How to drop a domain using a regular expression?

Question

Regular Expressions

Does anyone have a working regex that can get the address of a resource without taking into account the domain part? For example, if all of the lines below would return '/test-url':

https://www.domain.ru/test-url
http://www.domain.ru/test-url
https://domain.ru/test-url
http://domain.ru/test-url
https://localhost/test-url
http://localhost/test-url

Answer 1

https://regex101.com/r/7TW7vc/1

Answer 2

Regular Expression:
/\/([^\/]+)$/
But it's better to use built-in language functions (PHP has parse_url function for example):

<?php
$url = 'https://www.domain.ru/test-url';

// using preg_match
preg_match('/(\/[^\/]+)$/', $url, $matches);

echo $matches[1].PHP_EOL;


// using parse_url 
echo parse_url($url, PHP_URL_PATH);

PHP parse_url