How to extract file name from AWS url?

Question

I am trying to get the file name from URL and here I use basename to get the file name but it not works.

basename('https://test-assets.amazonaws.com/live%2Ftestsamantha-9-bucket%2Ff132eb33-f403-4881-80e2-de1cc3bdb43a.jpg');

Result is

live%2Ftestsamantha-9-bucket%2Ff132eb33-f403-4881-80e2-de1cc3bdb43a.jpg

How do I get

f132eb33-f403-4881-80e2-de1cc3bdb43a.jpg

what was your exact url? – Bhargav Chudasama Jul 15 '19 at 12:55 — Bhargav Chudasama, Jul 15 '19 at 12:55

score 3 · Accepted Answer · answered Jul 15 '19 at 12:51

You take the url just as you did, use urldecode on it, so slashes are plain slashes and nothing encoded. Then all you have to do, is explode by slash and take the last element of the created array. Voila.

$url = urldecode(basename('https://test-assets.amazonaws.com/live%2Ftestsamantha-9-bucket%2Ff132eb33-f403-4881-80e2-de1cc3bdb43a.jpg'));
$parts = explode('/', $url);

print end($parts); // Prints 'f132eb33-f403-4881-80e2-de1cc3bdb43a.jpg'

Or better, do it like the other poster did, its less code and he is right. While my solution works too, its more than needed.

score 2 · Answer 2 · answered Jul 15 '19 at 12:52

2

You need to decode the url first using the urldecode method:

basename(urldecode('https://test-assets.amazonaws.com/live%2Ftestsamantha-9-bucket%2Ff132eb33-f403-4881-80e2-de1cc3bdb43a.jpg'));

The urldecode method converts the url to https://test-assets.amazonaws.com/live/testsamantha-9-bucket/f132eb33-f403-4881-80e2-de1cc3bdb43a.jpg and then the basename method works properly.

answered Jul 15 '19 at 12:52

Claudio

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Take my upvote, you desever it more than me lmao – Manuel Mannhardt Jul 15 '19 at 13:09

How to extract file name from AWS url?

2 Answers2