Process output in Yii Framework - php

I have an ajax submit button as follows,
echo CHtml::ajaxSubmitButton(
'>', $this->createUrl('/shop/category/nextCategory&id=16&store=true&gift_store='.$_GET['gift_store'].'&startValue='.$start_value.'&endValue='.$endValue), array(
'type'=>'GET',
'update'=>'#test',
'beforeSend' => 'function(){
alert("beforeSend");
}',
'complete' => 'function(){
alert("complete");
}',
)
);
In my action in the controller i use reder partial as follows,
$this->renderPartial('view',array('model'=>$this->loadModel()),false,true);
as i have set processOutput to true, the ajax submit button works fine. But it keeps on executing(for the number of times that i have clicked). I want it to execute only once when the submit button is clicked.
It'll be great if anyone could help me out with this. I have been trying to fix this all day.

okay i solved the solution. What is happening is,
The ajax request are triggered based on the ID. Every time the div is reloaded through ajax a new delegate/event will be loaded pointing at the same ID of the one before him. So when you actually click on the button both delegates will do their job, because both match the clicked button.
there are two solutions to this problem. One is having a unique id each time the button is generated like as follows,
array('id' => 'next-category-'.uniqid());
The other way, and this is how i solved it,
array("id"=>"next-category", "live"=>false);
live: boolean, whether the event handler should be attached with live/delegate or direct style. If not set, liveEvents will be used.Giving this to the submit button solved it.

Better if you create normalizeUrl for ajax request.
echo CHtml::ajaxSubmitButton(
'>', CHtml::normalizeUrl(array('/shop/category/nextCategory',array('store'=>1,'gift_store'=>1,'startValue'=>1,'endValue'=>1))), array(
'type'=>'GET',
'update'=>'#test',
'beforeSend' => 'function(){
alert("beforeSend");
}',
'complete' => 'function(){
alert("complete");
}',
)
);
This code is perfect, i have checked. It seems that their may be some error somewhere in your js code .

With the help of jQuery, you can define the following code to "complete" function:
function() {
$('input[type=submit]").attr("disabled", "disabled");
}
So on successful completion of your button will be disabled for re-push, if I understand you correctly.
Don't forget to modify the css selector input[type=submit] to the CSS selector your button.
Solution 2: Wap your CHtml::ajaxButton into
if (!Yii::app()->request->isAjaxRequest) {
}
Then when you load content via ajax your button will not be displayed on the page

Related

Yii: a JavaScript callback on successful form submit

I want a JavaScript callback on successful save of an active form. Where can I put the JavaScript callback?
Do I need to make the form AJAX in order to do this or is is doable with regular (non-AJAX) forms?
In fact, I am going to put this form inside an iframe (inside a jQuery UI modal dialog).
When the form in the iframe is saved I need to do two things:
close the dialog;
update the list of things in a <select> of the main HTML document.
Try to put ajax handler on form submit, here is a live code from my project as example:
echo CHtml::ajaxSubmitButton('submit',
array('index'),
array(
'type'=>'POST',
'data' => array(),
'success' => 'js:function(){/*callback*/}',
));
Please, remember to give this button an unique (prefix_random) id to avoid inconvenionces
You can add button to your form:
<?=CHtml::ajaxSubmitButton('Save', Yii::app()->createUrl('some_url'),
array('success' => 'function(response){afterSubmitForm(response);}'),
array('class' => 'btn btn-primary'));
?>
Where afterSubmitForm() js function, which get server response as parameter.

Yii - about jquery ajax

I am writing a shop application, and I have a question.
As you know ajax in Yii looks like
<?php
echo CHtml::ajaxLink(
'',
array("cart/add/id/$item->id"),
array(
'update'=>'#cart',
),
array('class' => "button_basket")
);
?>
This code updates the div with id = cart. How can I update the other elements on the page? For example on the Cart page, I am removing via ajax one item and I need to update the Total price, what is the best way to do this?
For this you'll need to write some javascript, preferably a function that will read the response from the server, and make changes in the ui accordingly.
If you go through the documentation for ajaxLink();, you'll see that the third parameter is ajaxOptions which takes the options of jQuery's ajax method. Therefore you can write a callback function for the success event. And in this function you can make changes to your ui.
Sample code:
<?php
echo CHtml::ajaxLink(
'',
array("cart/add/id/$item->id"),
array(
// 'update'=>'#cart', this will be ignored when you have success callback
'success'=>'js:successFunctionName',
),
array('class' => "button_basket")
);
?>
Somewhere in the current view you can use registerScript to output the function to the view:
Yii::app->clientScript->registerScript('scriptname','
function successFunctionName(data, textStatus, jqXHR){
// write awesome javascript to change ui
}
');

Ajax Submit in Yii

I am using a CActive Form and, earlier i was having a normal submit button. Then I changed it to an Ajax Button like as follows,
CHtml::ajaxSubmitButton('Continue',CController::createUrl('//shop/order/create&store=true'),array('type'=>'POST'));
This goes to the url when u refresh the page. But i want it to go to the page without page refresh. Any idea how do it?
Thanks in advance.
try this, with some ajax success functionallity...
echo CHtml::ajaxSubmitButton('BottonLabel',
Yii::app()->createUrl('your/url',array('calltype' => 'system')),
array('type' => 'POST',
'cache' => true or false,
'success' => 'js:function(data) {yourJSfunction(data)}'),
array('id' => 'the_id'));
or follow this link to the class-reference for yii
ajaxSubmitButton
try this..
array('contoller_name/action','ajax'=>'somevalue'));
//In your case
echo CHtml::ajaxSubmitButton('Continue',array('contoller_name/action','ajax'=>'yw0'));
Just keep in mind that you are obviously setting auto created widget ids as params (yw0). You should assign your own ids to widgets, otherwise another widget could have the same id and that can be hard to debug..

CKEditor + Yii loaded with AJAX : $_POST doesn't contain the updated value

in short:
i'm using Yii Framework
i have a one Ckeditor window on my page ( php/ yii framework - works fine)
when i hit a button, a new CKeditor window is being generated and shown through AJAX call
THE PROBLEM: this new CKEditor window correctly displays the text stored in the database BUT : when i hit "Save" (an ajax button generated together with the rest of the form) the values from this new CKeditor window will not save : CKeditor sends back the old values that it got from the database.
When i remove the Ckeditor and leave the plain <textarea> : everything is ok so i know that the controller is fine.
Please, anybody went through something like this?
Sounds like a typical post-AJAX JS binding issue. :) There are a few possibilities for how to fix it, depending on what is going wrong.
This post in the Yii forum should be money for you, it's where I got most of these suggestions:
http://www.yiiframework.com/forum/index.php?/topic/9341-ckeditor-widget-in-a-cactiveform/
Use a widgetized Yii extension which has already solved this problem (NHCKEditor?)
Add an onClick callback to the submit button which saves the CKEditor content to the hidden 'textarea' ('onclick'=>'CKEDITOR.instances.TEXTAREA_ID.updateElement()',
Use jQuery to get the data from the CKEditor iFrame to use... wherever. AJAX validation, etc.
Good luck!
You can let CKEDITOR update the textarea before validating, and clientside/ajax validation will work as expected:
<?php $form = $this->beginWidget('CActiveForm', array(
'enableAjaxValidation' => true, // one or both
'enableClientValidation' => true, // one or both
'clientOptions' => array(
'validateOnSubmit' => true, // optional
'beforeValidate' => new CJavaScriptExpression('function(form) {
for(var instanceName in CKEDITOR.instances) {
CKEDITOR.instances[instanceName].updateElement();
}
return true;
}'),
),
)); ?>

Question about a dialog form and submit errors

I have the link below that executes the action message/new which shows a
form inside a jqueryui modal dialog.
<div id="myDialog">
</div>
echo jq_link_to_remote('Enviar mensaje', array(
'url' =>
'mensaje/new?receptor='.$miembro->getId().'&tipo=0&estado=0',
'update' => 'myDialog',
'complete' => "jQuery('#myDialog').dialog({ width:375,
height:220, top:123,
resizable:false,
modal:true, autoOpen: false });
jQuery('#myDialog').dialog('open')"
));
The point: if there are submit errors the form is not showed inside the
dialog form but in a empty page (mensaje/create)..
Any idea?
Javi
I would like to give you a code example, but it is extremely hard to follow what you are trying to do with the poorly formatted post above.
If you want to keep the user on the page you will need to submit your data via AJAX. Check out the jQuery AJAX docs. It appears you are using PHP, this tutorial for submitting via AJAX with jQuery and PHP may be useful as well.

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