Matcher doesn't return unique result

Question

This is my sample code:

public String testMethod() {
    String sampleString = "Hi <username>. Is <username> your name?. <username> rocks! <admin> wishes you well. Ask <admin> if you have any trouble!";
    String myRegex = "your regex here";

    Pattern pattern = Pattern.compile(myRegex);
    Matcher matcher = pattern.matcher(stringSample);
    int counter = 0;
    while (matcher.find()) {
        counter++;
    }

    return "Matched substring: " + counter;
}

First, I want to get tags with this pattern <([a-zA-Z0-9_]+)>. When I used the pattern, I get 5 as a result since there are 5 tags in sampleString. This works just fine but I want Matcher to return only unique match.

Based on the string in the sample code, the result would be 2 since there are 2 unique tags (<username> and <admin>). So I build my regex based on this answer and now I have this pattern <([a-zA-Z0-9_]+)>(?!.*\1). I tried the pattern on Regex101 and it works just fine. But when used with the sample code, the result is still 5.

Is there anything wrong with my pattern?

Edit: Just like the linked question, I want to avoid using Maps or Lists. And I want to emphasize that I'm asking why my regex doesn't work on Java when it's supposed to work (based on Regex101 result).

You're using regexes for things they're not designed for. Regexes for for finding patterns. `Set`s are the right tool for finding all the unique occurrences of something. Don't try to use regexes to solve everything--that's a common beginner mistake. — ajb, Jul 24 '17 at 05:02
Why do you want to avoid using `Map` or `List`? Is someone penalizing you $100 every time you use one? If not, what motivation do you have for avoiding what could be the right tool for the job? — ajb, Jul 24 '17 at 06:28
@ajb Ahahahaa. Good question. I'm just trying to learn what regex can do. — AceVez, Jul 24 '17 at 07:59
Java regexes have a lot of power. They can be used to find all sorts of complicated patterns. They also have the power to make your code unreadable unnecessarily. Use the power wisely. — ajb, Jul 24 '17 at 13:33

score 2 · Answer 1 · answered Jul 24 '17 at 04:57

2

Rather that coming up with a complex regex, you can use use simple regex <(\\w+)> and store your results in a Set to get unique matches only:

String sampleString = "Hi <username>. Is <username> your name?. <username> rocks! <admin> wishes you well. Ask <admin> if you have any trouble!";
String myRegex = "<(\\w+)>";

Pattern pattern = Pattern.compile(myRegex);
Matcher matcher = pattern.matcher(sampleString);

Set<String> tags = new HashSet<>();

while (matcher.find()) {
    tags.add(matcher.group(1));
}

System.out.printf("tags: %s, count: %d%n", tags, tags.size());

Output:

tags: [admin, username], count: 2

answered Jul 24 '17 at 04:57

anubhava

761,203
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I forgot to mention that I want to avoid using Maps or Lists. Question updated. – AceVez Jul 24 '17 at 05:20
Well `Set` is neither a `Map` and nor a `List` :) – anubhava Jul 24 '17 at 05:25
lol. You made my day. What do you call `Map`, `List` and `Set` then? Collection? – AceVez Jul 24 '17 at 05:47
Yes indeed, a `Collection` :) – anubhava Jul 24 '17 at 05:51

chengpohi · Accepted Answer · 2017-07-24T05:12:07.683

1

You should use <([a-zA-Z0-9_]+)>(?!.*\\1): \\1 for 1st capture group in Java Code not \1 .

Actual \1 is an octal value, see more about this:

What are all the escape characters in Java?

edited Jul 24 '17 at 05:12

answered Jul 24 '17 at 04:47

chengpohi

14,064
1
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Forgot about escape characters. Thank you, it works now. – AceVez Jul 24 '17 at 05:19

Matcher doesn't return unique result

2 Answers2