I have a same field in foreach loop like below
foreach ( $subCategoryData as $k => $val) {
<?= $form->field($model, 'sub_category', ['template' => '{input}'])->textInput(['maxlength' => 255, 'class' => 'form-control required section_name', 'name' => "Category[sub_category][$k][name]"]) ?>
} ?>
I have ajax validation with custom method it is working fine.
But it is Working with only first input. Because it has same ID.
But when I changed it with 'inputOptions' => ['id' => 'myCustomId'] and make it unique with below and my ajax validation is not called.
foreach ( $subCategoryData as $k => $val) {
<?= $form->field($model, 'sub_category', ['template' => '{input}','inputOptions' => ['id' => "category-sub_category_".$k]])->textInput(['maxlength' => 255, 'class' => 'form-control required section_name', 'name' => "Category[sub_category][$k][name]"]) ?>
}
I have seen this solution here
https://github.com/yiisoft/yii2/issues/7627
and also seen this https://stackoverflow.com/a/28460442/2286537
But nothing work
can anyone help me ?
Your question is different from the posts you introduced.
You should use loadMultiple.
Example:
if (\Yii::$app->request->isAjax) {
if (\yii\base\Model::loadMultiple($model,\Yii::$app->request->post())) {
\Yii::$app->response->format = Response::FORMAT_JSON;
echo json_encode(ActiveForm::validateMultiple($model));
\Yii::$app->end();
}
}
if ( \yii\base\Model::loadMultiple($model, Yii::$app->request->post()) && \yii\base\Model::validateMultiple($model)) {
foreach ($model as $models) {
$models->save(false);
}
in view:
<?php $form = ActiveForm::begin([
'enableAjaxValidation' => true,
]);
Related
I have a form like follow
<?php widgets\Pjax::begin(['id' => 'authors', 'enablePushState' => false]) ?>
<?php $form = ActiveForm::begin(['id' => $model->formName(),'enableAjaxValidation'=>true,'validationUrl'=>\yii\helpers\Url::toRoute(['paper-author/validation']), 'options' => ['data-pjax' => '']]); ?>
<?php echo $form->field($model, 'FirstName')->textInput()->input('string', ['placeholder' => Yii::t('app', ' FirstName...')])->label(false) ?>
<?php echo $form->field($model, 'LastName')->textInput()->input('string', ['placeholder' => Yii::t('app', ' LastName...')])->label(false) ?>
<?= Html::submitButton('Save', ['class' => 'btn btn-success', 'id' => 'btn1']) ?>
$widget = Yii::createObject([
'class' => 'yii\grid\GridView',
'dataProvider' => $dataprovider,
'columns' => [
'FirstName',
'LastName',
[
'class' => 'yii\grid\ActionColumn',
]
],
]
);
echo $widget->run();
<?php $form = ActiveForm::end(); ?>
<?php widgets\Pjax::end() ?>
and in my model I have some rules:
[['FirstName', 'LastName'],'required']
my controller:
public function actionValidation()
{
$model = new PaperAuthor();
$requestPost = Yii::$app->request->post();
if (Yii::$app->request->isAjax && $model->load($requestPost)) {
Yii::$app->response->format = 'json';
return ActiveForm::validate($model);
}
}
public function actionDelete($id)
{
$this->findModel($id)->delete();
echo Json::encode([
'success' => true,
'messages' => [
Yii::t('app', 'Successfully Deleted.')
],
]);
}
actionCreate works well. but I can't delete any recored from gridview, it throws an validation error(required validation).
what can I do?
thanks in advance.
I am very new to Ajax.In ajax function, I cannot show the variable $competition_id content.I am using following codes but it show the ID only.
function custom_fixture_template(){
$competition_id = $_POST[ 'competion' ];
echo $competition_id;
$args = array(
'post_type' => array( 'football_fixture' ), // profile and letter are CPTs
'posts_per_page'=>5,
'meta_key' => 'pb_match_date',
'orderby' => 'meta_value',
'order' => 'DESC',
'tax_query' => array(
array(
'taxonomy' => 'competition',
'field' => 'id',
'terms' => $competition_id
),
),
'meta_query' => array(
array(
'key' => 'pb_match_status',
'value' => 'fixture'
),
)
);
$my_query = null;
$my_query = new WP_Query($args); ?>
}
add_action("wp_ajax_my_custom_fixture_template", "custom_fixture_template");
add_action("wp_ajax_norpiv_my_custom_fixture_template", "custom_fixture_template");
Ajax Code :
jQuery(".fixture-competition").change(
function(){
jQuery.ajax({
url:"<?php echo admin_url();?>admin-ajax.php",
type:"post",
data:{"action":"my_custom_fixture_template",competion:jQuery(this).val()},
success:function(res){
jQuery('.competition-container').remove();
jQuery('.new-fixture').html('');
jQuery('.new-fixture').append(res);
}
});
});
HTML Codes:
<select name="fixture-competition" class="fixture-competition">
<?php
$args = array(
'type' => 'football_fixture',
'child_of' => 0,
'parent' => '',
'orderby' => 'name',
'order' => 'ASC',
'hide_empty' => 1,
'hierarchical' => 1,
'exclude' => '',
'include' => '',
'number' => '',
'taxonomy' => 'competition',
'pad_counts' => false );
$competition_fixture = get_categories($args); ?>
<option value="Select All">--Select--</option>
<?php
foreach ($competition_fixture as $competition) { ?>
<option value="<?php echo $competition->term_id; ?>">
<?php $url = get_term_link($competition);?>
<?php echo $competition->name; ?>
</option>
<?php
}
?>
</select>
Your php function doesn't return anything. So the jQuery script has nothing to append to .new-fixture.
You can do something like
$my_query = new WP_Query($args); // be carefull of typo (a php closing tag)
echo json_encode($my_query);
then in the ajax response, you'll be able to parse the json response.
success:function(res){
var json = $.parseJSON(res);
console.log(json);
...
}
Hope it gives you some hints...
I have solved it by myself using following codes:
<div class="team-league-name-top">
<?php
$competition_name = get_term( $competition_id, 'competition' );
echo $competition_name->name;
?>
</div>
I am working on a CakePHP 3 project which is having a Apply Coupon form.
I want to apply the coupon using Ajax.
The view of coupon form is
<?= $this->Form->create(null, [
'url' => ['controller' => 'Coupons', 'action' => 'checkCoupon'],
'name' => 'checkCoupon'
]) ?>
<?= $this->Form->input('coupon_code', [
'type' => 'text',
'placeholder' => 'Apply Coupon Code',
'label' => false
]) ?>
<?= $this->Form->submit('Apply Coupon') ?>
and the checkCoupon action in CouponsController is
public function checkCoupon()
{
$this->request->onlyAllow('ajax'); // No direct access via browser URL
if ($this->request->is('post')) {
$couponCode = $this->request->data['coupon_code'];
$couponCheck = $this->Coupons->find('all', [
'conditions' => [
'coupon_code' => $couponCode
]
]);
if ($couponCheck->count() === 1) {
$coupon = $couponCheck->first();
$valid_till = $coupon->valid_till;
$dt = new Time($valid_till);
$date = $dt->format('Y-m-d');
if ($date >= date('Y-m-d')) {
echo 'Coupon is Valid. Discount of '.$coupon->value.'has been applied';
} else {
echo 'Coupon is Expired';
}
} else {
echo 'This is not a valid coupon code';
}
}
}
I want the $coupon->value and $coupon->id to be retrieved and added to the checkout link as
<?= $this->Html->link(__('Confirm Checkout'), ['controller' => 'ServiceRequests', 'action' => 'confirmCheckout', $service->id, $primaryAddressId, $serviceArea->id, $coupon->id], ['class' => 'btn btn-block btn-success']) ?>
The Apply Coupon form is in checkout action of RequestsController
Also the form is working well. I have checked it by removing the onlyAllow('ajax') line and printing values in check_coupon.ctp view.
How could I do it using Ajax ?
Edit 2 : checkout.ctp
<div class="form-info coupon">
<?= $this->Form->create(null, [
'url' => ['controller' => 'Coupons', 'action' => 'ajax_checkCoupon'],
'name' => 'checkCoupon',
'id' => 'checkCoupon'
]) ?>
<?= $this->Form->input('coupon_code', [
'type' => 'text',
'placeholder' => 'Apply Coupon Code',
'label' => false
]) ?>
<label class="hvr-sweep-to-right">
<?= $this->Form->submit('Apply Coupon', ['id' => 'applyCoupon']) ?>
</label>
<label id="couponUpdate"></label>
<label id="loading" style="display:none;">Loading...</label>
<?php
$data = $this->Html->script('#checkCoupon')->serializeForm(['isForm' => true, 'inline' => true]);
$this->Html->script('#checkCoupon')->event(
'submit',
$this->Html->script(
[
'controller' => 'Coupons',
'action' => 'ajax_checkCoupon'
],
[
'update' => '#couponUpdate',
'data' => $data,
'async' => true,
'dataExpression' => true,
'before' => "$('#loading').fadeIn();$('#applyCoupon').attr('disabled','disabled');",
'complete' => "$('#loading').fadeOut();$('#applyCoupon').removeAttr('disabled');"
]
)
);
?>
</div>
Error : Call to a member function serializeForm() on string on line 18 in checkout.ctp
So basically I have a view action in my users controller where the user can modify his information(username,first name, last name, email..) this form sends to another update action which does the saving stuff, problem is that when I submit the form and one or more fields don't meet the validation rules it doesn't show underneath each fields but the data doesn't save and
$this->User->validationErrors
outputs the errors.
my update action (accessed after submiting the form on view.ctp)
view.ctp:
<?php
echo $this->Form->create('User', array(
'inputDefaults' => array(
'div' => 'form-group',
'wrapInput' => false,
'class' => 'form-control'
),
'class' => 'well',
'url'=>array('controller'=>'users','action'=>'update'),
'id'=>'info-form'
));
?>
<fieldset>
<legend>Personal Information</legend>
<?php
echo $this->Form->input('id', array('value' => $userinfo['User']['id']));
echo $this->Form->input('User.username', array(
'label' => 'Username',
'value'=>$userinfo['User']['username']
));
?>
<td><?php echo $this->Form->error('username'); ?></td>
<?php
echo $this->Form->input('User.email', array(
'label' => 'E-mail',
'value'=>$userinfo['User']['email']
));
?>
<?php
echo $this->Form->input('User.fname', array(
'label' => 'First name',
'value'=>$userinfo['User']['fname']
));
?>
<?php
echo $this->Form->input('User.lname', array(
'label' => 'Last name',
'value'=>$userinfo['User']['lname']
));
?>
<?php
echo $this->Form->submit('Update', array(
'div' => 'form-group',
'class' => 'btn btn-success'
));
?>
</fieldset>
<?php echo $this->Form->end(); ?>
update function:
function update() {
$this->autoRender = false;
$this->User->set($this->request->data);
if ($this->request->is('post')) {
if ($this->User->save($this->request->data)) {
$this->Session->setFlash(__('Information updated Successfuly.'), 'alert', array(
'plugin' => 'BoostCake',
'class' => 'alert-success'), 'success');
return $this->redirect('/users/view/' . $this->request->data['User']['id']);
} else {
// $errors = $this->User->validationErrors; var_dump($errors);die;
$this->Session->setFlash(__('An error occured'), 'alert', array(
'plugin' => 'BoostCake',
'class' => 'alert-danger'), 'danger');
return $this->redirect('/users/view/' . $this->request->data['User']['id']);
}
} else {
$this->Session->setFlash(__('Request was not of POST type.'), 'alert', array(
'plugin' => 'BoostCake',
'class' => 'alert-danger'), 'danger');
return $this->redirect('/users/index/');
}
It's because you're redirecting after - that will make it lose the validation warnings.
In YII views folder i have test module and admin.php file to manage contents are below and i render form here where i put the code of form and dropdown in it , i want that grid refresh value of status change in dropdown
Suppose i select "Approved" than Grid show the data where status is approved
<?php
Yii::app()->clientScript->registerScript('dropdown', "
$('.dropdown-form form').submit(function(){
$('#testimonial-grid').yiiGridView('update', {
data: $(this).serialize()
});
return false;
});
");
?>
<h1>Manage Testimonials</h1>
<div class="dropdown-form">
<?php $this->renderPartial('_dropdownform',array(
'model'=>$model,
)); ?>
</div><!-- search-form -->
<?php $this->widget('zii.widgets.grid.CGridView', array(
'id'=>'testimonial-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
'id',
'created_by',
'test_name',
'test_email',
'comments',
'created_at',
/*
'status',
'approved_on',
'approved_by',
*/
array(
'class'=>'CButtonColumn',
),
),
)); ?>
Form below is _dropdownform , it contain a form and dropdown from this dropdown i am choosing the value of status
<div class="wide form">
<?php
$form = $this->beginWidget('CActiveForm', array(
'action' => Yii::app()->createUrl($this->route),
'method' => 'get',
));
?>
<div class="row">
<?php
echo CHtml::dropDownList('status', '', array(0 => 'New', 1 => 'Approved', 2 => 'Declined'), array(
'prompt' => 'Select Status',
'ajax' => array(
'type' => 'POST',
'url' => Yii::app()->createUrl('testimonial/loadthedata'),
//or $this->createUrl('loadcities') if '$this' extends CController
'update' => '#testimonial-grid', //or 'success' => 'function(data){...handle the data in the way you want...}',
'data' => array('status' => 'js:this.value'),
)));
?>
</div>
<div class="row buttons">
<?php //echo CHtml::submitButton('Search'); ?>
</div>
<?php $this->endWidget(); ?>
</div><!-- search-form -->
AND THE CODE IN MY CONTROLLER OR URL GIVEN IN DROPDOWN TO FETCH DATA IS FOLLOWING ACTION BUT I DONT KNOW HOW TO FETCH DATA FROM THIS FUNCTION AND PASS TO GRID VIEW
public function actionloadthedata() {
if (isset($_POST['status'])) {
$status = $_POST['status'];
if($status==0){
$status='New';
}
if($status==1){
$status='Approved';
}
if($status==2){
$status='Declined';
}
Testimonial::model()->findByAttributes(array('status'=>$status));
}
}
You can use CGridView property filterCssClass to link the grid filter, for example
$this->widget('CGridView', array(
'id' => 'my-list',
'filterCssClass' => '#filterFormId .filter',
And there is filter form
<?php $form = $this->beginWidget('CActiveForm', array(
'id' => 'filter-fomr-id',
)); ?>
<div class="filter clearfix">
<?php echo $form->dropDownList($model, 'name', [0=>'all', '1'=>'some else']); ?>
</div>
Replace #filterFormId .filter on jquery selector specific to you form. In other words, set id attribute for filter form, then use "#THISID .row".
In your gridview file, make sure you have this code:
Yii::app()->clientScript->registerScript('search', "
$('.search-button').click(function(){
$('.search-form').toggle();
return false;
});
$('.search-form form').submit(function(){
$('#ad-grid').yiiGridView('update', {
data: $(this).serialize()
});
return false;
});
");
then in the CGridView definition:
$this->widget('zii.widgets.grid.CGridView', array(
'id'=>'testimonial-grid',
'dataProvider'=>$model->search(),
'filter'=>$model,
'columns'=>array(
...
array(
'name'=>'Status',
'filter'=>CHtml::dropDownList('YourModel[status]', $model->status, array(0 => 'New', 1 => 'Approved', 2 => 'Declined'), array('empty' => '--all--') ),
'value'=>'( $data->status == 0) ? "New": ( $data->status == 1) ? "Approved" : "Declined"',
'htmlOptions' => array(
'style' => 'width: 40px; text-align: center;',
),
),
...
array(
'class'=>'CButtonColumn',
),
),
));
// CGridView
In order to save if/else in the 'value' section, you can implement a method in your model that returns the string associated to the integer.
It works great, just with the default Yii admin.php view, which you can edit as much as you need.
Update
Added support for empty status, does not filter query results
Thanks #Alex for your help but i am successful to make filterdropdown for grid , code is following but will you please tell me that i want that grid show only values where status=New , how i can do when page load grid show values where status is New
but first i paste the working code of dropdown filter for grid
Here is my admin.php file
<?php
$this->breadcrumbs = array(
'Testimonials' => array('index'),
'Manage',
);
$this->menu = array(
array('label' => 'List Testimonial', 'url' => array('index')),
array('label' => 'Create Testimonial', 'url' => array('create')),
);
?>
<h1>Manage Testimonials</h1>
<!-----------drop down form------------->
<?php
Yii::app()->clientScript->registerScript('dropdownfilter', "
$('.dropdown-form form #staticid').change(function(){
$.fn.yiiGridView.update('testimonial-grid', {
data: $(this).serialize()
});
return false;
});
");
?>
<div class="dropdown-form">
<?php
$this->renderPartial('_dropdownfilter', array(
'model' => $model,
));
?>
</div><!-- search-form -->
<?php
$this->widget('zii.widgets.grid.CGridView', array(
'id' => 'testimonial-grid',
'dataProvider' => $model->search(),
// 'filter' => $model,
'columns' => array(
'id',
'created_by',
'test_name',
'test_email',
'comments',
'created_at',
'status',
array(
'class' => 'CButtonColumn',
),
),
));
?>
Here is my render partial form where i place the static drop dow
<div class="wide form">
<?php
$form = $this->beginWidget('CActiveForm', array(
'action' => Yii::app()->createUrl($this->route),
'method' => 'get',
));
?>
<div class="row">
<?php
echo CHtml::dropDownList('staticid', '', array('0' => 'New', '1' => 'Approved', '2' => 'Declined'), array(
// 'onChange' => 'this.form.submit()',
'ajax' => array(
'type' => 'POST', //request type
)));
?>
</div>
<?php $this->endWidget(); ?>
And Here is code of my adminaction in controller
public function actionAdmin() {
$model = new Testimonial('search');
$model->unsetAttributes(); // clear any default values
if (isset($_GET['staticid'])) {
$getStatus = $_GET['staticid'];
if ($getStatus == 0)
$status = 'New';
if ($getStatus == 1)
$status = 'Approved';
if ($getStatus == 2)
$status = 'Declined';
$model->status = $status;
}
if (isset($_GET['Testimonial']))
$model->attributes = $_GET['Testimonial'];
$this->render('admin', array(
'model' => $model,
));
}
Now i want that when i actionadmin triggered first time it show status=New values