2

I want to get all content of WYSIWYG content display on my popup. When the WYSIWYG content has an image and text added then the image doesn't display.

I want to get output like that. If I added Bold effect on text and an image is added in the WYSIWYG content, then it will display output like bold font and the image is displayed. If I added just an image, then the image will be displayed in the popup. If I added just Bold effect on textm then it will displayed as bold text.

How can do like this?

  • I also found solution : $objectManager = \Magento\Framework\App\ObjectManager::getInstance(); $filterManager = $objectManager->get('Magento\Cms\Model\Template\FilterProvider')->getPageFilter()->filter($content); echo $filterManager; – Rohan Hapani May 8 '17 at 7:08
  • 1
    yes it's right but don't use object manager directly as Magento 2 standards. – Prince Patel May 8 '17 at 8:02
  • Okey... @PrincePatel – Rohan Hapani May 8 '17 at 10:25
7

Let's say your popup has a block behind it to handle the logic, and a template to render the result (as it should have).
Let's say your block class is \Vendor\Module\Block\Popup

Add this to your block.

/**
 * @var \Zend_Filter_Interface
 */
protected $templateProcessor;
public function __construct(
    .....
    \Zend_Filter_Interface $templateProcessor, 
    ....
) {
    ....
    $this->templateProcessor = $templateProcessor;
    ....
}

Then create the method:

public function filterOutputHtml($string)
{
    return $this->templateProcessor->filter($string);
}  

In your di.xml file add this:

<type name="Vendor\Module\Block\Popup">
    <arguments>
        <argument name="templateProcessor" xsi:type="object">Magento\Widget\Model\Template\Filter</argument>
    </arguments>
</type>

Now, in your template you can use this:

<?php echo $block->filterOutputHtml($content);?>

where $content is the content you are trying to interpret as HTML.

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