Match first occurrence of semicolon in string, only if not preceded by '--'

Question

I'm trying to write a regular expression for Java that matches if there is a semicolon that does not have two (or more) leading '-' characters.

I'm only able to get the opposite working: A semicolon that has at least two leading '-' characters.

([\-]{2,}.*?;.*)

But I need something like

([^([\-]{2,})])*?;.*

I'm somehow not able to express 'not at least two - characters'.

Here are some examples I need to evaluate with the expression:

; -- a           : should match
-- a ;           : should not match
-- ;             : should not match
--;              : should not match
-;-              : should match
---;             : should not match
-- semicolon ;   : should not match
bla ; bla        : should match
bla              : should not match (; is mandatory)
-;--;            : should match (the first occuring semicolon must not have two or more consecutive leading '-')

How many semicolons can be in string? Is string like `-;--;` correct? — Pshemo, Jul 21 '14 at 15:04
Also do we want to forbid only leading `-`? What about strings like `x--;`? — Pshemo, Jul 21 '14 at 15:06
@Pshemo The first one has to match (updated my question accordingly). The second one must not match, just to keep things simple. Otherwise I would need to write a complete parser and thats not the intention of my small application. — Richard, Jul 21 '14 at 15:13

Pshemo · Answer 1 · 2014-07-21T20:08:09.180

It seems that this regex matches what you want

String regex = "[^-]*(-[^-]+)*-?;.*";

DEMO

Explanation: matches will accept string that:

[^-]* can start with non dash characters
(-[^-]+)*-?; is a bit tricky because before we will match ; we need to make sure that each - do not have another - after it so:
- (-[^-]+)* each - have at least one non - character after it
- -? or - was placed right before ;
;.* if earlier conditions ware fulfilled we can accept ; and any .* characters after it.

More readable version, but probably little slower

((?!--)[^;])*;.*

Explanation:

To make sure that there is ; in string we can use .*;.* in matches.
But we need to add some conditions to characters before first ;.

So to make sure that matched ; will be first one we can write such regex as

[^;]*;.*

which means:

[^;]* zero or more non semicolon characters
; first semicolon
.* zero or more of any characters (actually . can't match line separators like \n or \r)

So now all we need to do is make sure that character matched by [^;] is not part of --. To do so we can use look-around mechanisms for instance:

(?!--)[^;] before matching [^;] (?!--) checks that next two characters are not --, in other words character matched by [^;] can't be first - in series of two --
[^;](?<!--) checks if after matching [^;] regex engine will not be able to find -- if it will backtrack two positions, in other words [^;] can't be last character in series of --.

Adam Yost · Answer 2 · 2014-07-21T16:50:10.803

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You need a negative lookahead!

This regex will match any string which does not contain your original match pattern:

(?!-{2,}.*?;.*).*?;.*

This Regex matches a string which contains a semicolon, but not one occuring after 2 or more dashes.

Example: Regex Working

edited Jul 21 '14 at 16:50

answered Jul 21 '14 at 14:54

Adam Yost

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Which software did you use? – candlejack Jan 06 '18 at 21:20

score 0 · Answer 3 · edited Jun 20 '20 at 09:12

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How about using this regex in Java:

[^;]*;(?<!--[^;]{0,999};).*

Only caveat is that it works with up to 999 character length between -- and ;

Java Regex Demo

edited Jun 20 '20 at 09:12

Community

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answered Jul 21 '14 at 14:56

anubhava

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Thanks, @anubhava. I would prefer a solution without groups. But in general your solution works. – Richard Jul 21 '14 at 15:16
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Hi anubhava I have created a question almost dedicated to you! lol. I really like that tecnhique. Could you check it? http://stackoverflow.com/questions/24808793/regex-technique-to-disallow-variable-length-lookbehind-using-or/ – Federico Piazza Jul 21 '14 at 15:21

score 0 · Answer 4 · answered Jul 21 '14 at 15:07

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How about just splitting the string along -- and if there are two or more sub strings, checking if the last one contains a semicolon?

answered Jul 21 '14 at 15:07

Edwin Buck

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I would prefer a solution where I can call just .matches(), because all other statements in my class work this way. Just for reading purposes. – Richard Jul 21 '14 at 15:18
@RichardW. That's fine; however, if it is for readability's sake, some of these answers are far less readable (if you care about _understandability_) than two or three calls which make the task obvious. – Edwin Buck Jul 21 '14 at 15:56

Alan Moore · Answer 5 · 2014-07-21T16:44:28.127

I think this is what you're looking for:

^(?:(?!--).)*;.*$

In other words, match from the start of the string (^), zero or more characters (.*) followed by a semicolon. But replacing the dot with (?:(?!--).) causes it to match any character unless it's the beginning of a two-hyphen sequence (--).

If performance is an issue, you can exclude the semicolon as well, so it never has to backtrack:

^(?:(?!--|;).)*;.*$

EDIT: I just noticed your comment that the regex should work with the matches() method, so I padded it out with .*. The anchors aren't really necessary, but they do no harm.

Match first occurrence of semicolon in string, only if not preceded by '--'

5 Answers5

Java Regex Demo