Unique validator

Yii 2.0 General Discussions
1

Hi,

Is there a way to use user validation on ‘unique’ rules in forms ?

Thanks.

2

What do you mean? What would you like to do?

3

When a user input something in a field that is supposed to be unique, display a error message like ‘this username is already taken’ without submitting the form.

4

Submitting is required because only server has the info you need but you can use AJAX for this.

Check this section.

5

I did that and firebug show the correct JSON answer, but how do I display the error in my page?

6

Are you using ActiveForm widget? With the configuration and code snippet mentioned at the end of the this guide section error messages should be displayed automatically.

7

Yes,

here is the form :




<?php $form = ActiveForm::begin([

    	'id' => $model->formName(),

    	'validationUrl' => ['/commandes/ajaxvalid'],

    	'enableClientValidation' => true,

		'options' => ['class' => 'form-horizontal'],

		'fieldConfig' => [

            'template' => "{label}<div class=\"col-lg-6\">{input}</div><div class=\"col-lg-4\">{error}</div>",

            'labelOptions' => ['class' => 'col-lg-2 control-label'],

		]

	]); ?>


<?= $form->field($model, 'num_de_serie' ,['enableAjaxValidation' => true, 'validateOnChange' => false])->textInput(['maxlength' => '15' , 'id' => 'numero_de_serie']) ?>

The controller:




public function actionAjaxvalid ()

	{

		$model = new Commandes();

		if (Yii::$app->request->isAjax && $model->load(Yii::$app->request->post())) {

			Yii::$app->response->format = \yii\web\Response::FORMAT_JSON;

			return ActiveForm::validate($model, ['num_de_serie']);

		}

	}
8

You are overriding field ID (‘id’ => ‘numero_de_serie’) and I guess validation js looks for default ‘commandes-num_de_serie’ and this is probably why no error message is displayed.

Don’t override the field ID or configure everything to work with overriden one.

9

That worked.

Thanks a lot.