Validation Form in wiget









up vote
1
down vote

favorite












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














up vote
1
down vote

favorite












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












up vote
1
down vote

favorite









up vote
1
down vote

favorite











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

















up vote
1
down vote



accepted










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




















    Your Answer






    StackExchange.ifUsing("editor", function ()
    StackExchange.using("externalEditor", function ()
    StackExchange.using("snippets", function ()
    StackExchange.snippets.init();
    );
    );
    , "code-snippets");

    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "1"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );













    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53254568%2fvalidation-form-in-wiget%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote



    accepted










    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
























      up vote
      1
      down vote



      accepted










      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






















        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted






        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



























            draft saved

            draft discarded
















































            Thanks for contributing an answer to Stack Overflow!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f53254568%2fvalidation-form-in-wiget%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Top Tejano songwriter Luis Silva dead of heart attack at 64

            政党

            天津地下鉄3号線