Validation Form in wiget












1














I would like to have a form available on every page in my web app. So I created a widget with a form.



class AddNewURL extends Widget{

public $model;


public function init()
{
parent::init();
if ($this->model === null) {
$this->model = new ProductUrlForm();
}
}

public function run()
{
return $this->render('addNewURLForm', [
'model' => $this->model,
]);
}
}


and widget view



<div class="modal fade bs-example-modal-lg" id="newURL" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog modal-lg" role="document">

<div class="modal-content">
<?php Pjax::begin(); ?>
<?php $form = ActiveForm::begin([
'action' => Url::to(['site/new-url']),
'options' => ['data' => ['pjax' => true]],
]) ?>
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span>
</button>
<h4 class="modal-title" id="myModalLabel">Add new URL</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-md-12">
<?= $form->field($model, 'url', ['errorOptions' => ['class' => 'help-block', 'encode' => false]])
->textInput(['maxlength' => false])
?>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="submit" class="btn btn-primary">Add</button>
</div>

<?php ActiveForm::end() ?>
<?php Pjax::end(); ?>
</div>

</div>
</div>


and this how I use this widget in my layouts/main.php



<?= appwidgetsAddNewURL::widget() ?>


in my site/new-url action I have



public function actionNewUrl()
{
$model = new ProductUrlForm();
$model->status = STATUS_NEW;

if ($model->load(Yii::$app->request->post()) && $model->save()) {
//return success message
}

return $this->render('index', [
'model' => $model,
]);
}


and now when I submit form data goes to my action, the form is reset and that's all. How to show validation error or success message using pjax? In my model I have few custom validation rules.



Maybe there is better solution to use one form on all pages?










share|improve this question
























  • how and where are you calling this widget? add to your question.
    – Muhammad Omer Aslam
    Nov 12 at 13:41












  • <?= appwidgetsAddNewURL::widget() ?>
    – Roboto6_1on
    Nov 12 at 13:51
















1














I would like to have a form available on every page in my web app. So I created a widget with a form.



class AddNewURL extends Widget{

public $model;


public function init()
{
parent::init();
if ($this->model === null) {
$this->model = new ProductUrlForm();
}
}

public function run()
{
return $this->render('addNewURLForm', [
'model' => $this->model,
]);
}
}


and widget view



<div class="modal fade bs-example-modal-lg" id="newURL" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog modal-lg" role="document">

<div class="modal-content">
<?php Pjax::begin(); ?>
<?php $form = ActiveForm::begin([
'action' => Url::to(['site/new-url']),
'options' => ['data' => ['pjax' => true]],
]) ?>
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span>
</button>
<h4 class="modal-title" id="myModalLabel">Add new URL</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-md-12">
<?= $form->field($model, 'url', ['errorOptions' => ['class' => 'help-block', 'encode' => false]])
->textInput(['maxlength' => false])
?>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="submit" class="btn btn-primary">Add</button>
</div>

<?php ActiveForm::end() ?>
<?php Pjax::end(); ?>
</div>

</div>
</div>


and this how I use this widget in my layouts/main.php



<?= appwidgetsAddNewURL::widget() ?>


in my site/new-url action I have



public function actionNewUrl()
{
$model = new ProductUrlForm();
$model->status = STATUS_NEW;

if ($model->load(Yii::$app->request->post()) && $model->save()) {
//return success message
}

return $this->render('index', [
'model' => $model,
]);
}


and now when I submit form data goes to my action, the form is reset and that's all. How to show validation error or success message using pjax? In my model I have few custom validation rules.



Maybe there is better solution to use one form on all pages?










share|improve this question
























  • how and where are you calling this widget? add to your question.
    – Muhammad Omer Aslam
    Nov 12 at 13:41












  • <?= appwidgetsAddNewURL::widget() ?>
    – Roboto6_1on
    Nov 12 at 13:51














1












1








1







I would like to have a form available on every page in my web app. So I created a widget with a form.



class AddNewURL extends Widget{

public $model;


public function init()
{
parent::init();
if ($this->model === null) {
$this->model = new ProductUrlForm();
}
}

public function run()
{
return $this->render('addNewURLForm', [
'model' => $this->model,
]);
}
}


and widget view



<div class="modal fade bs-example-modal-lg" id="newURL" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog modal-lg" role="document">

<div class="modal-content">
<?php Pjax::begin(); ?>
<?php $form = ActiveForm::begin([
'action' => Url::to(['site/new-url']),
'options' => ['data' => ['pjax' => true]],
]) ?>
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span>
</button>
<h4 class="modal-title" id="myModalLabel">Add new URL</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-md-12">
<?= $form->field($model, 'url', ['errorOptions' => ['class' => 'help-block', 'encode' => false]])
->textInput(['maxlength' => false])
?>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="submit" class="btn btn-primary">Add</button>
</div>

<?php ActiveForm::end() ?>
<?php Pjax::end(); ?>
</div>

</div>
</div>


and this how I use this widget in my layouts/main.php



<?= appwidgetsAddNewURL::widget() ?>


in my site/new-url action I have



public function actionNewUrl()
{
$model = new ProductUrlForm();
$model->status = STATUS_NEW;

if ($model->load(Yii::$app->request->post()) && $model->save()) {
//return success message
}

return $this->render('index', [
'model' => $model,
]);
}


and now when I submit form data goes to my action, the form is reset and that's all. How to show validation error or success message using pjax? In my model I have few custom validation rules.



Maybe there is better solution to use one form on all pages?










share|improve this question















I would like to have a form available on every page in my web app. So I created a widget with a form.



class AddNewURL extends Widget{

public $model;


public function init()
{
parent::init();
if ($this->model === null) {
$this->model = new ProductUrlForm();
}
}

public function run()
{
return $this->render('addNewURLForm', [
'model' => $this->model,
]);
}
}


and widget view



<div class="modal fade bs-example-modal-lg" id="newURL" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog modal-lg" role="document">

<div class="modal-content">
<?php Pjax::begin(); ?>
<?php $form = ActiveForm::begin([
'action' => Url::to(['site/new-url']),
'options' => ['data' => ['pjax' => true]],
]) ?>
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">&times;</span>
</button>
<h4 class="modal-title" id="myModalLabel">Add new URL</h4>
</div>
<div class="modal-body">
<div class="row">
<div class="col-md-12">
<?= $form->field($model, 'url', ['errorOptions' => ['class' => 'help-block', 'encode' => false]])
->textInput(['maxlength' => false])
?>
</div>
</div>
</div>
<div class="modal-footer">
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
<button type="submit" class="btn btn-primary">Add</button>
</div>

<?php ActiveForm::end() ?>
<?php Pjax::end(); ?>
</div>

</div>
</div>


and this how I use this widget in my layouts/main.php



<?= appwidgetsAddNewURL::widget() ?>


in my site/new-url action I have



public function actionNewUrl()
{
$model = new ProductUrlForm();
$model->status = STATUS_NEW;

if ($model->load(Yii::$app->request->post()) && $model->save()) {
//return success message
}

return $this->render('index', [
'model' => $model,
]);
}


and now when I submit form data goes to my action, the form is reset and that's all. How to show validation error or success message using pjax? In my model I have few custom validation rules.



Maybe there is better solution to use one form on all pages?







forms yii2 widget pjax






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 13:54

























asked Nov 12 at 0:19









Roboto6_1on

100112




100112












  • how and where are you calling this widget? add to your question.
    – Muhammad Omer Aslam
    Nov 12 at 13:41












  • <?= appwidgetsAddNewURL::widget() ?>
    – Roboto6_1on
    Nov 12 at 13:51


















  • how and where are you calling this widget? add to your question.
    – Muhammad Omer Aslam
    Nov 12 at 13:41












  • <?= appwidgetsAddNewURL::widget() ?>
    – Roboto6_1on
    Nov 12 at 13:51
















how and where are you calling this widget? add to your question.
– Muhammad Omer Aslam
Nov 12 at 13:41






how and where are you calling this widget? add to your question.
– Muhammad Omer Aslam
Nov 12 at 13:41














<?= appwidgetsAddNewURL::widget() ?>
– Roboto6_1on
Nov 12 at 13:51




<?= appwidgetsAddNewURL::widget() ?>
– Roboto6_1on
Nov 12 at 13:51












1 Answer
1






active

oldest

votes


















1














Action actionNewUrl() must render view addNewURLForm so Pjax can replace old form with new html recieved from the action.



It is better to place your widget, controller and form model to separate module.






share|improve this answer





















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    Action actionNewUrl() must render view addNewURLForm so Pjax can replace old form with new html recieved from the action.



    It is better to place your widget, controller and form model to separate module.






    share|improve this answer


























      1














      Action actionNewUrl() must render view addNewURLForm so Pjax can replace old form with new html recieved from the action.



      It is better to place your widget, controller and form model to separate module.






      share|improve this answer
























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        Action actionNewUrl() must render view addNewURLForm so Pjax can replace old form with new html recieved from the action.



        It is better to place your widget, controller and form model to separate module.






        share|improve this answer












        Action actionNewUrl() must render view addNewURLForm so Pjax can replace old form with new html recieved from the action.



        It is better to place your widget, controller and form model to separate module.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 13 at 5:27









        Anton Rybalko

        672816




        672816






























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