How to find matches with a regular expression

Short answer

Compile the pattern into a Pattern, create a Matcher for the text and call find():

void main() {
    Matcher matcher = Pattern.compile("#\\w+").matcher("Learning #java and #regex today");
    while (matcher.find()) {
        IO.println(matcher.group());
    }
}

Output:

#java
#regex

Groups

Parentheses create groups — parts of the match you can get by number:

void main() {
    Matcher matcher = Pattern.compile("(\\d{4})-(\\d{2})-(\\d{2})").matcher("Due: 2026-03-15");
    if (matcher.find()) {
        IO.println("year " + matcher.group(1) + ", month " + matcher.group(2));
    }
}

Output:

year 2026, month 03

Named groups

A group name (?<name>...) makes the code clearer than numbers:

void main() {
    var pattern = Pattern.compile("(?<key>\\w+)=(?<value>\\w+)");
    pattern.matcher("host=localhost port=8080").results()
            .forEach(m -> IO.println(m.group("key") + " -> " + m.group("value")));
}

Output:

host -> localhost
port -> 8080

results() returns MatchResult objects. They support named groups since Java 20.

Replacing with groups

In the replacement string $1, $2 refer to groups:

void main() {
    IO.println("2026-03-15".replaceAll("(\\d{4})-(\\d{2})-(\\d{2})", "$3.$2.$1"));
}

Output:

15.03.2026

Common mistake

Using matches() to look for part of the text. matches() requires the whole string to match, while find() searches inside it:

void main() {
    Matcher matcher = Pattern.compile("\\d+").matcher("room 42");
 
    IO.println(matcher.matches());
    IO.println(matcher.reset().find());
}

Output:

false
true

Thanks for reading!

Short answer

Groups

Named groups

Replacing with groups

Common mistake

Functions used in this recipe

Thanks for reading!