Lua match start from the end

Question

I am trying to extract url with match. I'm trying to find the filename after the "/" character. But url is always variable so i have to start from end

123, 123.py, 123.dat

url://xxxxx/yyyyy/123
url://xxxxx/yyyyy/123.py
url://xxxxx/yyyyy/123.dat

I tried url:match("^.+(%..+)$") but I can't access files that don't start with a ..

Why not use `url:gsub(".*/", "")` if you need `123`, `123.py`, `123.dat`? — Wiktor Stribiżew, Oct 19 '22 at 22:05
@WiktorStribiżew I didn't think it would be that simple. great answer! Thank you. — sweetngx, Oct 19 '22 at 22:08
I posted the [answer](https://stackoverflow.com/a/74132450/3832970) below also providing the solution to find directories, as you asked in a now deleted comment. — Wiktor Stribiżew, Oct 20 '22 at 10:49

Wiktor Stribiżew · Answer 1 · 2022-10-19T22:17:07.423

1

You need

url:gsub(".*/", "")

The pattern matches

.* - zero or more chars as many as possible
/ - a / char.

To find the directory, you can use

url:gsub("/[^/]*$", "")

Here,

/ - a / char
[^/]* - zero or more chars other than /
$ - end of string.

edited Oct 19 '22 at 22:17

answered Oct 19 '22 at 22:09

Wiktor Stribiżew

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score 0 · Answer 2 · answered Oct 20 '22 at 06:22

Here's the alternative using match:

url:match"[^/]+$"

Explanation: Match one or more non-slash characters anchored to the end of string $.

This will also match just a filename like foo.bar without any slashes. If you want to force slashes to precede your filename, use /(.-)$ instead.

Lua match start from the end

2 Answers2