What would be the regex expression for this situation?

Question

I have a regex challenge for you (I need you help)

I'm trying to make a regex that finds "bbb" only if "aaa" is not present between the beginning of the string and the "bbb" word.

So I built the following expression: ^(?!.*(aaa)).*?(bbb)

It works but I noticed that "bbb" is not found when "aaa" is located after "bbb"... I need the expression to find "bbb" even if "aaa" is located between "bbb" and the end of the string.

So:

Must find "bbb" in:

"ddd bbb eee"
"bbb fff aaa"

Must not find "bbb" in:

"aaa bbb eee ddd" 
"aaa ccc bbb"

Ps: See this link (select Java) http://www.regexplanet.com/cookbook/ahJzfnJlZ2V4cGxhbmV0LWhyZHNyEwsSBlJlY2lwZRiAgIDii-WPCgw/index.html to test the regex. I simplified the test strings for my case but just so you know, in real life those strings will be actual various sentences.

Thank you

EDIT:

Awwhh man! I always forget something in my regex questions! Loll

I forgot to mention that "bbb" needs to be replaced (replaceAll) by another word (ie: "zzz"). So the result of the expression needs to be a modified string and not a boolean.

So we should have these results:

"ddd bbb eee" => "ddd zzz eee"
"bbb fff aaa" => "zzz fff aaa"
"bbb ggg bbb" => "zzz ggg zzz"
"aaa bbb eee ddd" => "aaa bbb eee ddd"
"aaa ccc bbb" => "aaa ccc bbb"
"fff aaa eee bbb jjj bbb" => "fff aaa eee bbb jjj bbb"

Sorry I could not update the test link from my mobile.

Further to @Thefourthbird's comments, the "Tempered Greedy Token" is discussed [here](https://www.rexegg.com/regex-quantifiers.html#tempered_greed). — Cary Swoveland, Jun 22 '20 at 07:44
@Thefourthbird, if the string may contain newlines and the restriction applies only when `aaa` and `bbb` are in the same line, a slight variant may be appropriate: `^(?:(?!\baaa\b|\n).)*\bbbb\b`. — Cary Swoveland, Jun 22 '20 at 07:53
I forgot to mention something, please see my edited question. — Don Madrino, Jun 22 '20 at 08:26
Use a capturing group `^((?:(?!\baaa\b).)*?)\bbbb\b` and in the replacement use `$1zzz` See https://regex101.com/r/DsVBVO/1 — The fourth bird, Jun 22 '20 at 08:29

score 0 · Answer 1 · edited Jun 22 '20 at 05:00

0

It's almost correct!!

Jus a small correction: String pattern = "^(?!.*(aaa).*?(bbb))";

public static void main(String[] args)
    {
        String[] lines =  {"ddd bbb eee",
                           "bbb fff aaa",
                           "aaa bbb eee ddd",
                           "aaa ccc bbb"};

        String pattern = "^(?!.*(aaa).*?(bbb))";

        Pattern p = Pattern.compile(pattern);


        for (String line: lines)
        {
            Matcher matcher = p.matcher(line);
            System.out.println(matcher.find());
        }
    }

O/p:

True
True
False
False

Hope this works..

edited Jun 22 '20 at 05:00

answered Jun 22 '20 at 02:38

Sabareesh Muralidharan

636
4
11

2

You have my vote. But please try to use word boundaries as it makes the regex more failproof. Something like `^(?!.*?(\baaa\b).*?(\bbbb\b))` – Jun 22 '20 at 04:54
1

Your regex fails to match the string `bbb aaa bbb`. [ref](https://regex101.com/r/nvxuP9/1/) – Cary Swoveland Jun 22 '20 at 06:39
I'm not sure in that case whether it should allow or not. :| It neither satisfies not dissatisfies the req. – Sabareesh Muralidharan Jun 22 '20 at 07:45
I forgot to mention something, please see my edited question. – Don Madrino Jun 22 '20 at 08:26

What would be the regex expression for this situation?

1 Answers1