php - Updating Yii Dropdownlist After Insert New Element [SOLVED] -

i'm totally new in yii, so, excuse me if answer question little bit trivial.

here's code:

<p>   <?php echo $form->labelex($model,'phone_type'); ?>   <span class="field">   <?php echo $form->dropdownlist($model,'phone_type',       chtml::listdata(phonestypes::model()->findall(),  'id','type' )); ?>        <?php echo $form->error($model,'phone_type'); ?> </span>                                  </p> 

there button register new phone types. so, after submition of form, inside of cjuidialog, wish above dropdownlist updated new type, without refresh page.

i google lot, find things related "dependent dropdowns" in yii.

what's better approach solve problem? there $.fn.cgridview.update?

here's dialog code:

<?php  $this->endwidget('zii.widgets.jui.cjuidialog');    $this->beginwidget('zii.widgets.jui.cjuidialog', array(    'id'=>'dialog-crud',    'options'=>array(     'title'=>'create new phone type',     'autoopen'=>false,     'modal'=>true,     'width'=>1080,     'height'=>820,     'resizable'=>false     ),   ));  ?>  <iframe src="http://myapp/phone_types/create"  width="100%" height="100%"></iframe>  <?php $this->endwidget(); ?> 

and code of controller, trivial create function:

public function actioncreate(){  $model = new phonetype;  if(isset($_post['phonetype'])){      $model->attributes = $_post['phonetype'];    if( $model->save() ){      //----> suggestion here? echo chtml::script("");     yii::app()->end();    }  } } 

---

so, below code of solution. in view:

<?php $this->beginwidget('zii.widgets.jui.cjuidialog', array(   'id'=>'dialog',   'options'=>array(     'title'=>'phone types',     'autoopen'=>false,     'modal'=>true,     'width'=>1080,     'height'=>820,     'resizable'=>false   ),   ));  ?>  <iframe src="phonetypes/create" id="cru-frame" width="100%" height="100%"></iframe>  <?php $this->endwidget(); ?> 

in phonetypescontroller:

public function actioncreate(){      $model = new phonetypes;      if(isset($_post['phonetypes'])){          $model->attributes = $_post['phonetypes'];          if($model->save()){                           echo chtml::script("                 window.parent.$('#dialog').dialog('close');                 window.parent.$('#phone_types_id').append('<option value=".$model->id." >'+'".$model->type."'+'</option>');             ");              yii::app()->end();                         }        }      $this->render('create',array(         'model'=>$model,     )); } 

thanks help!

you have action adds phone type (for example, let's call phonetype/create).

when send ajax request action create phone type, action should return newly created phone types information. can add dropdown using jquery.

look @ example:

<?php // in protected/controllers/phonetypecontroller.php public function actioncreate($phonetype) {     $phonetype = new phonetype;     $phonetype->phone_type = $phonetype;     if ($phonetype->save())     {         echo cjson::encode(array('value' => $phonetype->id, 'label' => $phonetype->phone_type)); // echos {"value":5,"label":"test"}     } } ?> 

for rest of example, i'm going assume original model (the 1 have phone_type field) called company (this has impact on jquery code below i'm selecting phone_type dropdown).

you can use output in success callback of ajax function add new option select (i.e. dropdown):

jquery.get( // 'data' , 'url' here     success: function(data) {         $('#company_phone_type').append('<option value="' + data.value + '">' + data.label + '</option>'); }); 

for more information on how in jquery, refer adding options select using jquery/javascript.