How to get parameter only, if full url inside variable in php?

Question

My code:

$url = "https://www.example.com/reset_password/6f88e62e492ed288200b99b09f6d8132";
$para = $this->uri->segment($url);
echo $para;

In this code I only want the parameter that is passed in the url i.e 6f88e62e492ed288200b99b09f6d8132 but it's not working. How can I get only the parameter in php?

Does this answer your question? [REST-style URLS and PHP](https://stackoverflow.com/questions/2121069/rest-style-urls-and-php) — Leo, Dec 04 '19 at 20:32

score 0 · Answer 1 · edited Dec 04 '19 at 20:38

0

Use explode function.

$url = $_SERVER["REQUEST_URI"];
$exploded_url = explode("/",$url);
$code = $exploded_url[1];

edited Dec 04 '19 at 20:38

GrumpyCrouton

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answered Dec 04 '19 at 20:05

Marty1452

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Sorry i typing it on iPhone and there is no this quotes – Marty1452 Dec 04 '19 at 20:34
Thanks very much – Marty1452 Dec 04 '19 at 20:42

How to get parameter only, if full url inside variable in php?

1 Answers1