How to extract the first letter and the following three digits from a string using regex java

Question

I have strings like

S103 FYUN031800644100

5 S103 LKAAN031801261400

32 S103 AKJA031804534600

From the above strings, I have to extract the bolded part from each of the above strings using regex, anyone with an idea on how to go about it?

So far I have tried this

String acc = "S103L5AN031801261400";
            String regex = ".*S(.{3})";
            Pattern pattern = Pattern.compile(regex);
            Matcher matcher = pattern.matcher(acc);
            while(matcher.find()){
                Logger.debug("Match:", matcher.group());
            }

But still, I can't get it right.

Answer 1

You can use the regex, [A-Za-z]\\d{3} which means an alphabet followed by 3 digits.

If you want the substring to start with S or s only, use [Ss]\d{3} or (?i)s\\d{3} instead. Note that (?i) makes the pattern case-insensitive.

If you want the substring to start with S only, use S\d{3} instead.

import java.util.regex.Matcher;
import java.util.regex.Pattern;
import java.util.stream.Stream;

public class Main {
    public static void main(String[] args) {
        Stream.of(
                    "S103FYUN031800644100",
                    "5S103LKAAN031801261400",
                    "32S103AKJA031804534600"
                ).forEach(s -> System.out.println(s + " => " + getSubstr(s)));
    }

    static String getSubstr(String s) {
        Matcher matcher=Pattern.compile("[A-Za-z]\\d{3}").matcher(s);
        String substr="";
        if(matcher.find()) {
            substr = matcher.group();
        }
        return substr;
    }
}

Output:

S103FYUN031800644100 => S103
5S103LKAAN031801261400 => S103
32S103AKJA031804534600 => S103