So I basically thought that when the ajax.php file is called, there's no need to write session_start(); again. Because ajax.php was called inside a PHP file that had session_start(); in it already. So, I was wrong!
Adding session_start(); to the beginning of my code in ajax.php simply fixed the issue.

refresh page with joomla component

I have a simple form in my tmpl/default.php:
<form id='AddForm' action="<?php echo JRoute::_('index.php?option=com_mycomponent&task=addcam'); ?>" >
<p>
<label for="CamName">Name:
</label>
<input type="text" id="CamName" name="cam_name" />
</p>
<button type='submit' class='submit_cam' name='addcam' value='Add'>Add</button>
<button type='reset' class='cancel_changes' name='cancel_changes' value='Cancel'>Cancel</button>
</form>
In my controller.php file I'm trying to process the values:
function addcam()
{
$add_name=JRequest::getString('cam_name');
$model = &$this->getModel();
$model->AddWebcam($add_name); //send to model to add to DB
}
In my model I just return the result of the query. With this implementation I just get routed to an empty page. I'd like to have it refresh the current page. Typically you do this with action="" but in my case I need it to route to the function called addcam in the controller. Or is there a better way to do this?
A common technique in Joomla when directing to the task is to have that function do a full redirect to a view at the end. This prevents a page refresh from trying to resubmit the data and leads to a cleaner url for the client. To do this, try the following:
function addcam()
{
$add_name=JRequest::getString('cam_name');
$model = &$this->getModel();
$model->AddWebcam($add_name); //send to model to add to DB
JFactory::getApplication()->redirect(JRoute::_(index.php?option=com_mycomponent&view=whatever));
}
Obviously, update the JRoute bit to the url you actually need. You can also include a message if you would like (like "Saved!"): http://docs.joomla.org/JApplication::redirect/1.6

Update multilingual content with AJAX in Yii

Refer to my code below, when user click on en button, the content will be changed to English, while clicking tw button, the content will be changed to Chinese.
However, the page will be refreshed each time when user click either en or tw button. I want to ask how can I implement AJAX content update in this case?
The result is when user click either en or tw button, the page won't be refreshed to change the content language.
Thanks
I have refer to Yii docs here, but seem that it is not appropriate for my case
C:\wamp\www\website\protected\views\site\index.php
<?php
$lang = isset($_GET["lang"]) ? $_GET["lang"] : "en_uk";
$lang = $lang == "en" ? "en_uk" : "zh_tw";
Yii::app()->setLanguage($lang);
?>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="get">
<input type="submit" value="en" name="lang" />
<input type="submit" value="tw" name="lang" />
</form>
<div class="main">
<?php echo Yii::t(Yii::app()->controller->id, "Causeway Bay"); ?>
</div>
Best practice is to reload the page in these cases, because usually you have to update so much, that it is just not worth it.
That said, CHtml's ajaxSubmitButton is the cleanest way to implement this, because you can map every event of your call very easily. It looks something like this:
<?php
echo CHtml::ajaxSubmitButton('en', CHtml::normalizeUrl(array('site/changeLanguage')),
array(
'error'=>'js:function(){
alert("error");
}',
//if you add a return false in this, it will not submit.
'beforeSend'=>'js:function(){
alert("beforeSend");
}',
'success'=>'js:function(data){
alert("success, data from server: "+data);
}',
'complete'=>'js:function(){
alert("complete");
}',
//'update'=>'#where_to_put_the_response',
)
);
?>
You don't have to use every parameter of course. The update parameter can update a HTML tag instantly.
EDIT:
This can be done easily if you use the controller's renderPartial method, for instance in your site controller if you have the action responsible for the index.
public function actionIndex(){
//get variables, etc
if(Yii::app()->request->isAjaxRequest) {
$lang = $_POST['nameOfSubmit'];
}else {
//...
}
//if the 3rd parameter is true, the method returns the generated HTML to a variable
$page = $this->renderPartial('_page', array(/*parameters*/ ), true);
echo $page;
}
And then, in your view file you can simply have
<?php echo CHtml::ajaxSubmitButton('en', CHtml::normalizeUrl(array('site/index')),
array('update'=>'#content_div',));?>
and
<?php echo CHtml::ajaxSubmitButton('tw', CHtml::normalizeUrl(array('site/index')),
array('update'=>'#content_div',));?>

how to create an function using jquery live?

I am writing a function that well keep the user in lightbox images while he adds to cart.
When you click any image it well enlarge using lightbox v2, so when the user clicks the Add image, it will refresh the page. When I asked about it at jcart support they told me to use jquery live, but I dont know how to do that. T tried this code but still nothing is happening:
jQuery(function($) {
$('#button')
.livequery(eventType, function(event) {
alert('clicked'); // to check if it works or not
return false;
});
});
I also used
jQuery(function($) {
$('input=[name=addto')
.livequery(eventType, function(event) {
alert('clicked'); // to check if it works or not
return false;
});
});
yet nothing worked.
for code to create those images http://pasite.org/code/572
I also tried:
function adding(form){
$( "form.jcart" ).livequery('submit', function() {var b=$(this).find('input[name=<?php echo $jcart['item_id']?>]').val();var c=$(this).find('input[name=<?php echo $jcart['item_price']?>]').val();var d=$(this).find('input[name=<?php echo $jcart['item_name']?>]').val();var e=$(this).find('input[name=<?php echo $jcart['item_qty']?>]').val();var f=$(this).find('input[name=<?php echo $jcart['item_add']?>]').val();$.post('<?php echo $jcart['path'];?>jcart-relay.php',{"<?php echo $jcart['item_id']?>":b,"<?php echo $jcart['item_price']?>":c,"<?php echo $jcart['item_name']?>":d,"<?php echo $jcart['item_qty']?>":e,"<?php echo $jcart['item_add']?>":f}
});
return false;
}
and it seems to add to jcart but yet it still refreshes
.live() is to assign handlers to future creating elements. On your site, however, you are re-loading the page so .live would have no bearing. (you are submitting a form)
It sounds like you want to make an ajax request to add the item to the cart and update that display on the site? That would be in the submit of the form and if jcart is dynamically created then yes, live is the answer.
$('.jcart').live('submit', function() {
// aggregate form elements into object and send via ajax
// update the cart on the page, since we haven't reloaded the page the light box is still displayed
});
Regarding comment:
When you send an ajax request, jquery takes an object as an argument. Such as $.post('urlToPostTo.php', { title: 'title of whatever', id: 5 } );
The server sees this the same as:
<form id="myForm" action="uroToPostTo.php" method="POST" >
<input type="text" name="title" value="title of whatever" />
<input type="hidden" name="id" value="5" />
<input type="submit" name="submit" value="submit" />
</form>
So if you were to aggregate the form inputs into an object, there's a few ways (even some jquery plugins to help you out). The primitive way would be:
var $form = $('#myForm'); // instead of finding myForm over and over, cache it as a variable to use
var objToSend = {};
objToSend.title = $form.find('input[name=title]').val();
objTosend.id = $form.find('input[name=id]').val();
$.post( 'urlToPostTo.php', objToSend );
A more Elegant solution is to have something loop through all form elements and put them into an object for you. Plugins like http://docs.jquery.com/Plugins:Forms make that a bit easier.
The end result is the form elements are stuffed into an object to send to your script.

Codeigniter: How to redirect properly with form validation

I understand how to do it w/ a plain form w/o existing values, but let's say I have a view that I can call via http://domain.com/account/settings. let's say I have two fields, username, password and city, which are all pulled from the DB (except for password of course). So, if a user tries to submit the form and fails validation for whatever reason, where should I "redirect" them to? Right now, I have it showing the same view but the problem is, it pulls the info from the DB again. Should I be creating two different views?
The second view would essentially show the information they tried to enter along w/ the error message.
You do not need two separate views. Check out Form Helper's functions set_value(), set_select(), set_checkbox() and set_radio(). These re-populate form after its submission and validation. So in your case, you should specify the fields this way:
<input type="text"
name="username"
value="<?php echo set_value('username', $user['username']); ?>" />
<input type="text"
name="city"
value="<?php echo set_value('city', $user['city']); ?>" />
By default, the input will have $user['city'] value. But after failed validation it will be re-populated with previously entered values (including incorrect ones).
Just remember that all fields you want to re-populate need to be passed through form_validation library:
$this->form_validation->set_rules('username', 'Username', 'required');
$this->form_validation->set_rules('city', 'City', '');
On the same controller you could have something like this:
if ($this->form_validation->run('create_comment') === TRUE)
{
$this->comments_model->name = $this->input->post('name', TRUE);
$this->comments_model->email = $this->input->post('email', TRUE);
$this->comments_model->website = $this->input->post('website', TRUE);
$this->comments_model->comment = $this->input->post('comment', TRUE);
$this->comments_model->create_comment();
redirect(current_url());
}
$this->load->view('your_view');
That's all there is to it.
The idea is to have it redirect to itself or wherever you want, when the validation returns 'true' so that we kind of refresh the page, hence, update the page.
If the validation returns 'false' then you won't have to do anything.
Redirect to the same form.
And in your view give error information to the visitor.
There are two ways you can do this.
Use this error in your view. This will show validation error info.
echo validation_errors('<p class="error">','</p>');
Or you can use flashdata()
In your controller
...
...
$this->session->set_flashdata('msg', 'All fields are required. or other useful info here. Please try again!');
redirect('yourcontroller');
And in your view, you need to show it.
<?php
if ($this->session->flashdata('msg')){ //change!
echo "<div class='message'>";
echo $this->session->flashdata('msg');
echo "</div>";
}
?>
Had the same problem and discovered that a redirection makes you lose the data that would have been provided by form_error(...) or validation_errors(), except you store such data in a session or in an array being passed into the loaded view.
The point to note is that you should redirect only if the data you want passed around is in session, else you should just load a view. The latter ensures that you have your validation errors intact when you reach the loaded view.
Just load same view if form validation failed
controller
$userData=array(
'username'=NULL,
'password'=NULL
);
#set form validation rules
$this->form_validation->set_rules('username', 'Username', 'required');
$this->form_validation->set_rules('password', 'Password', 'required');
#get all posted data ,this helps you in two ways,
# 1. Get all form data and can be use here server side
# 2. repopulating the form data by passing to view page
$userData=$this->input->post(NULL, TRUE);
#check form validation result
if ($this->form_validation->run() == TRUE) {
//do the operation
redirect(URL);
}else{
$this->load->view($view, $userData);
}
View page
<form method=post action='URL'>
<input type='text' name='username'value='<?php echo $username?>'/>
<?php echo (form_error('username')) ? form_error('username', "<div style='color:red'>", "</div>") : ""; ?>
<input type='text' name='password' value='<?php echo $password?>'/>
<?php echo (form_error('username')) ? form_error('username', "<div style='color:red'>", "</div>") : ""; ?>
<input type='submit' value='submit'/>
</form>
This code display form errors and repopulate the form

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