How to store the result from an external reference into a variable?

Question

PHP

There is an external resource that issues an array using a link
https://api.deezer.com/search?q=LUCKY%20LUKE%20HYP...
It is required to save the contents of "cover_medium" to a variable
Please tell me how to implement this.

Answer 1

I do not like this way, but the essence will be clear:

<?php
$payload = \file_get_contents('https://api.deezer.com/search?q=LUCKY%20LUKE%20HYPNOTIZING');
$payload = \json_decode($payload, true, 512, JSON_THROW_ON_ERROR);
$media   = $payload['data'][0]['album']['cover_medium'] ?? 'media_not_found.jpg';

If you need to pass headers and other magic, then I advise you to use cURL or a good Guzzle wrapper