Zend Framework - using jquery dialog for popup form in a Zend controller/action
I am editing to try to put my primary question at the beginning of this post: 1) I am sitting at my customers/preferences page updating info in the form 2) I want to add a widget to my list of available widgets, so I need to open the widgets/addWidget page/form to add a new widget (note I am still working in my current customers/preferences page/form, so I need to continue with this page after adding a new widget in the widgets/addWidget page/form. 3) I select a link to go to widgets/addWidget, and I have jQuery dialog open this link in a dialog with the addWidget form and submit button 4) after adding a newWidget in the dialog popup form, I want to submit the newWidget (the code for addWidget is in the widgets/addWidget controller/action, not in the customers/preferences action)
Problem: the addWidget form action goes to widgets/addWidget, so when I submit the addWidget form in the dialog, it takes me away from my current customers/preferences page
Question 1) what is the proper/best way given this scenario to popup a form in another controller/action to submit that form, but then resume my current controller/action? Question 2) should I move all of my form and code from the widgets/addWidget controller action into my existing customers/preferences action? (that seems to be the wrong approach)
I have researched jQuery documentation, Zend documentation, and searched other threads with trusty Google and stackoverflow, but I am still struggling to figure out the right way to get this to work with Zend Framework and popup forms such as jQuery dialog.
I have a working Zend action/controller that using a Zend Form and view to display and process my form. I have also been able to use jQuery dialog to popup a window when I click the link to that controller/action.
My issue is with the proper way to get the dialog to display the form and still be able to submit and process the page. If I only load the #content tag in the dialog the form and submit button appear in the dialog box, but the submit button no longer works. If I let the dialog open the full page (not just the #content) now the form will process properly, but the form submit is set to go to my action page for processing, so the form submits and takes me away from the original page and to the real controller/action page instead.
In my customerController I have a preferencesAction where a customer can choose a widget from a list of widgets. When the customer needs to add a new widget to the list, I want to open the addWidgetAction from the WidgetsController in a popup form. I want to add the widget in the 开发者_Go百科popup form, submit the form to the addWidgetAction, and come back to the customer/preferences page I was already working in (with the newly added widget available in my list to select from).
//CustomerController
public function preferencesAction(){
//this is where I click a link to /widgets/addWidget and use jQuery dialog for form
}
//Customer preferences.phtml
<script>
$(document).ready(function() {
$('#add-link').each(function() {
var $link = $(this);
var $dialog = $('<div></div>')
.load($link.attr('href'))
.dialog({
autoOpen: false,
title: $link.attr('title'),
width: 600,
height: 500,
buttons: {
"Add Widget": function(){
$("#AddWidget").submit();
},
"Cancel": function(){
$(this).dialog("close");
}
}
});
$link.click(function() {
$dialog.dialog('open');
return false;
});
});
});
</script>
<a id="add-link" href='<?php echo $this->url(array('controller' => 'widgets',
'action' => 'addwidget')); ?>'>Add a Widget...</a>
//WidgetsController
public function addWidgetAction(){
// display form to add widget
// process form, validate, add to widgets table
$baseUrl = $this->getRequest()->getBasePath();
$action = $baseUrl . '/widgets/addWidget';
$form = new Form_Widget();
$form->setName('Add Widge')
->setAction($action);
}
I need to understand how to get the dialog to load my Zend action page for form processing, but without requiring the entire layout with header, footer, etc. I also need to understand how to process the form in the dialog popup without moving away from my original page where the popup was linked from.
Thank you for your assistance!
Well, I am not sure if i understand your question but you can try this
You can load your form in UI dialog box, to disable layout do this on you can do this
public function addwidgetAction(){
$this->_helper->layout()->disableLayout();
//just diable layout and do eveything normal
}
on your preferences.phtml file load the content of url baseurl/contrller/addwidget
to jquery modal dialog
UPDATE
One way is use a jqueryplugin fancybox (external src) check the forgot password link
The other way is to use ajax
UPDATE
Well now i read your question well, though don't understand so clearly, i understand that you are trying to do something that is quite ambitious. And you cannot achieve through normal load method. You need to use ajax
I have to admit that I don't like jquery ui modal dialog so i don't know much about it, and i usually use jquery fancy box where it would give me an ability to load external page. Maybe you can do this through jquery ui modal dialog box.
I guess the problem that you are facing right now is you open a dialog box, you get the form in dialog, and you submit the form, it sends the normal post request, so dialog is reloaded, that's what happens in the link i have above, just click on forgot password link
Well there's solution for that too, the solution is ajax, and i hope you are quite familiar with it. you send ajax request and get response back, and depending on your response, you can exit dialog box as well update the the parent (preferences controller page) but that is quite ..... I tried to so same with few months back, but not with zf, i don't know how much i accomplished but it sure gave me a nasty headache
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