Java regex matches to arraylist

Question

I have this code:

String line = "There were bear#67 with dog#1323 and cat#5475 in the forest";
String pattern = ".* ([^ ]+)#(\\d{4}).*";

Pattern r = Pattern.compile(pattern);
Matcher m = r.matcher(line);
if (m.find( )) {
    System.out.println("Found value: " + m.group(0) );
    System.out.println("Found value: " + m.group(1) );
    System.out.println("Found value: " + m.group(2) );
}  else {
    System.out.println("NO MATCH");
}

It prints

Found value: There were bear#67 with dog#1323 and cat#5475 in the forest
Found value: cat
Found value: 5475

However I need to get all matches as ArrayList:

dog#1323, cat#5475

How would I do that?

What did you mean by However I need to get all matches as ArrayList ? — Youcef LAIDANI
– Youcef LAIDANI, Commented Oct 30, 2020 at 16:52
your regex should be String pattern = "[^ ]+#\\d{4}"; and add matcher.group(0) to a list inside the loop — anubhava
– anubhava, Commented Oct 30, 2020 at 16:54
@YCF_L That resulting arraylist need to consist of two elements: String dog#1323, and String cat#5475 — parsecer
– parsecer, Commented Oct 30, 2020 at 16:56

anubhava · Accepted Answer · 2020-10-30 17:00:31Z

1

Converting my comment to answer so that solution is easy to find for future visitors.

You may use a simpler regex without any capture group:

[^ ]+#\\d{4}

Which matches 1+ of a non-space characters before # and then 4 digits.

Code:

final String line = "There were bear#67 with dog#1323 and cat#5475 in the forest";

final Pattern r = Pattern.compile("[^ ]+#\\d{4}");    
Matcher m = r.matcher(line);

List<String> arr = new ArrayList<>();

while (m.find( )) {
   arr.add(m.group());
}

System.out.println(arr);
//=> [dog#1323, cat#5475]

answered Oct 30, 2020 at 17:00

anubhava

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Comments

Youcef LAIDANI · Accepted Answer · 2020-10-30 16:59:32Z

1

If you want to put all matches in a list, then you have to create one like this :

String pattern = "[^ ]+#\\d{4}"; // or "[a-z]+#\\d{4}" if you want only alphabets
//..
List<String> values = new ArrayList<>();
while (m.find()) {
    values.add(m.group(0));
}

Note: you need to use while loop and not an if statement(which match only one value).

Output

[dog#1323, cat#5475]

answered Oct 30, 2020 at 16:59

Youcef LAIDANI

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Comments

Arvind Kumar Avinash · Accepted Answer · 2020-10-30 18:06:32Z

Using the Stream API, you can do it with a single statement.

Demo:

import java.util.List;
import java.util.regex.MatchResult;
import java.util.regex.Pattern;
import java.util.stream.Collectors;

public class Main {
    public static void main(String[] args) {
        List<String> list = Pattern.compile("(?<=\\s?)\\p{L}+#\\d{4}")
                                .matcher("There were bear#67 with dog#1323 and cat#5475 in the forest")
                                .results()
                                .map(MatchResult::group)
                                .collect(Collectors.toList());
        
        System.out.println(list);
    }
}

Output:

[dog#1323, cat#5475]

Explanation of the regex:

(?<=\s?) specifies the positive lookbehind for optional whitespace
\p{L}+ specifies one or more letter(s)
# specifies the char, #
\d{4} specifies 4 digits

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