I have a Yii form which calls a render partial from another model (team has_many team_members). I want to call via ajax a partial view to add members in team/_form. All works (call, show, save) except for ajax validations (server and client side). If i submit form, member's model isn't validating, even in client side, it's not validating the required fields.
Any clue?
//_form
<?php $form=$this->beginWidget('CActiveForm', array(
'id'=>'team-form',
'enableAjaxValidation'=>true,
'enableClientValidation'=>true,
'clientOptions'=>array(
'validateOnSubmit'=>true,
'validateOnChange'=>true
),
'htmlOptions' => array('enctype' => 'multipart/form-data'),
)); ?>
//Controller
public function actionMember($index)
{
$model = new TeamMember();
$this->renderPartial('_member',array(
'model'=> $model, 'index'=> $index
)
,false,true
);
}
public function actionCreate()
{
$model=new Team;
$members = array();
if(isset($_POST['Team']))
{
$model->attributes=$_POST['Team'];
if(!empty($_POST['TeamMember'])){
foreach($_POST['TeamMember'] as $team_member)
{
$mem = new TeamMember();
$mem->setAttribute($team_member);
if($mem->validate(array('name'))) $members[]=$mem;
}
}
$this->redirect(array('team/create','id'=>$model->id,'#'=>'submit-message'));
}
$members[]=new TeamMember;
$this->performAjaxMemberValidation($members);
$this->render('create',array(
'model'=>$model,'members'=>$members
));
}
//_member
<div class="row-member<?php echo $index; ?>">
<h3>Member <?php echo $index+1; ?></h3>
<div class="row">
<?php echo CHtml::activeLabel($model, "[$index]name",array('class'=>'member')); ?>
<?php echo CHtml::activeTextField($model, "[$index]name",array('class'=>'member')); ?>
<?php echo CHtml::error($model, "[$index]name");?>
</div>
</div>
ProcessOutput was set to true. No dice.
Switch renderPartial() to render(). No dice.
If you will look at the CActiveForm::run:
$cs->registerCoreScript('yiiactiveform');
//...
$cs->registerScript(__CLASS__.'#'.$id,"jQuery('#$id').yiiactiveform($options);");
Then you will understand that you validation will not work, because you render partial and not the whole page. And these scripts show up at the bottom of the page. So you should solve this by execute these scripts.
After you partial is rendered, try to get activeform script which should be stored at the scipts array:
$this->renderPartial('_member',array('model'=> $model, 'index'=> $index));
$script = Yii::app()->clientScript->scripts[CClientScript::POS_READY]['CActiveForm#team-form'];
after, send it with rendered html to page:
echo "<script type='text/javascript'>$script</script>"
Also remember before you will append recieved html on the page you should include jquery.yiiactiveform.js, if you not already did it(by render another form, or registerCoreScript('yiiactiveform')), on the page from calling ajax request. Otherwise javascript error will raised.
Hope this will help.
Edit:
Sorry I'm not understood that you are render part of form and not the whole. But you validation will not work exactly with the same issue. Because jQuery('#$id').yiiactiveform($options); script was not created for the field.
The actual problem is that the ActiveForm saves its attributes to be validated in the "settings" data attribute. I see you are already using indexes so what you need to add the new elements to this settings object in order for the validation to work. After the ajax response this is what must be done:
//Get the settings object from the form
var settings = $("#form").data('settings');
//Get all the newly inserted elements via jquery
$("[name^='YourModel']", data).each(function(k, v) {
//base attribute skeleton
var base = {
model : 'YourModel',
enableAjaxValidation : true,
errorCssClass : 'error',
status : 1,
hideErrorMessage : false,
};
var newRow = $.extend({
id : $(v).attr('id'),
inputID : $(v).attr('id'),
errorID : $(v).attr('id') + '_em_',
name : $(v).attr('name'),
}, base);
//push it to the settings.attribute object
settings.attributes.push(newRow);
});
//update the form
$("#form").data('settings', settings);
```
This way the ActiveForm will be aware of the new fields and will validate them.
Well, setting processOutput to true in renderPartial (in order to make client validation works on newly added fields) will not help in this case since it will only work for CActiveForm form and you don't have any form in your _member view (only input fields).
A simple way to deal with this kind of problem could be to use only ajax validation, and use CActiveForm::validateTabular() in your controller to validate your team members.
Related
I would like a help to solve this problem. I'm using Yii 1.1 and trying to create a dynamic CheckBox list in which its elements change depending on the value selected in a DropDown element.
I created this DropDown element with the arguments in Ajax to perform the function update in the Controller. The function in the Controller is doing the lookup in the table according to the value passed.
Up to this point, the code is working fine, generating an array with the values that should be displayed. The problem is that I'm not able to configure the return of these data and I can't even display them in the View.
Below is the code data:
View - DropDown element:
echo CHtml::activeDropDownList($model, 'standard',
array(CHtml::listData(standard::model()->findAll(), 'name','name')),
array('empty'=>'',
'ajax'=>array(
'type'=>'POST',
'url'=>CController::createUrl('/getTeam',array('form'=>'Team','field'=>'standard')),
'update'=>'#Audit_team'),
)
);?>
Controller:
public function actionGetTeam($form,$field) {
$post = $_POST;
$standard = $post[$form][$field];
$Lists = TblAuditStandard::model()->findAll("standard = $standard");
foreach ($Lists as $List){
$AuditTeam[] = $List->name." - ".$List->login;
asort($AuditTeam);
foreach ($AuditTeam as $id=>$value )
echo CHtml::tag('option',array('value'=>$id),$value,true);
}
}
View - Checkbox element:
<div class="row">
<?php echo $form->labelEx($model,'team'); ?>
<?php echo CHtml::activeCheckBoxList($model,'team',array()); ?>
<?php echo $form->error($model,'team'); ?>
</div>
I hope someone can help me solve this problem. Thanks.
I was wondering if it is possible to repopulate a dynamic dropdown(specifically the option of a select) after form validation fails which is generated by another dropdown at on change event.
My jQuery is working good on populating dynamic option of a select as well as at server-side when fetching the data, my only problem is when the form is submitted and validation fails, so basically the dynamic select option/s will reset.
Can somebody help me with this issue?
You have 2 options here:
1) You manually set the 2x select fields with CI and repopulate/set them. You would construct these based on the POST values that were incorrect.
Assuming that list A populates list B which populates list C you may not want to do that. In which case you could define some hidden variables such as this:
var field1 = '<?php $_POST['field1']; ?>';
var field2 = '<?php $_POST['field2']; ?>';
var field3 = '<?php $_POST['field3']; ?>';
$(document).ready(function(){
$('#field1').val(field1).trigger('change'); // or whatever on() query event you use
$('#field2').val(field1).trigger('change'); // or whatever on() query event you use
$('#field3').val(field1).trigger('change'); // or whatever on() query event you use
})
Might help? You would do the trigger to then load whichever lists/ajax calls populate each select box.
Yes it is possible,
this example works with Codeigniter 4 but I'm quite sure it works also with Codeigniter 3. Here I wanted to do a multiple dropdown field with Select2
<div class="form-group">
<label>Categories</label>
<?php
$parameters = array('class' => 'form-control select2_cat', "multiple" => "multiple");
$options = array();
foreach ($categories as $cat) {
$options[$cat['name']] = $cat['name'];
}
echo form_dropdown(
'categories[]',
$options,
set_value('categories[]') != "" ? set_value('categories[]') : 0,
$parameters
);
?>
</div>
You see from the code, I'm using set_value() . I'm checking also if set_value is empty string.
You could find more info here: https://www.py4u.net/discuss/37330
I have a problem repopulating my form after validation fails. Problem is my url contains an additional uri which I access by clicking a link. This is what it looks like:
http://www.example.com/admin/trivia/add/5
At first trouble was that the segment 4 of the uri completely disappeared, so even though the validation errors showed and the form was repopulated, I lost my added uri.
Then I found in another question that the solution was to set form open like this:
echo form_open(current_url());
Problem is now it isn't showing any validation errors and the form is not repopulated. Is there a way to achieve this?
This is what my controller looks like:
function add()
{
$data = array('id' => $this->uri->segment(4));
if($_POST)
{
$this->_processForm();
}
$this->load->view('admin/trivia_form', $data);
}
And inside _processForm() I got all the validation rules, error message and redirecting in case success.
[edit] Here is my _processForm() :
function _processForm()
{
$this->load->library('form_validation');
//validation rules go here
//validation error messages
$this->form_validation->set_rules($rules);
$this->form_validation->set_error_delimiters('<div style="color:red">', '</div>');
if ($this->form_validation->run())
{
//get input from form and assign it to array
//save data in DB with model
if($this->madmin->save_trivia($fields))
{
//if save is correct, then redirect
}
else
{
//if not show errors, no redirecting.
}
}//end if validation
}
To keep the same url, you can do all things in a same controller function.
In your controller
function add($id)
{
if($this->input->server('REQUEST_METHOD') === 'POST')// form submitted
{
// do form action code
// redirect if success
}
// do your actual stuff to load. you may get validation error in view file as usual if validation failed
}
to repopulate the form fields you are going to need to reset the field values when submitting it as exampled here and to meke it open the same page you can use redirect() function as bellow:
redirect('trivia/add/5','refresh');
i don't know what you are trying to do, but try this to repopulate the form with the values user entered
<?php
echo form_input('myfield',set_value('myfield'),'placeholder="xyz"');
?>
I have form in yii that validates the form field. When I submit the form it shows the errors.
But when the value of the field with the validation error is updated, the error still present.
I want the message to clear. How should i clear the validation error?
Below the form widget code
<?php $form = $this->beginWidget('CActiveForm', array(
'id'=>'user-form',
'enableAjaxValidation'=>true
)); ?>
In my form I echo the validation error like the code below:
<?php echo $form->error($model, 'firstname'); ?>
I tried the solution from this problem
Trigger Yii field validation onchange of another field
$('#user-form').change(function(){
var settings = $(this).data('settings');
$.each(settings.attributes, function () {
this.status = 2; // force ajax validation
});
$(this).data('settings', settings);
// trigger ajax validation
$.fn.yiiactiveform.validate($(this), function (data) {
$.each(settings.attributes, function () {
$.fn.yiiactiveform.updateInput(this, data, $(this));
});
});
});
But the error message is still not cleared. I have confirmed that the ajax request is sent and there are response as its showed on the firebug console.
[EDIT]
It seems those validation errors for "select" fields are the ones that are not updated/cleared only.
[EDIT]
All the validation errors that are printed/echo after the form is submmitted will not disappear even if the value is supplied or change to satisfy the validation rules.
Place this just above the // trigger ajax validation comment:
$('.errorSummary, .errorMessage').hide();
This should reset the errors before they get re-validated.
In my case I added the code below on the page to remove the red highlight on input field on the form.
$('#user-form select, #user-form input').change(function(){
field = $(this).attr('id');
if($('#'+field+'_em').text() == ''){
$(this).removeClass('error');
}
});
I also add the updateInput function on framework/web/js/source/jquery.yiiactiveform.js so it will remove the validation error for certain field.
if(hasError == false){
$error.toggle(hasError);
$el2 = form.find('#' + attribute.id);
$el2.removeClass(attribute.errorCssClass);
}
I am not sure if this the proper solution but it works for me.
With your form widget code, i don't think ajax validation won't work.
To enable Ajax validation on form you should configure your widget as below,
<?php
$form = $this->beginWidget('CActiveForm', array(
'id'=>'user-form',
'enableAjaxValidation' => true,
'clientOptions' => array(
'validateOnSubmit' => true,
),
));
?>
With this configuration, your form will be validated when form field lost the focus(blur).
Try it, It will give the solution for your problem.
Usually, I just set a $feedback var or array and then check for that to display in my views.
However, it occurred to me I should perhaps use flashdata instead.
The problem is sometimes - for say an edit record form, I may simply want to reload the form and display feedback - not redirect. when i use flashdata, it shows but then it shows on the next request as well.
What would be the best practice to use here?
CodeIgniter supports "flashdata", or session data that will only be available for the next server request, and are then automatically cleared.
u use hidden field for that
I would use the validation errors from the Form validation class and load those directly to the view in its 2nd argument.
$this->form_validation->set_error_delimiters('<p>', '</p>');
$content_data = array();
if (!$this->form_validation->run()) {
$content_data['errors'] = validation_errors();
}
$this->load->view('output_page', $content_data);
Then check in your view whether $errors isset.
Controller:
$data['message'] = 'some message you want to see on the form';
$this->load->view('yourView', $data);
View:
if (isset ($message)) : echo $message; endif;
...