import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String[] args) {
final String regex = ".*Azure\\s+v(.*?)\\s+.*";
final String string = "N/A 9.0 Yes Yes N/A N/A N/A N/A 8.8 Yes Yes N/A N/A N/A N/A 8.7 Yes Yes N/A N/A N/A N/A 9.2 Yes Yes Yes Yes Yes Yes 9.1 Yes Yes Yes Yes Yes Yes 9.0 Yes Yes Yes Yes Yes Yes 8.8 Yes Yes Yes Yes Yes Yes 8.7 Yes Yes Yes Yes Yes Yes 22.04 Yes Yes Yes Yes Yes Yes AKS Microsoft Azure v1.27 EKS Amazon v1.27 Amazon Linux 2 is certified";
final String subst = "\\1";
final Pattern pattern = Pattern.compile(regex);
final Matcher matcher = pattern.matcher(string);
// The substituted value will be contained in the result variable
final String result = matcher.replaceAll(subst);
System.out.println("Substitution result: " + result);
}
}
Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Java, please visit: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html