CODEWITHSUNDEEP
phpajaxvalidationyii
alwaysLearn
alwaysLearn

Reputation: 6950

Ajax message for yii unique validator

I am trying to validate a form against the model in Yii keeing ajax validation on. I want to keep a field unique in my database . The problem I am facing is the ajax message is not getting displayed for the unique validator but for all other rules it is working fine.

Please let me know where I am wrong . Relevant code is posted below

MODEL RULES

  public function rules()
{
    // NOTE: you should only define rules for those attributes that
    // will receive user inputs.
    return array(
       -----------
        ------------
        array('user_code', 'length', 'max'=>20),
        array('user_code','unique','message'=>'This  code already exists. Please try different code', 'className' => 'User',
    'attributeName' => 'user_code',),

    );
}

FORM

<?php $form=$this->beginWidget('CActiveForm', array(
                    'id'=>'frm-useraccount',
                    'enableAjaxValidation'=>true,
                    'enableClientValidation'=>true,
                    'action'=>Yii::app()->createUrl('dashboard/index').'#user',
                    'clientOptions'=>array(
                            'validateOnSubmit'=>true,

                    ),
                ));
                ?>

CONTROLLER

$objMdl                   = new User;
      $isUserSetUpFormSubmitted     = (isset($postedData['User']))?true:false;
      // handle form submit
      if($isUserSetUpFormSubmitted)
      {
         //validate the model 
          $isMdlValidated = $objMdl->validate($postedData['user_code']);
          if($isMdlValidated)
          {
            //handle insert
          }
          else
         {
            //display model errors
          }
      }

Upvotes: 0

Views: 469

Answers (2)

Mihai P.
Mihai P.

Reputation: 9357

what he said ^^^. In adition to what Deele said, you are creating a new object without assigning it ever any values. $objMdl->validate($postedData['user_code']); means validate the attribute $postedData['user_code'] of the model objMdl. It does NOT mean validate the attribute user_code but whatever random stuff you inserted into $postedData['user_code'].

Upvotes: 0

Deele
Deele

Reputation: 3684

To display errors for specific field, you either need error summary or error field in your form.

Error summary

<?php echo $form->errorSummary($model); ?>

Error field

<?php echo $form->error($model,'user_code'); ?>

In addition to that, you need to set up your controller to return ajax validation result

/**
 * Provides output to controller action request "/dashboard/index"
 */
public function actionIndex() {

    // Prepare login form
    $model = new User('update');

    // Perform AJAX validation if required
    $this->performAjaxValidation($model, 'frm-useraccount');

    // Perform basic validation
    if ($model->attributes = $req->getPost('User')) {
        if ($model->validate()) {

            // Save in DB
            $model->save(false);

            // Show confirmation for user
            Yii::app()->user->setFlash(
                'success',
                Yii::t(
                    'site',
                    'Save successfull.'
                )
            );

            // Refresh
            $this->refresh();
        }
        else {
            Yii::app()->user->setFlash(
                'error',
                CHtml::errorSummary($model)
            );
        }
    }
}

/**
 * Creates data validation output for ajax requests with data from given form ID
 */
protected function performAjaxValidation($model, $formId) {

    /**
     * @var $app CWebApplication
     * @var $req CHttpRequest
     */
    $app = Yii::app();
    $req = $app->getRequest();

    if ($req->getIsAjaxRequest() && $req->getPost('ajax') === $formId) {
        echo CActiveForm::validate($model);
        $app->end();
    }
}

In addition to your model and rules, this should be enough, to display error you require.

Upvotes: 1

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