如果url包含某个目录，则返回url的结尾

Question

I need to search thru a sql file and find the filename (the end of the url) where the url contains /assets/

Example url:

https://www.example.com/assets/screenshots/this-is-an-example/2016-12-20_15-42-29.png

It has /assets/ in it, so I want to get the last part of the url, being:

2016-12-20_15-42-29.png

Another example:

https://www.example.com/images/screenshots/this-is-an-example/2016-12-20_15-42-29.png

Nothing would return because /assets/ was not found.

*The number of directories deep it goes will be different, so it could be /assets/dir1/dir2/filename.jpg, and the next could be /assets/dir1/dir2/dir3/dir4/filename.jpg

Hoping this can be accomplished using regular expression.

Answer 1

You may use the following regex:

^(?=.*\/assets\/).*\/\K.*$

• ^ Assert position beginning of string.
• (?=.*\/assets\/) Positive lookahead to ensure that /assets/ substring is present in the string.
• .* Match anything greedily.
• \/ Match the last /.
• \K Reset operator, resets match.
• .*$ Match anything until end of string $.

You can try it here.

---

Php snippet:

<?php

$url = 'https://w...content-available-to-author-only...e.com/assets/screenshots/this-is-an-example/2016-12-20_15-42-29.png';
preg_match('/^(?=.*\/assets\/).*\/\K.*$/',$url,$matches);

print_r($matches[0]);

Prints:

2016-12-20_15-42-29.png

You can try it here.