Regex as Lua pattern

Question

I wrote (?<=;)[^\^]* regex but I have problem with transforming it to use it in Lua. I want to transform

^#57df44;Cobblestone^white;

into

Cobblestone

but if the string does not contain any color code then it should be returned "as is".

Could someone help me with this?

s = s:gsub("^.-;(.-)^.*$", "%1")

Egor Skriptunoff
– Egor Skriptunoff

2017-08-29 14:43:01 +00:00
Commented Aug 29, 2017 at 14:43 — Egor Skriptunoff
– Egor Skriptunoff, Commented Aug 29, 2017 at 14:43

Wiktor Stribiżew · Accepted Answer · 2017-08-29 14:55:11Z

2

Use a capturing group:

local s = "^#57df44;Cobblestone^white;"
res = s:match(";([^^]*)") or s
print( res )
-- Cobblestone

See the Lua demo.

Here,

; - matches the first ;
([^^]*) - Capturing group 1 matches any 0+ chars other than ^ into Group 1

The string.match will only return the captured part if the capturing group is defined in the pattern.

More details

As is mentioned in comments, you might use a frontier pattern %f[^:] instead of (?<=:) in the current scenario.

The frontier pattern %f followed by a set detects the transition from "not in set" to "in set".

However, the frontier pattern is not a good substitute for a positive lookbehind (?<=:) because the latter can deal with sequences of patterns, while a frontier pattern only works with single atoms. So, %f[^:] means place in string between : and a non-:. Howerver, once you need to match any 0+ chars other than ^ after city=, a frontier pattern would be a wrong construct to use. So, it is not that easily scalable as a string.match with a capturing group.

edited Aug 29, 2017 at 14:55

answered Aug 29, 2017 at 14:08

Wiktor Stribiżew

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7 Comments

Wiktor Stribiżew Over a year ago

Here is a demo for multiple occurrences. But I suspect only the first match is what is necessary, so string.match is what you need.

user2463506 Over a year ago

It mostly works, but it fails if text has no color codes (returns nil). This minor issue can be however fixed with simple or. s:match(";([^^]*)") or s;

Wiktor Stribiżew Over a year ago

@user2463506 Sorry, I overlooked that part of the question. Yes, looks like that will do. I added the or bit to the answer.

moteus Over a year ago

string.match(s, '%f[^;]([^^]*)')

Egor Skriptunoff Over a year ago

@moteus - You can omit parentheses: s:match"%f[^;][^^]*"

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