I have a large Dijit-based form with many Dijits in collapsible TitlePanes.
When the form validates, any invalid items hidden inside closed TitlePanes (obviously) cannot be seen. So it appears as though the form is just dead and won't submit, though, unbeknownst to the user, there's actually an error hidden in a closed TitlePane which is preventing the form processing.
What's the solution here? Is there an easy way to simply open all TitlePanes containing Dijits that are in an error state?
If validation is done by following, it will work:-
function validateForm() {
var myform = dijit.byId("myform");
myform.connectChildren();
var isValid = myform.validate();
var errorFields = dojo.query(".dijitError");
errorFields.forEach(fieldnode){
var titlePane = getParentTitlePane(fieldnode);
//write a method getParentTitlePane to find the pane to which this field belongs
if(titlePane) {
titlePane.set('open',true);
}
}
return isValid;
}
function getParentTitlePane(fieldnode) {
var titlePane;
//dijitTitlePane is the class of TitlePane widget
while(fieldnode && fieldnode.className!="dijitTitlePane") {
fieldnode= fieldnode.parentNode;
}
if(fieldnode) {
mynode = dijit.getEnclosingWidget(fieldnode);
}
return titlePane;
}
Lets say if the following is the HTML and we call the above validateForm on submit of form.
<form id="myform" data-dojo-type="dijit/form/Form" onSubmit="validateForm();">
......
</form>
Here's what I ended up doing (I'm not great with Javascript, so this might sucked, but it works -- suggestions for improvement are appreciated):
function openTitlePanes(form) {
// Iterate through the child widgets of the form
dijit.registry.findWidgets(document.getElementById(form.id)).forEach(function(item) {
// Is this a title pane?
if (item.baseClass == 'dijitTitlePane') {
// Iterate the children of this title pane
dijit.registry.findWidgets(document.getElementById(item.id)).forEach(function(child) {
// Does this child have a validator, and -- if so -- is it valid?
if (!(typeof child.isValid === 'undefined') && !child.isValid()) {
// It's not valid, make sure the title pane is open
item.set('open', true);
}
});
}
});
}
I'm basically trying to create a multi-step form using the CActiveForm class in Yii. The idea is I want to use the built-in functionality to achieve this in the simplest way possible. The requirement I have is as follows:
A multi step ONE PAGE form (using DIVs that show/hide with jQuery)
AJAX validation on EACH step (validate step-specific attributes only)
The validation MUST work using the validateOnChange() and validateOnSubmit() methods
This is a half-working solution I have developed so far:
View:
<div class="form">
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'listing-form',
'enableClientValidation'=>false,
'enableAjaxValidation'=>true,
'clientOptions'=>array(
'validateOnChange'=>true,
'validateOnSubmit'=>true,
'afterValidate'=>'js:validateListing',
),
)); ?>
<?php echo $form->errorSummary($model); ?>
<div class="step" id="step-1">
// model input fields
<?php echo CHtml::submitButton('Next Step', array('name'=>'step1')); ?>
</div>
<div class="step" id="step-2" style="display: none;">
// model input fields
<?php echo CHtml::submitButton('Next Step', array('name'=>'step2')); ?>
</div>
<div class="step" id="step-3" style="display: none;">
// model input fields
<?php echo CHtml::submitButton('Submit', array('name'=>'step3')); ?>
</div>
<?php $this->endWidget(); ?>
</div>
JavaScript:
function validateListing(form, data, hasError)
{
if(hasError)
{
// display JS flash message
}
else
{
if($('#step-1').css('display') != 'none')
{
$('#step-1').hide();
$('#step-2').show();
}
else if($('#step-2').css('display') != 'none')
{
$('#step-2').hide();
$('#step-3').show();
}
else if($('#step-3').css('display') != 'none')
{
return true; // trigger default form submit
}
}
}
Controller:
public function actionCreate()
{
$model = new Listing;
// step 1 ajax validation
if(isset($_POST['step1']))
{
$attributes = array('name', 'address1', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// step 2 ajax validation
if(isset($_POST['step2']))
{
$attributes = array('category', 'type', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// step 3 ajax validation
if(isset($_POST['step3']))
{
$attributes = array('details', 'source', 'etc');
$this->performAjaxValidation($model, $attributes);
}
// process regular POST
if(isset($_POST['Listing']))
{
$model->attributes = $_POST['Listing'];
if($model->validate()) // validate all attributes again to be sure
{
// perform save actions, redirect, etc
}
}
$this->render('create', array(
'model'=>$model,
));
}
protected function performAjaxValidation($model, $attributes=null)
{
if(isset($_POST['ajax']) && $_POST['ajax']==='listing-form')
{
echo CActiveForm::validate($model, $attributes);
Yii::app()->end();
}
}
To summarise. Basically what I have is a form with 3 submit buttons (one for each step). In my controller I check which submit button was pressed and I run AJAX validation for the attributes specific to that step.
I use a custom afterValidate() function to show/hide the steps upon submit. On step 3, the default form submit is triggered, which posts all the form attributes to the controller.
This works well, except it won't work with validateOnChange() (since the submit button doesn't get posted). Also I was wondering whether this is actually the best way to do this, or if anyone knows of a better way?
Thanks.
I'd suggesting using scenarios to turn on and off the appropriate rules. Adjust the model scenario based on what is sent to your controller.
Note: this may also be a really good place to use a CFormModel instead of a CActiveRecord, depending on what is in your form.
Edit: can you add a hidden field to each div section that contains the info about what step you are on? Seems like that should work instead of your submit buttons.
OPTION 1
When you do not receive a button, why not validate the entire form, why do you need to validate only specific attributes? Yii will validate the entire model, send back all the errors but only that particular error will be shown by the active form because that is how it works already.
OPTION 2
You can have 3 forms (not 1 like you have now), 1 on each step. Also create 3 scenarios 1 for each step.
Each form has a hidden field that gets posted with the form, it can actually be the scenario name just validate it when it comes in. Validate the model using this hidden field to set the scenario you are on.
You can cache parts on the model when the form is submitted successfully and at the end you have the complete model.
you can always have custom validation and it won't break your normal form validation
in your model
private $step1 = false;
private $step2 = false;
private $all_ok = false;
protected function beforeValidate()
{
if(!empty($this->attr1) && $this->attr2) // if the fields you are looking for are filled, let it go to next
{
$this->step1 = true;
}
if($this->step1)
{
... some more validation
$this->step2 = true;
}
if($this->step2)
{
... if all your logic meets
$this->all_ok = true;
}
// if all fields that your looking for are filled, let parent validate them all
// if they don't go with their original rules, parent will notify
if($this->all_ok)
return parent::beforeValidate();
$this->addError($this->tableSchema->primaryKey, 'please fillout the form correctly');
return false;
}
I think better create specific class for each step of validation and use scenarios with rules. Below is small example.
//protected/extensions/validators
class StepOneMyModelValidator extends CValidator
{
/**
* #inheritdoc
*/
protected function validateAttribute($object, $attribute)
{
/* #var $object YourModel */
// validation step 1 here.
if (exist_problems) {
$object->addError($attribute, 'step1 is failed');
}
...
Create other classes(steps) for validation...
// in your model
public function rules()
{
return array(
array('attr', 'ext.validators.StepOneMyModelValidator', 'on' => 'step1'),
...
How to use in controller:
$model = new Listing();
$steps = array('step1', 'step2', /* etc... */);
foreach($_POST as $key => $val) {
if (in_array($key, $steps)) {
$model->setScenario($key);
break;
}
}
$model->validate();
echo '<pre>';
print_r($model->getErrors());
echo '</pre>';
die();
Or we can validate all steps in one validator.
I made a "My bookmarks" tab on the user profile page using Views. The tab shows nodes the user has flagged.
However - "My bookmarks" should only be visible on the user's own profile page and at the moment the "My bookmarks" tab is visible on every profile a user visits. How do I check whether the current user matches the profile being viewed? I tried that from the View interface, but the access permissions don't have any options that work.
EDIT:
I think it is this code, but I still need some guidelines as to how to implement that:
<?php
global $user;
if (arg(0) == 'user' && $user->uid == arg(1)){
return TRUE;
}
else {
return FALSE;
}
?>
I also found this module, I think it helps a lot Views Access Callback
I managed to solve this using the code and module from above.
The custom module contains this code
<?php
function MYMODULE_views_access_callbacks() {
return array(
'MYCALLBACK_user_has_access' => t('User can only see tab on his own profile'));
}
function MYCALLBACK_user_has_access() {
global $user;
if (arg(0) == 'user' && $user->uid == arg(1)){
return TRUE;
}
else {
return FALSE;
}
}
?>
The Views Access Callback module adds your callback to the Views interface and from there, you can use it for your own view.
I am a newbie interested to know are there any extension to count views on click of a button as to know no. of registered users or visiters to web page to know the view count on click of a image is there any extension.
Plz let me know if any
thanx :)
I think , there is no need of any extension. Make a Ajax call on click button or image you are interested.
Improved:
I supposed you have Site as controller and index as action. then, please keep this code on views/site/index.php .
Yii::app()->clientScript->registerScript('logo_as_image_script', '$(document).ready(function() {
$("#logo_as_image").click(function() {
$.post("'.Yii::app()->createAbsoluteUrl('site/index').'",
{
clicked: "1"
},
function(data, status) {
alert("Data: " + data + "\nStatus: " + status);
});
});
});');
Yii::app()->clientScript->registerCoreScript('jquery');
echo CHtml::image(Yii::app()->baseUrl . '/images/logo.png', 'Logo as Image', array('id' => 'logo_as_image'));
And, keep this code on SiteController.php .
public function actionIndex()
{
// keep record of data ; do more filtering ; other manupulation
if(isset($_POST['clicked'])){
$nextCount = Yii::app()->user->getState('clickCount')+1;
Yii::app()->user->setState('clickCount',$nextCount );
echo $nextCount;
Yii::app()->end();
}
#other codes here.
$this->render('index');
}
Lets assume that you want to store how many registered users have accessed the page at :
www.something.com/something/someaction
then visit the controller and add the code like so :
public function actionSomeAction()
{
$model = new CountDbModel();
if(!Yii::app()->user->isGuest){
$model->page = 'This page name here.';
$model->user_id = Yii::app()->user->id;
$model->count = #Add the value here.
#You other code here....
$this->render('whateverView',array('model'=>$blah));
}
}
I hope it helped.
I have created a module and assigned that to custom layout (Route - product/category).
In this page i need to show only module contents.
So is there any way to do check if module is assigned to current layout ?
SOmething like below,
if ($current module = "Product_list") {
// Dont display products
}
else {
// Else display products
}
If you're coding this into the module code itself, you just need to check the current route
if(!empty($this->request->get['route']) && $this->request->get['route'] == 'product/category') {
... Code ...
} else {
... Code ...
}