Regular expression to match any hyphenated word in a url in PHP

Question

I'm trying to replace a URL with CloudFront URL.

For example. the URL could be

https://s3.amazonaws.com/bucket/folder/d05c229a73b5c6f169d599652015b1.png

https://s3.amazonaws.com/bucket-with-hyphen/folder/folder1/d05c229a73b5c6f169d599652015b1.png

https://s3.amazonaws.com/bucket-with-hyphen/folder-with-hyphen/folder1/d05c229a73b5c6f169d599652015b1.png

And then I'll use:

preg_replace($pattern, $replacement, $string)

My $replacement would be $1cloudfront.net/$4';

Here's my $pattern so far

^(https?:\/\/)(s3.amazonaws.com)\/(\w+-\w+-\w+)\/(<I want the rest of the url>)

It will only match bucket-with-hyphen but not bucket-with & I don't know how to capture the rest of the URL.

Answer 1

Here is your updated regex that captures "bucket", "bucket-with" or "bucket-with-anything", but won't match "bucket-with-something-else". If you need to capture the 4th option, change {0,2} to * .

I only added optional non-capturing groups (0 or 2 occurrences):

^(https?:\/\/)(s3.amazonaws.com)\/(\w+(?:-\w+){0,2})\/(.*)
                                       ^ ^ ^    ^

See the regex demo and IDEONE demo

Answer 2

You can use the following to match:

(?<=\/)(?:\w+-)+\w+(?=\/)   //no need to capture whole url here

And replace with cloudfront.net

See DEMO