Regex Negate Pattern

Question

I'm having trouble getting the other side of this regex.

Stranger Number is %19. Friend Number is %30.

My current regex is this (%[0-9]+) So when I run this regex. The only highlighted are the %30 and %19.

Strange Number is %19. Friend Number is %30.

I just want it the other way around where %19 and %30 is not the highlighted one and everything else is highlighted.

I have tried. this one [^%](?![0-9]) but im not getting my expected output.

thanks for the help!

So can the string you want include a %? Or only a % followed by a number must be excluded? Same with digits: Exclude all digits, or only digits following a %? — markspace, Nov 06 '18 at 19:53
Yes it can be. Any number. As long it is satisfied by % + any number. @mavriksc — Dranym, Nov 06 '18 at 20:00
Couldn't you write your program in a way, that all the things that are not matched by `%[0-9]+` are used instead of the matches? — Socowi, Nov 06 '18 at 20:00
What is the exact logic behind why only `%19` gets matched but not `%30` ? — Tim Biegeleisen, Nov 06 '18 at 20:00
Right, but why do you not want %30 to be highlighted? Is it just that it's the second number in the line? — markspace, Nov 06 '18 at 20:02
@markspace I want both %19 and %30 not to be highlighted... and the rest is highlighted.. — Dranym, Nov 06 '18 at 20:03
OK so you want only the string parts highlighted, everything except the %number bits. — markspace, Nov 06 '18 at 20:06
Your requirement makes no sense. The regex engine _has_ to match all content in order to continue parsing the input string. Can you tell us what your expected output is? Do you just want to match the text minus the numbers in percentage? — Tim Biegeleisen, Nov 06 '18 at 20:06
@TimBiegeleisen I think that's it, just the text part minus the number and %. — markspace, Nov 06 '18 at 20:07
This would be really easy with a simple parser. Regex is harder because the digit length can be anything. Is a parser OK or are you required to use regex? — markspace, Nov 06 '18 at 20:08
Also, what about the final result? What format to you want to extract the result in? Must it be a regex match, or would an array of strings be OK? — markspace, Nov 06 '18 at 20:10
@markspace As Socowi already mentioned it would just be enough to "subtract" the positive matches from the whole span to create negative matches. — Michael Butscher, Nov 06 '18 at 20:12
@markspace an array of strings will be ok. as i am going to use it in java. — Dranym, Nov 06 '18 at 20:12

score 1 · Answer 1 · answered Nov 06 '18 at 20:12

1

Depending you what you want to do, you don't have to find an inverted regex for (%[0-9]+).

For instance, if you planned to extract all substrings matching the inverted regex, you could use yourString.split("%[0-9]+").

If you planned to extract all the %... by splitting the string with yourString.split(invertedRegex) you could use a matcher instead.
_{Code taken from this answer.}

String[] matches = Pattern.compile("%[0-9]+").matcher(yourString)
                          .results().map(MatchResult::group)
                          .toArray(String[]::new);

answered Nov 06 '18 at 20:12

Socowi

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I think this is the one. Let me try it. – Dranym Nov 06 '18 at 20:19
Don't forget the `import java.util.regex.Pattern;` and `import java.util.regex.MatchResult;`. – Socowi Nov 06 '18 at 20:20
This should work but im not yet on Java 9. :( Im using the one above :) Can I make 2 solutions correct? lol – Dranym Nov 06 '18 at 20:29
@Dranym Thank you for your reply. You can only accept one answer. But when you have (I think) 15 points you can upvote the other answer. By the way, java 8 should be sufficient. – Socowi Nov 06 '18 at 20:31
I have upvoted this answer but the system says my vote wont get publicized unless i reached 15 points. – Dranym Nov 06 '18 at 20:33

score 1 · Accepted Answer · answered Nov 06 '18 at 20:27

1

For completeness, and those on earlier versions of Java, you can just use the Matcher and a while loop to extract strings that match a pattern.

   public static void main( String[] args ) {
      String test = "Number %1.  And number %2 also.";
      Matcher matcher = Pattern.compile( "%[\\d]+" ).matcher( test );
      ArrayList<String> results = new ArrayList<>();
      while( matcher.find() )
         results.add( matcher.group() );
      System.out.println( results );
   }

answered Nov 06 '18 at 20:27

markspace

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Both works but I'm not yet on java 9 so this is the one code for me. Thanks! – Dranym Nov 06 '18 at 20:32
Streams where added in Java 8, I thought. – markspace Nov 06 '18 at 20:32

Regex Negate Pattern

2 Answers2