How to make a ContactForm in a modal window?

Question

Yii

Good afternoon. I would like to make a contact form in a modal window, the link to which is in the main menu (i.e. in layout main).
If you take and put in SiteIndex in actionIndex

public function actionIndex() {
            $support = new ContactForm();
   
           if ($support->load(Yii::$app->request->post()) && $support->contact(Yii::$app->params['adminEmail'])) {
            Yii::$app->session->setFlash('contactFormSubmitted');
            
            return $this->refresh();
        }
        return $this->render('index', [
                    'support' => $support
        ]);     
    }

and in main (layout) I put it in a modal window

<?php $form2 = ActiveForm::begin(['id' => 'contact-form']); ?>

<?= $form2->field($support, 'name')->textInput(['autofocus' => true]) ?>

<?= $form2->field($support, 'email') ?>

<?= $form2->field($support, 'subject') ?>

<?= $form2->field($support, 'body')->textArea(['rows' => 6]) ?> 

<div class="form-group">
<?= Html::submitButton('Submit', ['class' => 'btn btn-primary', 'name' => 'contact-button']) ?>
</div>

<?php ActiveForm::end(); ?>

then I get an error

Call to a member function formName() on a non-object

Is it somehow necessary to transfer the form not to the actionIndex, but to the layout? but how to do it?
$this->view->params['support'] = $support;Does not help

Answer 1

Override the renderContent function . Better yet, create a base controller with implemented renderContent and inherit from it.
Something like

<?php

class BaseController extends \yii\web\Controller
{
    private $_support;

    public function beforeAction($action)
    {
        if (!parent::beforeAction($action)) {
            return false;
        }

        $support = new ContactForm();
        $this->_support = $support;
   
        if ($support->load(Yii::$app->request->post()) 
            && $support->contact(Yii::$app->params['adminEmail'])) {
                
            Yii::$app->session->setFlash('contactFormSubmitted');
            return $this->refresh();
        }

        return true;
    }
    
    public function renderContent($content)
    {
        $layoutFile = $this->findLayoutFile($this->getView());
        if ($layoutFile !== false) {
            $layoutParams = [
                'content' => $content,
                'support' => $this->_support
            ];
            return $this->getView()->renderFile($layoutFile, $layoutParams, $this);
        } else {
            return $content;
        }
    }
}

Answer 2

It is necessary to lay down in /views/index/index.php
For what you are trying to forward from the controller will be visible only in ITS template, which you set in the render () method.
There is another option, put the form code into a widget and connect it anywhere.