Regex with optional number and leading char ? (lua match)

Question

I'm a beginner with regex and I'd like (for string.match() under lua) a regex that would recognize a positive or negative number prefixed by a special character (example : "!"). Example :

"!1" -> "1"
"!-2"  ->  "-2"
"!+3"  ->  "+3"
"!"  ->  ""

I tried

!(^[-+]?%d+)

but it does not work...

your solutions does not match "!" alone (last case of my example in the main post). I don't know if it's calssical, but in lua string.match can return NIL if nothing is found, but can return "" if something is found but nothing in it ;-) — DaveInDev, Oct 27 '22 at 15:36
This is not possible, in Lua patterns, there is no optional group support. — Wiktor Stribiżew, Oct 27 '22 at 15:37
OK, so no possibility in one match to solve my example above (with the "!" alone) ? Then I'll do a separate find("!") after a failed match. — DaveInDev, Oct 27 '22 at 15:41
It will just find any `!`, but if you use an `if` condition, thus sorting out those that match `!([-+]?%d+)`, then it should work. — Wiktor Stribiżew, Oct 27 '22 at 15:42
@MikeV. Why the surrounding `.-`? That's superfluous and appears to me as if it would likely slow down `gmatch`. Just `s:gmatch"!([-+]?%d+)"` is probably the most efficient it gets. — Luatic, Oct 27 '22 at 16:26

score 2 · Accepted Answer · answered Oct 27 '22 at 15:15

2

Your pattern only contains minor mistakes:

The start-of-string anchor ^ is misplaced (and thus treated as a character); the pattern anchor for end-of-string $ is missing (thus the pattern only matching part of the string would be allowed).
You're not allowing an empty number: %d+ requires at least one digit.

Fixing all of these, you get ^!([-+]?%d*)$. Explanation:

^ and $ anchors: Match the full string, from start to end.
!: Match the prefix.
(: Start capture.
[-+]?: Match an optional sign.
%d*: Match zero or more digits.
): End capture.

Note that this pattern will also accept !+ or !-.

answered Oct 27 '22 at 15:15

Luatic

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Thanks. - I forgot to say that I'm searching this pattern anywhere in a big string. So maybe I do not need any ^ or $ ? - your solution is almost perfect, but it recognizes "!-" or "!+" which should not be allowed... Is there a way to write something like (expr1) or (expr2) ? So it could be "!([-+]?%d+)" OR "!()". And the second would return an empty string... – DaveInDev Oct 27 '22 at 15:24
You **must** remove the anchors `^` and `$` if you're searching in a string, yes. Unfortunately Lua patterns have no "or" (alternation/choice, usually denoted by the pipe `|`) operator. You'll have to add additional Lua code to check for these patterns, such as calling `string.match` two times with two different patterns. – Luatic Oct 27 '22 at 16:25

score 1 · Answer 2 · edited Oct 28 '22 at 18:20

1

I ended up doing a match("!([-+]?%d+)") and if there is no result, I do a find("!") to handle the lonely "!" case.

Too bad that Lua does not have | for or...

edited Oct 28 '22 at 18:20

andrewJames

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answered Oct 27 '22 at 17:33

DaveInDev

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Regex with optional number and leading char ? (lua match)

2 Answers2