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I have a link field, which I can populate with both internal and external links:

/node/123

I'm building an API that parses the node and returns the URL from the link field:

$node->my_field->uri

Which returns internal:/node/123 which makes this URL useless.

If I try using use Drupal\Core\Url; and Url::fromUri($node->my_field->uri), I get a 500 error!

So what is the correct approach for getting a formatted URL, that can parse both internal and external links?

1 Answer 1

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just kidding. To help future devs with this issue, check the logs *duh and you'll see this:

PHP message: Uncaught PHP Exception InvalidArgumentException: "The URI '' is invalid.

So, check for the uri value first!

$node->my_field->uri ? Url::fromUri($node->my_field->uri)->toString() : null,

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    You can also use $node->get('my_field')->get(0)->getUrl(). you still need to check it for not being empty though
    – Berdir
    Feb 19, 2018 at 20:36

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