how do i get the user's input value and submit it to ajax then to controller? Right now, in my controller it says that id is an undefined index in $_POST['id']
This input textfield is actually within another form, kind of like a form inside a form.
<?php echo $form->textField($model,'email',array('id'=>'email')); ?>
<?php echo $form->error($model,'email'); ?>
<?php echo CHtml::ajaxButton ( 'Request Code',
CHtml::normalizeUrl(array('site/requestResetCode', 'render'=>true),
array (
'type' => 'POST',
//'data'=> array('id'=> 'js:$("#ResetPasswordForm_email").val()'),
'data'=> array('id'=> 'js:$("#email").val()'),
'success'=>'function(data){
if(data.status == "sent")
bootbox.alert("Code is sent. Please check your email.");
else (data.status == "failed")
bootbox.alert("Request Failure");
}',
'error'=> 'function (xhr, ajaxOptions, thrownError) {
alert(xhr.status);
alert(thrownError);
}',
)
))
?>
controller:
public function actionRequestResetCode()
{
$id = $_POST['id'];
//stuff..
your POST array is empty because value is not being sent. You need to make a javascript function like
<script type="text/javascript">
function getVal()
{
var value= $("#myEmail").attr('value');
alert(value);
return value;
}
</script>
check if the alert shows correct value then remove this alert statement. If it does not show the value then there is a problem with the id of your textfield.
If you get it working then use it like
'data'=> array('id'=> 'js:getVal()'),
Related
Using the beforeSubmit function of yii.activeForm.js, how can I make it perform a normal form submit when validation passes?
I have tried the following:
$('.ajax-form').on('beforeSubmit', function (event) {
var form = $(this);
var url = form.attr('action');
var type = form.attr('method');
var data = form.serialize();
$.ajax({
url: url,
type: type,
data: data,
success: function (result) {
if (result.errors.length != 0) {
form.yiiActiveForm('updateMessages', result.errors, true);
}
else if (result.confirmed == true) {
$('.confirm-panel').show();
}
else {
return true;
}
},
error: function() {
alert('Error');
}
});
// prevent default form submission
return false;
});
Controller:
public function actionProcess()
{
$model = $this->findModel($id);
if (Yii::$app->request->isAjax) {
$return_array = [
'errors' => [],
'confirmed' => false,
];
Yii::$app->response->format = Response::FORMAT_JSON;
$return_array['errors'] = ActiveForm::validate($model);
if ($model->confirm == 1) {
$return_array['confirmed'] = true;
}
return $this->asJson($return_array);
}
if ($model->load(Yii::$app->request->post()) && $model->save()) {
return $this->redirect(['success']);
}
return $this->render('process', [
'model' => $model,
]);
}
As you can see I am also trying to return additional data in my AJAX response. The problem I am having is the return true in the ajax success isn't working. I can't seem to break out of the function. I have also tried form.submit() here but this just does a submit loop via AJAX.
By the way I am not using enableAjaxValidation because I have some additional custom validation that happens in my controller. So this is why I have created my own custom handler for this.
First of all, you can't return true from within an ajax success function to continue form submission as it is javascript and the last line return true is already executed before the response is received so the form ain't going to submit by returning true inside the success function.
You need to use the event afterValidate if you want to submit your page manually after successful ajax validation rather than using beforeSubmit as it will go into an infinite loop if you try to submit the form using $("form").submit() inside the ajax success function. so change your line
$('.ajax-form').on('beforeSubmit', function (event) {
to
$('.ajax-form').on('afterValidate', function (event) {
and then change your success function to
success: function (result) {
if (result.errors.length != 0) {
form.yiiActiveForm('updateMessages', result.errors, true);
}
else if (result.confirmed == true) {
$('.confirm-panel').show();
}
else {
form.submit();
}
},
Hope it helps you out.
Input validation should be made in models no matter if it's built in or custom. Then you can easily use the default ajax validation.
For creating custom validators check http://www.yiiframework.com/doc-2.0/guide-input-validation.html#creating-validators
you need not to write any such code for this propose. Yii can handle ajax validations it-self. only thing that you need to do is enable it in active form like.
php $form = ActiveForm::begin([
'id' => 'contact-form',
'enableAjaxValidation' => true,
]); ?>
and placing this code in controller after initialize $model.
if (Yii::$app->request->isAjax) {
Yii::$app->response->format = Response::FORMAT_JSON;
return = ActiveForm::validate($model);
}
I newbie in Laravel API. There is an update function which only allows users to update their own post. It worked. When users try to update other user's post, it alswo worked, but it shows the error like this image. Actually i want it show in response json.
I want to show message like this
{
"status": "error",
"message": "This action is unauthorized",
}
This is my code for PostController.
public function update(Request $request, Post $post)
{
$this->authorize('update', $post);
//this will check the authorization of user but how to make if else statement, if the post belong to the user it will show this json below but if the post belong to other, it will show error message(response json)
$post->content = $request->get('content', $post->content);
$post->save();
return fractal()
->item($post)
->transformWith(new PostTransformer)
->toArray();
}
This code for PostPolicy
public function update(User $user, Post $post)
{
return $user->ownsPost($post);
}
This is code for User model
public function ownsPost(Post $post)
{
return Auth::user()->id === $post->user->id;
}
This code for AuthServiceProvider
protected $policies = [
'App\Post' => 'App\Policies\PostPolicy',
];
Hope anyone can help me.
I'm using Laravel 5.4
In the app/Exceptions/Handler.php class you can change the render function like so
public function render($request, Exception $exception)
{
$preparedException = $this->prepareException($exception);
if ($preparedException instanceof HttpException) {
return response(
[
'message' => sprintf(
'%d %s',
$preparedException->getStatusCode(),
Response::$statusTexts[$preparedException->getStatusCode()]
),
'status' => $preparedException->getStatusCode()
],
$preparedException->getStatusCode(),
$preparedException->getHeaders()
);
}
return parent::render($request, $exception);
}
Or if you look further in the rendering, overriding the renderHttpException might be a little safer. This will remove the custom error pages in views/errors
protected function renderHttpException(HttpException $e)
{
return response(
[
'message' => sprintf(
'%d %s',
$e->getStatusCode(),
Response::$statusTexts[$e->getStatusCode()]
),
'status' => $e->getStatusCode()
],
$e->getStatusCode(),
$e->getHeaders()
);
}
I have a form in which contains a dropdown field.The options for the dropdown is based on the radio button selection.
Radio Button
<?php
$list=array("1"=>"Assign to Department","2"=>"Assign to user");
echo $form->field($model, 'option')->radioList($list,
[
'template' => '{input}{label}', // put the label behind
]);
?>
Script
<script>
$('input[type=radio]').click(function(){
var Flag=this.value;
if(Flag=='1') {
'<?php $items= ArrayHelper::map($company->Users, 'id', 'fullname');?>';
} else if(Flag=='2') {
'<?php $items= ArrayHelper::map(Department::find()->all(), 'dep_id', 'name'); ?>';
}
alert(Flag);
});
</script>
Drodown
<?php
echo $form->field($model, 'assignee')
->dropDownList(
$items, // Flat array ('id'=>'label')
['prompt'=>''] // options
)->label('');
?>
Expected response is the change of option values according to the drop down selection.But it is not changing even though the flag value is different for radio buttons and changes on click.
Any ideas?
I have tried this for a sample and it works -
<?php
$items1 = range(0, 5); //PHP Array 1
$items2 = range(5, 10); //PHP Array 2
?>
<script>
var arr1 = <?php echo json_encode($items1); ?>; //Javascript Array 1
var arr2 = <?php echo json_encode($items2); ?>; //Javascript Array 2
$(document).ready(function() {
$('input[type=radio]').click(function(){
var Flag = this.value;
setDropdown(Flag);
});
});
function setDropdown(Flag) {
var elem = $('#Skill_order_no'); //Your dropdown element name.
if(Flag == '1') {
elem.empty();
$.each(arr1, function(index, value) {
elem.append($('<option>').text(value).attr('value', value));
});
} else if(Flag == '2') {
elem.empty();
$.each(arr2, function(index, value) {
elem.append($('<option>').text(value).attr('value', value));
});
}
}
</script>
I have parent page and there are two child view _sectionhead.php and _classteacher.php i need to renderPartial to those child view when user select from dropdwon how can do this
this is my create from
IF there is any other way i like to know it.i need quick help
In view:
<?php Yii::app()->clientScript->registerScript('some-name', "
$('select').on('change', function (event) {
var url = $(event.currentTarget).val();
if (url != 0)
$.get(url, function (data) {
$('#result').html(data);
});
});", CClientScript::POS_READY) ?>
<?php echo
CHtml::dropDownList("param", "select here", [
'Select here',
$this->createUrl('/testing/action1') => "Add class teacher",
$this->createUrl('/testing/action2') => "Add class techer"
]) ?>
<div id="result"></div>
Assuming you are doing in HTML ,Add below in HTML code
<option onClick="fnCallPage('classteacher')">class teacher</option>
<option onClick="fnCallPage('Section')">Section</option>
Add below in Javascript tag :
function fnCallPage(param){
//Javascript code for sending
}
Based on #Fraz solution:
HTML:
<option onClick="fnCallPage('_classteacher')">class teacher</option>
<option onClick="fnCallPage('_sectionhead')">Section</option>
<div id="divContainerPartial"></div> <!-- View rendered container -->
Javascript:
function fnCallPage(param){
$.ajax({
url: '<?php echo Yii::app()->createUrl('yourModel/renderPageByAjax');?>/' + param,
type: 'GET',
success: function(html){
$('#divContainerPartial').html(html);
},
error: function(data){
alert('Error loading a '+param+' view.');
}
});
}
Controller:
/**
* #param $view String : View name to render.
*/
public function actionRenderPageByAjax($view)
{
$params = array(); // Variables to view.
$this->renderPartial('//myFolderView/'.$view, array(
'params' => $params,
), false, true);
}
I want to add a new ZF2' Zend\Form\Element within jQuery ajax. But while I use the Zend Form , I don't know how to make it. This is the add.phtml file.
<script type="text/javascript">
$(document).ready(function(){
$(".pid").change(function(){
var id = $('.pid').val();
var $_this = $(this);
$.ajax({
type:"POST",
url:"<?php echo $this->url('Element/default',array('action'=>'change'));?>",
data:"pid="+id,
dataType:'json',
async:false,
success:function(data){
if(data.response){
//here I want to add a new select component after select conponent "pid" using like "$this->formRow($form->get('mid'))" or else .
}
}
});
});
});
</script>
the following is the remaining part of the html.
<?php
$title = 'add';
$this->headTitle($title);
?>
<h1><?php echo $this->escapeHtml($title); ?></h1>
<?php
$form = $this->form;
$form->setAttribute('action', $this->url(
'Element/default',
array(
'action' => 'add'
)
));
$form->prepare();
echo $this->form()->openTag($form);
echo $this->formRow($form->get('pid'));
echo $this->formRow($form->get('name'));
echo $this->formRow($form->get('desc'));
echo $this->formSubmit($form->get('submit'));
echo $this->form()->closeTag();
How do I add a new zend form element within jquery ajax ? Thanks.
You have script to receive data using ajax and view/html should be received by this script. You need controller/action to rendering data and return as response.
use Zend\View\Model\JsonModel;
//some controller
public function changeAction(){ // your requested action
//1. get partial helper to rendering html;
$partial = $this->getServiceLocator()->get('ViewHelperManager')->get('partial');
$form = new Form();// your form
//2. render html
$html = $partial('path/to/your/file/phtml',['form'=>$form]);
//3. return data as JSON because in your ajax configuration dataType: 'json'
return new JsonModel([
'html' => $html,
]);
}
in your js success function should be:
success:function(data){
if(data.html){
$_this.find('blockToAppend').append(data.html);
}
}