Regex, substitute part of a string always at the end

Question

I am trying to substitute a string so a part of this url always goes to the end

google.com/to_the_end/faa/
google.com/to_the_end/faa/fee/
google.com/to_the_end/faa/fee/fii

Using this

(google\.com)\/(to_the_end)\/([a-zA-Z0-9._-]+)

$1/$3/$2

It works for the first example, but I need something a bit more versatile so no matter how many folders it always moves to_the_end as the last folder in the url string

Desired output

google.com/faa/to_the_end
google.com/faa/fee/to_the_end/
google.com/faa/fee/fii/to_the_end/

I assume it would be as simple as using `(.*)` as last group. What exactly is not working with your current approach? — knittl, Jun 06 '22 at 10:49
`(google\.com)\/(to_the_end)\/(.*[^\/])\/?$` will work even if there is a `/` at the end ([demo](https://regex101.com/r/mGm46l/2)). — Wiktor Stribiżew, Jun 06 '22 at 10:51
@knittl using that gives me this for the first results `google.com/foo//to_the_end` Would have to do `$1/$3$2` — Álvaro, Jun 06 '22 at 10:51

score 2 · Accepted Answer · answered Jun 06 '22 at 10:58

You can use

(google\.com)\/(to_the_end)\/(.*[^\/])\/?$

See the regex demo.

Details:

(google\.com) - Group 1: google.com
\/ - a / char
(to_the_end) - Group 2: to_the_end
\/ - a / char
(.*[^\/]) - Group 3: any zero or more chars other than line break chars as many as possible and then a char other than a / char
\/? - an optional / char
$ - end of string.

Regex, substitute part of a string always at the end

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