I have a node with a field called "field_paragraph" which is a paragraph reference. Referencing a paragraph type which has a field called "field_image"

I'm trying to get the url of the field_image in a custom module and so far i have this. But i don't seem to get it work.

use Drupal\paragraphs\Entity\Paragraph; 
use Drupal\node\Entity\Node;
$node = Node::load($nid);
$paragraph_field = $node->get("field_paragraph")->getValue();
$paragraph = Paragraph::load($paragraph_field[0]['target_id']);

Here is where the problem occurs

$uri = $paragraph->field_image->entity->getFileUri();

The above returns nothing.

I used kint to print "$paragraph->field_image->entity" and it returns null. What I'm I doing wrong? I tried all answers I could find on here to no avail.

After getting the uri I think calling the below line would be enough.

$url = file_create_url($uri);

Question: How do I get the url of an image field in paragraphs programmatically?

  • There seems to be nothing wrong with the code. The first step is very verbose and the second a much shorter version. But both steps are basically the same, both are reference fields, to a paragraph and a file. So you can use in the second step the verbose version you used in the first step and debug the array if there is content in the image field, that means a target_id to the image file.
    – 4uk4
    Commented Jun 19, 2017 at 16:40
  • Linking drupal.stackexchange.com/questions/224294/… which deals with a very similar question from a PHP point of view. Commented Nov 30, 2021 at 11:02

1 Answer 1


Your pretty much there but when calling the field name use the method get()

$uri = $paragraph->get('field_image')->entity->getFileUri();

Then once you have the uri you'll be able to get the full url

$url = file_create_url($uri);

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